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Sascha Spors, Professorship Signal Theory and Digital Signal Processing, Institute of Communications Engineering (INT), Faculty of Computer Science and Electrical Engineering (IEF), University of Rostock, Germany # Tutorial Digital Signal Processing **Uniform Quantization, Dithering, Noiseshaping**, Winter Semester 2021/22 (Master Course #24505) - lecture: https://github.com/spatialaudio/digital-signal-processing-lecture - tutorial: https://github.com/spatialaudio/digital-signal-processing-exercises Feel free to contact lecturer [email protected] # Fundamentals of Quantization ## Packages / Functions We import the required packages first and put some functions here that we will frequently use. ```python # most common used packages for DSP, have a look into other scipy submodules import numpy as np import matplotlib as mpl import matplotlib.pyplot as plt from scipy import signal # audio write and play stuff import soundfile as sf # requires 'pip install soundfile' # last tested with soundfile-0.10.3 ``` ```python def my_xcorr2(x, y, scaleopt='none'): r""" Cross Correlation function phixy[kappa] -> x[k+kappa] y input: x input signal shifted by +kappa y input signal scaleopt scaling of CCF estimator output: kappa sample index ccf correlation result """ N = len(x) M = len(y) kappa = np.arange(0, N+M-1) - (M-1) ccf = signal.correlate(x, y, mode='full', method='auto') if N == M: if scaleopt == 'none' or scaleopt == 'raw': ccf /= 1 elif scaleopt == 'biased' or scaleopt == 'bias': ccf /= N elif scaleopt == 'unbiased' or scaleopt == 'unbias': ccf /= (N - np.abs(kappa)) elif scaleopt == 'coeff' or scaleopt == 'normalized': ccf /= np.sqrt(np.sum(x**2) * np.sum(y**2)) else: print('scaleopt unknown: we leave output unnormalized') return kappa, ccf ``` ```python def uniform_midtread_quantizer(x, deltaQ): r"""uniform_midtread_quantizer from the lecture: https://github.com/spatialaudio/digital-signal-processing-lecture/blob/master/quantization/linear_uniform_quantization_error.ipynb commit: b00e23e note: we renamed the second input to deltaQ, since this is what the variable actually represents, i.e. the quantization step size input: x input signal to be quantized deltaQ quantization step size output: xq quantized signal """ # [-1...1) amplitude limiter x = np.copy(x) idx = np.where(x <= -1) x[idx] = -1 idx = np.where(x > 1 - deltaQ) x[idx] = 1 - deltaQ # linear uniform quantization xq = deltaQ * np.floor(x/deltaQ + 1/2) return xq ``` ```python def my_quant(x, Q): r"""Saturated uniform midtread quantizer input: x input signal Q number of quantization steps output: xq quantized signal Note: for even Q in order to retain midtread characteristics, we must omit one quantization step, either that for lowest or the highest amplitudes. Typically the highest signal amplitudes are saturated to the 'last' quantization step. Then, in the special case of log2(N) being an integer the quantization can be represented with bits. """ tmp = Q//2 # integer div quant_steps = (np.arange(Q) - tmp) / tmp # we don't use this # forward quantization, round() and inverse quantization xq = np.round(x*tmp) / tmp # always saturate to -1 xq[xq < -1.] = -1. # saturate to ((Q-1) - (Q\2)) / (Q\2), note that \ is integer div tmp2 = ((Q-1) - tmp) / tmp # for odd N this always yields 1 xq[xq > tmp2] = tmp2 return xq ``` ## Quantization Process and Error Quantization generates signals that have discrete values $x_q[k]$, $x_q(t)$ from signals with continuous values $x[k]$, $x(t)$. For quantization, the signals can be both, discrete and continuous in time. However, a signal that is discrete in time **and** discrete in value is termed a **digital** signal. Only digital signals can be processed by computers. Here the quantization of discrete-time signals is treated due to practical importance. To describe quantization analytically, the model in the figure below is used. The input and output signal differ by the so called quantization error (quantization noise) $e[k]$, that is defined as \begin{equation} e[k] = x_q[k] - x[k], \end{equation} so that the error constitutes an additive superposition \begin{equation} x[k] + e[k] = x_q[k] \end{equation} To use this error model, some assumption have to be made. The quantization noise shall be uniformly distributed, which then can be modeled with the probability density function (PDF) $p_e(\theta) = \frac{1}{\Delta Q} \mathrm{rect}(\frac{\theta_e}{\Delta Q})$, where $\Delta Q$ denotes the quantization step size and $\theta_e$ the amplitudes of the quantization error signal. This PDF is shown below. ```python plt.figure(figsize=(4, 2)) plt.plot((-1, -1/2, -1/2, +1/2, +1/2, +1), (0, 0, 1, 1, 0, 0), lw=3) plt.xlim(-1, 1) plt.ylim(-0.1, 1.1) plt.xticks((-0.5, +0.5), [r'-$\frac{\Delta Q}{2}$', r'+$\frac{\Delta Q}{2}$']) plt.yticks((0, 1), [r'0', r'+$\frac{1}{\Delta Q}$']) plt.xlabel(r'$\theta_e$') plt.ylabel(r'$p_e(\theta)$') plt.title( r'$p_e(\theta) = \frac{1}{\Delta Q} \mathrm{rect}(\frac{\theta_e}{\Delta Q})$') plt.grid(True) ``` Furthermore, it is assumed that $e[k]$ is not correlated with $x[k]$. That this is not necessarily the case can be demonstrated with the help of some straightforward examples. ```python Q = 9 # odd, number of quantization steps N = 100 k = np.arange(2*N) x = np.sin(2*np.pi/N*k) xq = my_quant(x, Q) e = xq-x # actually stem plots would be correct, for convenience we plot as line style plt.plot(k, x, 'C2', lw=3, label=r'$x$') plt.plot(k, xq, 'C0o-', label=r'$x_q$') plt.plot(k, e, 'C3', label=r'$e=x_q-x$') plt.plot(k, k*0+1/(Q-1), 'k:', label=r'$\frac{\Delta Q}{2}$') plt.xlabel(r'$k$') plt.legend() plt.grid(True) ``` A sine signal is quantized with $Q=9$ quantization steps. A periodicity of the quantization noise can be easily identified. For odd $Q$, the maximum amplitude of the quantization error can be estimated to $$\frac{\Delta Q}{2}=\frac{\frac{2}{Q-1}}{2}=\frac{1}{Q-1}=\frac{1}{8}=0.125.$$ The auto-correlation function of the error signal $e[k]$ is presented next. ```python kappa, acf = my_xcorr2(e, e, 'unbiased') plt.plot(kappa, acf) plt.xlim(-175, +175) plt.xlabel(r'$\kappa$') plt.ylabel(r'$\phi_{ee}[\kappa]$') plt.title('ACF of quantization error') plt.grid(True) ``` If $e[k]$ would be exactly following the probability density function $p_e(\theta) = \frac{1}{\Delta Q} \mathrm{rect}(\frac{\theta_e}{\Delta Q})$, the auto-correlation function $\phi_{ee}[\kappa]=\delta[\kappa]$ results. However, this is not observable in this example! Instead, from the above plot, we can deduce that $e[k]$ is correlated to itself, i.e. it exhibits periodicity each 100 samples in phase, and each 50 sample out of phase. The sine period is precisely 100 samples, thus the input signal and the quantization error are somewhat linked and not independent. Thus, the error model assumption is violated. That is bad, since the sine signal allows for otherwise comparable simple analytical calculus. The links between the signals can be furthermore confirmed with the help of the cross-correlation functions. Their oscillating characteristics reveal that quantization error is highly correlated. ```python plt.figure(figsize=(9, 3)) plt.subplot(1, 2, 1) kappa, acf = my_xcorr2(e, x, 'unbiased') plt.plot(kappa, acf) plt.xlim(-170, +170) plt.xlabel(r'$\kappa$') plt.ylabel(r'$\phi_{e,x}[\kappa]$') plt.title('CCF quantization error and input signal') plt.grid(True) plt.subplot(1, 2, 2) kappa, acf = my_xcorr2(e, xq, 'unbiased') plt.plot(kappa, acf) plt.xlim(-170, +170) plt.xlabel(r'$\kappa$') plt.ylabel(r'$\phi_{e,xq}[\kappa]$') plt.title('CCF quantization error and quantized signal') plt.grid(True) ``` Therefore, the special case of sine signals is in fact not suited for the quantization model above. Because of the simplicity of the involved calculation it is common practice to conduct this analysis for sine signals nevertheless, and signal-to-noise ratios in the data sheets of A/D converters are mostly stated for excitation with sine signals. For random signals, the quantization model is only valid for high levels in the quantizer. For more information see [Udo Zölzer, Digital Audio Signal Processing, Wiley](https://onlinelibrary.wiley.com/doi/book/10.1002/9780470680018) (might be available as free access in your uni network) - Task: Increase the (odd) number of quantization steps $Q$ and check what happens with the shape and amplitudes of the correlations functions. Hint: closer look to the amplitudes of the correlation signals. ## Quantization Modeling / Mapping The mapping of the infinitely large continuous set of values to a discrete number of amplitude steps is realized with a transfer characteristic. The height of the amplitude steps is $\Delta Q$. From the lecture, we know that the following mapping is used in order to quantize the continuous amplitude signal $x[k]$ towards \begin{equation} x_Q[k] = g( \; \lfloor \, f(x[k]) \, \rfloor \; ), \end{equation} where $g(\cdot)$ and $f(\cdot)$ denote real-valued mapping functions, and $\lfloor \cdot \rfloor$ a rounding operation (**not necessarily the plain floor operation**). ### Uniform Saturated Midtread Quantization Characteristic Curve With the introduced mapping, the uniform saturated midtread quantizer can be discussed. This is probably the most important curve for uniform quantization due to its practical relevance for coding quantized amplitude values as bits. In general, the uniform midtread quantizer can be given as the mapping \begin{equation} x_Q[k] = \frac{1}{Q \backslash 2} \cdot \lfloor (Q \backslash 2) \cdot x[k]\rfloor, \end{equation} where for $\lfloor \cdot \rfloor$ a rounding operation might used and $\backslash$ denotes integer division. So the mapping functions $g$ and $f$ are simple multiplications. At the beginning of this notebook, the function `my_quant` is implemented that realizes quantization based on this mapping. The approach uses `numpy`'s `round` operation. When asking for rounding, care has to be taken, which [approach](https://en.wikipedia.org/wiki/Rounding) shall be used. Numpy rounds to the nearest **even** integer in contrast to e.g. Matlab's rounding to nearest integer. Detailed analysis for `my_quant`: - the quantization should be properly performed only for $-1 \leq x < 1$ - thus, it always saturates $x<-1$ towards $x_q = -1$ - in the case of an **odd** number of quantization steps $Q$, it saturates $x>+1$ towards $x_q = +1$. The quantization step size is $\Delta Q = \frac{2}{Q-1}$. - In the case of an **even** number of quantization steps $Q$, it saturates $x>\frac{Q - 1 - \frac{Q}{2}}{\frac{Q}{2}} = 1-\frac{2}{Q}$ towards $x_q = \frac{Q - 1 - \frac{Q}{2}}{\frac{Q}{2}}=1-\frac{2}{Q}$. The quantization step size is $\Delta Q = \frac{2}{Q}$. ### AD / DA Converter Convention The case of **even** $Q$ is practically used for virtually all analog/digital (AD) and digital/analog (DA) converters. When additionally to the above statements \begin{equation} \log_2(Q)\in\mathbb{N} \end{equation} holds, it is meaningful to code the even and power of two $Q$ possible quantization steps with bits. With $B\in\mathbb{N}$ denoting the number of bits, the number range convention for AD and DA converters is \begin{equation} \begin{split} &-1\leq x \leq 1-2^{-(B-1)}\\ &-1\leq x \leq 1-\frac{2}{Q} \end{split} \end{equation} using \begin{equation} Q=2^B \end{equation} quantization steps. Values of $x$ outside this range will be saturated to the minimum $-1$ and maximum $1-\frac{2}{Q}$ quantization values in the quantization process. For example, $B = 16$ bits are used to code [PCM data for CD](https://en.wikipedia.org/wiki/Compact_disc) audio quality. Then we get the following quantities. ```python B = 16 # number of bits Q = 2**B # number of quantization steps # for even Q only: deltaQ = 2/Q # maximum quantize value: xqmax = 1-2**(-(B-1)) # or more general for even Q: xqmax = 1-deltaQ print(' B = %d bits\n quantization steps Q = %d\n quantization step size %e' % (B, Q, deltaQ)) print(' smallest quantization value xqmin = -1') print(' largest quantization value xqmax = %16.15f' % xqmax) # B = 16 bits # quantization steps Q = 65536 # quantization step size 3.051758e-05 # smallest quantization value xqmin = -1 # largest quantization value xqmax = 0.999969482421875 ``` So called high definition audio uses 24 Bit. Video and photo typically uses 8-12 Bit quantization per color channel. ### Plotting the Midtread Curve We now can visualize the characteristic curve for a simple, made up input signal, i.e. a monotonic increasing signal between $x_{max} = -x_{min}$ using an equidistant increment $\Delta x$ over sample index $k$. Here, we use $x_{max} = 1.25$ and $\Delta x=0.001$ and assume that we start with $x_{min} = -1.25$ at $k=0$. If $\Delta x$ is sufficiently small, the signal's amplitude can be interpreted as continuous straight line. This straight line is degraded in a quantization process. Plotting the quantization result over the input, results in the characteristic curve, in our example in the curve of the uniform saturated midtread quantizer. Let us plot this. **Please note:** The quantizer `uniform_midtread_quantizer` known from lecture and `my_quant` yield the same results besides a slight detail: `uniform_midtread_quantizer` always exhibits an **even** number of quantization steps $Q$. So, only for even $Q$ results are exactly identical. We might verify this in the next plots as well. ```python x = np.arange(-1.25, +1.25, 1e-3) plt.figure(figsize=(4, 2)) plt.plot(x) # actually a stem plot is correct plt.ylim(-1.25, +1.25) plt.xlabel(r'$k$') plt.ylabel(r'$x[k]$') plt.grid(True) ``` ```python Q = 9 # number of quantization steps, odd or even deltaQ = 1/(Q//2) # quantization step size, even/odd Q xq = my_quant(x, Q) # used in exercise xumq = uniform_midtread_quantizer(x, deltaQ) # as used in lecture plt.figure(figsize=(6, 6)) plt.plot(x, xumq, 'C0', lw=2, label='uniform_midtread_quantizer()') plt.plot(x, xq, 'C3', label='my_quant()') plt.xticks(np.arange(-1, 1.25, 0.25)) plt.yticks(np.arange(-1, 1.25, 0.25)) plt.xlabel(r'input amplitude of $x$') plt.ylabel(r'output ampliude of $x_q$') plt.title( r'uniform saturated midtread quantization, Q={0:d}, $\Delta Q$={1:3.2f}'.format(Q, deltaQ)) plt.axis('equal') plt.legend() plt.grid(True) ``` The following exercises used to be a homework assignment as exam's prerequisite. # Exercise 1: Uniform Saturated Midtread Characteristic Curve of Quantization ## Task Check this quantizer curve for $Q=7$ and $Q=8$. Make sure that you get the idea of the midtread concept (the zero is always quantized to zero) and saturation (for even $Q$) largest quantization step is saturated). ```python def check_my_quant(Q): N = 5e2 x = 2*np.arange(N)/N - 1 xq = my_quant(x, Q) e = xq - x plt.plot(x, x, color='C2', lw=3, label=r'$x[k]$') plt.plot(x, xq, color='C3', label=r'$x_q[k]$') plt.plot(x, e, color='C0', label=r'$e[k] = x_q[k] - x[k]$') plt.xticks(np.arange(-1, 1.25, 0.25)) plt.yticks(np.arange(-1, 1.25, 0.25)) plt.xlabel('input amplitude') plt.ylabel('output amplitude') if np.mod(Q, 2) == 0: s = ' saturated ' else: s = ' ' plt.title( 'uniform'+s+'midtread quantization with Q=%d steps, $\Delta Q$=%4.3e' % (Q, 1/(Q//2))) plt.axis('equal') plt.legend(loc='upper left') plt.grid(True) ``` ```python Q = 7 # number of quantization steps deltaQ = 1 / (Q//2) # general rule deltaQ = 2 / (Q-1) # for odd Q only plt.figure(figsize=(5, 5)) check_my_quant(Q) ``` ```python Q = 8 # number of quantization steps deltaQ = 1 / (Q//2) # general rule deltaQ = 2 / Q # for even Q only plt.figure(figsize=(5, 5)) check_my_quant(Q) ``` # Exercise 2: Quantization and Signal-to-Noise Ratio From theory the **6dB / Bit rule of thumb** is well known for uniform quantization. It states that the signal-to-noise ratio increases by 6 dB for every additional bit that is spent to quantize the input data. Hence, \begin{equation} \text{SNR in dB} = 6 \cdot B + \gamma, \end{equation} where $\gamma$ is a offset value in dB that depends on the PDF of the signal to be quantized. Note that this rule of thumb assumes that the quantization error exhibits uniform PDF and is not correlated with the quantized signal. We can see that this rule of thumb is inaccurate when quantizing a sine signal with small number of bits or an amplitude in the range of the quantization step. Then, the mentioned assumptions are not fulfilled. We will observe this in exercise 3. We plot the function SNR of bits below for uniform, normal and Laplace PDF noises and a sine signal. We should observe the slope of always 6 dB per Bit. We should note the different absolute values of the SNR depending on the varying $\gamma$. The `dBoffset` values are discussed in the lecture and in the textbook [Udo Zölzer, Digital Audio Signal Processing, Wiley](https://onlinelibrary.wiley.com/doi/book/10.1002/9780470680018). ```python def check_quant_SNR(x, dBoffset, title): print('std: {0:f}, var: {1:f}, mean: {2:f} of x'.format(np.std(x), np.var(x), np.mean(x))) Bmax = 24 SNR = np.zeros(Bmax+1) SNR_ideal = np.zeros(Bmax+1) for B in range(1, Bmax+1): # start at 1, since zero Q is not meaningful xq = my_quant(x, 2**B) SNR[B] = 10*np.log10(np.var(x) / np.var(xq-x)) SNR_ideal[B] = B*20*np.log10(2) + dBoffset # 6dB/bit + offset rule plt.figure(figsize=(5, 5)) plt.plot(SNR_ideal, 'o-', label='theoretical', lw=3) plt.plot(SNR, 'x-', label='simulation') plt.xticks(np.arange(0, 26, 2)) plt.yticks(np.arange(0, 156, 12)) plt.xlim(2, 24) plt.ylim(6, 148) plt.xlabel('number of bits') plt.ylabel('SNR / dB') plt.title(title) plt.legend() plt.grid(True) print('maximum achievable SNR = {0:4.1f} dB at 24 Bit (i.e. HD audio)'.format(SNR[-1])) ``` ```python N = 10000 k = np.arange(N) ``` ```python np.random.seed(4) x = np.random.rand(N) x -= np.mean(x) x *= np.sqrt(1/3) / np.std(x) dBoffset = 0 check_quant_SNR(x, dBoffset, 'Uniform PDF') ``` ```python Omega = 2*np.pi * 997/44100 # use a rather odd ratio: e.g. in audio 997 Hz / 44100 Hz sigma2 = 1/2 dBoffset = -10*np.log10(2 / 3) x = np.sqrt(2*sigma2) * np.sin(Omega*k) check_quant_SNR(x, dBoffset, 'Sine') ``` ```python np.random.seed(4) x = np.random.randn(N) x -= np.mean(x) x *= np.sqrt(0.0471) / np.std(x) dBoffset = -8.5 # from clipping propability 1e-5 check_quant_SNR(x, dBoffset, 'Normal PDF') ``` ```python np.random.seed(4) x = np.random.laplace(size=N) pClip = 1e-5 # clipping propability sigma = -np.sqrt(2) / np.log(pClip) x -= np.mean(x) x *= sigma / np.std(x) dBoffset = -13.5 # empircially found for pClip = 1e-5 check_quant_SNR(x, dBoffset, 'Laplace PDF') ``` # Exercise 3: Dithering The discrete-time sine signal - $x[k]=A \cdot\sin(\frac{2\pi f_\text{sin}}{f_s}k)$ for - $0\leq k<96000$ with - sampling frequency $f_s=48\,\text{kHz}$ and - $f_\text{sin}=997\,\text{Hz}$ shall be quantized with the saturated uniform midtread quantizer for $-1\leq x_q \leq 1-\Delta Q$ using $B$ bits, i.e. $Q=2^B$ number of quantization steps and quantization step size of $\Delta Q = \frac{1}{Q\backslash 2}$. We should discuss different parametrizations for signal amplitude $A$ and number of bits $B$. Before quantizing $x[k]$, a dither noise signal $d[k]$ shall be added to $x[k]$ according figure below. This dither signal with small amplitudes aims at de-correlating the quantization error $e[k]$ from the quantized signal $x_q[k]$, which is especially important for small amplitudes of $x[k]$. This technique is called **dithering**. For $d[k]=0$ no dithering is applied. Since the quantization error may be in the range $-\frac{\Delta Q}{2}\leq e[k]\leq \frac{\Delta Q}{2}$ (assuming uniform distribution), it appears reasonable to use a dither noise with a probability density function (PDF) of \begin{equation} p_\text{RECT}(d)=\frac{1}{\Delta Q}\,\text{rect}\left(\frac{d}{\Delta Q}\right), \end{equation} i.e. a **zero-mean, uniformly distributed noise** with maximum amplitude $|d[k]|=\frac{\Delta Q}{2}$. It can be shown that this dither noise improves the quality of the quantized signal. However, there is still a noise modulation (i.e. a too high correlation between $x_q[k]$ and $e[k]$) that depends on the amplitude of the input signal. The noise modulation can be almost completely eliminated with a **zero-mean noise** signal exhibiting a **symmetric triangular PDF** \begin{equation} p_\text{TRI}(d)=\frac{1}{\Delta Q}\,\text{tri}\left(\frac{d}{\Delta Q}\right) \end{equation} with maximum amplitude $|d[k]|=Q$. By doing so, an almost ideal decorrelation between $x_q[k]$ and $e[k]$ is realized. In audio, this technique is called TPDF-Dithering (Triangular Probability Density Function Dithering) and can be applied in the mastering process of audio material that is to be distributed e.g. on a CD or via streaming. ## Task To get an impression on how dithering may be implemented and what quantized signals sound like, the following exercises shall be performed. - Generate the sine signal $x[k]$ defined above. - Generate the dither noise $d_\text{RECT}[k]$ according to the PDF $p_\text{RECT}(d)$. Check the resulting amplitude and distribution carefully. The length of $d_\text{RECT}[k]$ and $x[k]$ must be equal. - Generate the dither noise $d_\text{TRI}[k]$ according to the PDF $p_\text{TRI}(d)$. Check the resulting amplitude and distribution carefully. The length of $d_\text{TRI}[k]$ and $x[k]$ must be equal. - Add each dither noise $d_\text{RECT}[k]$ and $d_\text{TRI}[k]$ individually to $x[k]$. Together with the case of no dithering we now have three signals to be quantized. - Quantize these signals individually using `my_quant(x,Q)` with $Q$ quantization steps. - Plot the midtread characteristic curve. - Plot the histogram of the dither noises as estimate of its PDF. - Plot the histogram of the error noises as estimate of its PDF. - Plot the sine signal, the dithered signal, the quantized signal and the quantization error signal in one diagram for all three cases. - Calculate and plot the CCF of the signals $x_q[k]$ and $e[k]$ for all three cases. - Interpret the obtained graphics. - For each case, render WAV files from $x[k]$, $x[k]+d[k]$, $x_q[k]$ und $e[k]$ and listen to them. **Be careful! Do not harm your ears!** Pay special attention to the sound of the quantization error, how it is correlated with the quantized signal and how loud it appears. - Consider the 5 cases 1. $B=16$ Bit, $A=1-\Delta Q$ 2. $B=16$ Bit, $A=\Delta Q$ 3. $B=3$ Bit, $A=1-\Delta Q$ 4. $B=3$ Bit, $A=\Delta Q$ 5. $B=3$ Bit, $A=\frac{\Delta Q}{2}$ In the last case the signal has amplitude even below the quantization step size $\Delta Q$. You might verify by listening that the sine is still perceivable if dithering is applied, but not if no dithering is applied. **Again: Be careful! Do not harm your ears!** The signal amplitude $A$ and chosen $B$ is directly related to the playback level! **Warning again: start with very very low playback level, find the loudest signal first and then increase volume to your convenience** ## Solution The task asks for repeated steps. This is perfectly handled by a little function that solves the repeating subtasks. ```python fs = 48000 N = 2*fs k = np.arange(0, N) fsin = 997 ``` ```python def check_dithering(x, dither, Q, case): deltaQ = 1 / (Q//2) # general rule # dither noise pdf_dither, edges_dither = np.histogram(dither, bins='auto', density=True) xd = x + dither # quantization xq = my_quant(xd, Q) e = xq-x pdf_error, edges_error = np.histogram(e, bins='auto', density=True) # write wavs sf.write(file='x_'+case+'.wav', data=x, samplerate=48000, subtype='PCM_24') sf.write(file='xd_'+case+'.wav', data=xd, samplerate=48000, subtype='PCM_24') sf.write(file='xq_'+case+'.wav', data=xq, samplerate=48000, subtype='PCM_24') sf.write(file='e_'+case+'.wav', data=e, samplerate=48000, subtype='PCM_24') # CCF kappa, ccf = my_xcorr2(xq, e, scaleopt='biased') plt.figure(figsize=(12, 3)) if case == 'nodither': plt.subplot(1, 2, 1) # nothing to plot for the zero signal # the PDF would be a weighted Dirac at amplitude zero else: # plot dither noise PDF estimate as histogram plt.subplot(1, 2, 1) plt.plot(edges_dither[:-1], pdf_dither, 'o-', ms=5) plt.ylim(-0.1, np.max(pdf_dither)*1.1) plt.grid(True) plt.xlabel(r'$\theta$') plt.ylabel(r'$\hat{p}(\theta)$') plt.title('PDF estimate of dither noise') # plot error noise PDF estimate as histogram plt.subplot(1, 2, 2) plt.plot(edges_error[:-1], pdf_error, 'o-', ms=5) plt.ylim(-0.1, np.max(pdf_error)*1.1) plt.grid(True) plt.xlabel(r'$\theta$') plt.ylabel(r'$\hat{p}(\theta)$') plt.title('PDF estimate of error noise') # plot signals plt.figure(figsize=(12, 3)) plt.subplot(1, 2, 1) plt.plot(k, x, color='C2', label=r'$x[k]$') plt.plot(k, xd, color='C1', label=r'$x_d[k] = x[k] + dither[k]$') plt.plot(k, xq, color='C3', label=r'$x_q[k]$') plt.plot(k, e, color='C0', label=r'$e[k] = x_q[k] - x[k]$') plt.plot(k, k*0+deltaQ, ':k', label=r'$\Delta Q$') plt.xlabel('k') plt.title('signals') plt.xticks(np.arange(0, 175, 25)) plt.xlim(0, 150) plt.legend(loc='lower left') plt.grid(True) # plot CCF plt.subplot(1, 2, 2) plt.plot(kappa, ccf) plt.xlabel(r'$\kappa$') plt.ylabel(r'$\varphi_{xq,e}[\kappa]$') plt.title('CCF betwen xq and e=xq-x') plt.xticks(np.arange(-100, 125, 25)) plt.xlim(-100, 100) plt.grid(True) ``` Chose one of the 5 cases and evaluate no dither, uniform PDF dither and triangular PDF dither noises below. ```python # case 1 B = 16 # Bit Q = 2**B # number of quantization steps deltaQ = 1 / (Q//2) # quantization step size x = (1-deltaQ) * np.sin(2*np.pi*fsin/fs*k) # largest positive amplitude ``` ```python # case 2 B = 16 Q = 2**B deltaQ = 1 / (Q//2) x = deltaQ * np.sin(2*np.pi*fsin/fs*k) # smallest amplitude ``` ```python # case 3 B = 3 Q = 2**B deltaQ = 1 / (Q//2) x = (1-deltaQ) * np.sin(2*np.pi*fsin/fs*k) ``` ```python # case 4 this is the default case when running the whole notebook B = 3 Q = 2**B deltaQ = 1 / (Q//2) x = deltaQ * np.sin(2*np.pi*fsin/fs*k) ``` ```python # case 5 if False: B = 3 Q = 2**B deltaQ = 1 / (Q//2) # amplitude below quantization step! x = deltaQ/2 * np.sin(2*np.pi*fsin/fs*k) ``` ```python plt.figure(figsize=(4, 4)) check_my_quant(Q) ``` ### No Dither Noise ```python # no dither check_dithering(x=x, dither=x*0, Q=Q, case='nodither') ``` **Be very careful! Do not harm your ears!** | Signal | Audio Player | | ----------------- | :------------ | | $x[k]$ | <audio type="audio/wave" src="x_nodither.wav" controls></audio> | | $x_q[k]$ | <audio type="audio/wave" src="xq_nodither.wav" controls></audio> | | $e[k]$ | <audio type="audio/wave" src="e_nodither.wav" controls></audio> | ### Uniform PDF Dither Noise ```python # uniform dither with max amplitude of deltaQ/2 np.random.seed(1) dither_uni = (np.random.rand(N) - 0.5) * 2 * deltaQ/2 check_dithering(x=x, dither=dither_uni, Q=Q, case='unidither') ``` **Be very careful! Do not harm your ears!** | Signal | Audio Player | | ----------------- | :------------ | | $x[k]$ | <audio type="audio/wave" src="x_unidither.wav" controls></audio> | | $x_d[k]$ | <audio type="audio/wave" src="xd_unidither.wav" controls></audio> | | $x_q[k]$ | <audio type="audio/wave" src="xq_unidither.wav" controls></audio> | | $e[k]$ | <audio type="audio/wave" src="e_unidither.wav" controls></audio> | ### Triangular PDF Dither Noise ```python np.random.seed(1) # uniform PDF for amplitudes -1...+1: dither_uni1 = (np.random.rand(N) - 0.5) * 2 dither_uni2 = (np.random.rand(N) - 0.5) * 2 # triangular PDF with max amplitude of deltaQ dither_tri = (dither_uni1 + dither_uni2) * deltaQ/2 check_dithering(x=x, dither=dither_tri, Q=Q, case='tridither') ``` **Be very careful! Do not harm your ears!** | Signal | Audio Player | | ----------------- | :------------ | | $x[k]$ | <audio type="audio/wave" src="x_tridither.wav" controls></audio> | | $x_d[k]$ | <audio type="audio/wave" src="xd_tridither.wav" controls></audio> | | $x_q[k]$ | <audio type="audio/wave" src="xq_tridither.wav" controls></audio> | | $e[k]$ | <audio type="audio/wave" src="e_tridither.wav" controls></audio> | # **Copyright** The notebooks are provided as [Open Educational Resources](https://en.wikipedia.org/wiki/Open_educational_resources). Feel free to use the notebooks for your own purposes. The text is licensed under [Creative Commons Attribution 4.0](https://creativecommons.org/licenses/by/4.0/), the code of the IPython examples under the [MIT license](https://opensource.org/licenses/MIT). Please attribute the work as follows: *Frank Schultz, Digital Signal Processing - A Tutorial Featuring Computational Examples* with the URL https://github.com/spatialaudio/digital-signal-processing-exercises

69a5735a69c5eefe3674ac76502d340b7de666f4

ipynb

Jupyter Notebook

quantization/quantization.ipynb

spatialaudio/digital-signal-processing-exercises

0e16bc05cb8ed3dee0537371dbb5826db21c86b3

2019-10-24T14:27:43.000Z

2022-02-22T02:14:43.000Z

quantization/quantization.ipynb

spatialaudio/digital-signal-processing-exercises

0e16bc05cb8ed3dee0537371dbb5826db21c86b3

2019-11-05T12:51:46.000Z

2021-12-17T19:46:19.000Z

quantization/quantization.ipynb

spatialaudio/digital-signal-processing-exercises

0e16bc05cb8ed3dee0537371dbb5826db21c86b3

2019-10-24T14:27:51.000Z

2021-08-06T17:33:24.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Lecture 2 - Introduction to Probability Theory > Probability theory is nothing but common sense reduced to calculation. P. Laplace (1812) ## Objectives + To use probability theory to represent states of knowledge. + To use probability theory to extend Aristotelian logic to reason under uncertainty. + To learn about the **pruduct rule** of probability theory. + To learn about the **sum rule** of probability theory. + What is a **random variable**? + What is a **discrete random variable**? + When are two random variable **independent**? + What is a **continuous random variable**? + What is the **cumulative distribution function**? + What is the **probability density function**? ## Readings Before coming to class, please read the following: + [Chapter 1 of Probabilistic Programming and Bayesian Methods for Hackers](http://nbviewer.ipython.org/github/CamDavidsonPilon/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers/blob/master/Chapter1_Introduction/Chapter1.ipynb) + [Chapter 1](http://home.fnal.gov/~paterno/images/jaynesbook/cc01p.pdf) of (Jaynes, 2003). + [Chapter 2](http://home.fnal.gov/~paterno/images/jaynesbook/cc02p.pdf) of (Jaynes, 2003) (skim through). ## The basic desiderata of probability theory It is actually possible to derive the rules of probability based on a system of common sense requirements. Paraphrasing [Chapter 1](http://home.fnal.gov/~paterno/images/jaynesbook/cc01p.pdf) of \cite{jaynes2003}), we would like our system to satisfy the following desiderata: 1) *Degrees of plausibility are represented by real numbers.* 2) *The system should have a qualitative correspondance with common sense.* 3) *The system should be consistent in the sense that:* + *If a conclusion can be reasoned out in more than one way, then every possible way must lead to the same result.* + *All the evidence relevant to a question should be taken into account.* + *Equivalent states of knowledge must be represented by equivalent plausibility assignments.* ## How to speak about probabilities? Let + A be a logical sentence, + B be another logical sentence, and + and I be all other information we know. There is no restriction on what A and B may be as soon as none of them is a contradiction. We write as a shortcut: $$ \mbox{not A} \equiv \neg, $$ $$ A\;\mbox{and}\;B \equiv A,B \equiv AB, $$ $$ A\;\mbox{or}\;B \equiv A + B. $$ We **write**: $$ p(A|BI), $$ and we **read**: > the probability of A being true given that we know that B and I is true or (assuming knowledge I is implied) > the probability of A being true given that we know that B is true or (making it even shorter) > the probability of A given B. $$ p(\mbox{something} | \mbox{everything known}) = \mbox{probability samething is true conditioned on what is known}. $$ $p(A|B,I)$ is just a number between 0 and 1 that corresponds to the degree of plaussibility of A conditioned on B and I. 0 and 1 are special. + If $$ p(A|BI) = 0, $$ we say that we are certain that A is false if B is true. + If $$ p(A|BI) = 1, $$ we say that we are certain that A is false if B is false. + If $$ p(A|BI) \in (0, 1), $$ we say that we are uncertain about A given that B is false. Depending on whether $p(A|B,I)$ is closer to 0 or 1 we beleive more on one possibiliy or another. Complete ignorance corresponds to a probability of 0.5. ## The rules of probability theory According to [Chapter 2](http://home.fnal.gov/~paterno/images/jaynesbook/cc02m.pdf) of \cite{jaynes2003} the desiderata are enough to derive the rules of probability. These rules are: + The **obvious rule** (in lack of a better name): $$ p(A | I) + p(\neg A | I) = 1. $$ + The **product rule** (also known as the Bayes rule or Bayes theorem): $$ p(AB|I) = p(A|BI)p(B|I). $$ or $$ p(AB|I) = p(B|AI)p(A|I). $$ These two rules are enough to compute any probability we want. Let us demonstrate this by a very simple example. ### Example: Drawing balls from a box without replacement Consider the following example of prior information I: > We are given a box with 10 balls 6 of which are red and 4 of which are blue. The box is sufficiently mixed so that so that when we get a ball from it, we don't know which one we pick. When we take a ball out of the box, we do not put it back. Let A be the sentence: > The first ball we draw is blue. Intuitively, we would set the probability of A equal to: $$ p(A|I) = \frac{4}{10}. $$ This choice can actually be justified, but we will come to this later in this course. From the "obvious rule", we get that the probability of not drawing a blue ball, i.e., the probability of drawing a red ball in the first draw is: $$ p(\neg A|I) = 1 - p(A|I) = 1 - \frac{4}{10} = \frac{6}{10}. $$ Now, let B be the sentence: > The second ball we draw is red. What is the probability that we draw a red ball in the second draw given that we drew a blue ball in the first draw? Just before our second draw, there remain 9 bals in the box, 3 of which are blue and 6 of which are red. Therefore: $$ p(B|AI) = \frac{6}{9}. $$ We have not used the product rule just yet. What if we wanted to find the probability that we draw a blue during the first draw and a red during the second draw? Then, $$ p(AB|I) = p(A|I)p(B|AI) = \frac{4}{10}\frac{6}{9} = \frac{24}{90}. $$ What about the probability o a red followed by a blue? Then, $$ p(\neg AB|I) = p(\neg A|I)p(B|AI) = \left[1 - p(A|I) \right]p(B|\neg AI) = \frac{6}{10}\frac{5}{9} = \frac{30}{90}. $$ ### Other rules of probability theory All the other rules of probability theory can be derived from these two rules. To demonstrate this, let's prove that: $$ p(A + B|I) = p(A|I) + p(B|I) - p(AB|I). $$ Here we go: \begin{eqnarray*} p(A+B|I) &=& 1 - p(\neg A \neg B|I)\;\mbox{(obvious rule)}\\ &=& 1 - p(\neg A|\neg BI)p(\neg B|I)\;\mbox{(product rule)}\\ &=& 1 - [1 - p(A |\neg BI)]p(\neg B|I)\;\mbox{(obvious rule)}\\ &=& 1 - p(\neg B|I) + p(A|\neg B I)p(\neg B|I)\\ &=& 1 - [1 - p(B|I)] + p(A|\neg B I)p(\neg B|I)\\ &=& p(B|I) + p(A|\neg B I)p(\neg B|I)\\ &=& p(B|I) + p(A\neg B|I)\\ &=& p(B|I) + p(\neg B|AI)p(A|I)\\ &=& p(B|I) + [1 - p(B|AI)] p(A|I)\\ &=& p(B|I) + p(A|I) - p(B|AI)p(A|I)\\ &=& p(A|I) + p(B|I) - p(AB|I). \end{eqnarray*} ### The sum rule Now consider a finite set of logical sentences, $B_1,\dots,B_n$ such that: 1. One of them is definitely true: $$ p(B_1+\dots+B_n|I) = 1. $$ 2. They are mutually exclusive: $$ p(B_iB_j|I) = 0,\;\mbox{if}\;i\not=j. $$ The **sum rule** states that: $$ P(A|I) = \sum_i p(AB_i|I) = \sum_i p(A|B_i I)p(B_i|I). $$ We can prove this by induction, but let's just prove it for $n=2$: $$ \begin{array}{ccc} p(A|I) &=& p[A(B_1+B_2|I]\\ &=& p(AB_1 + AB_2|I)\\ &=& p(AB_1|I) + p(AB_2|I) - p(AB_1B_2|I)\\ &=& p(AB_1|I) + p(AB_2|I), \end{array} $$ since $$ p(AB_1B_2|I) = p(A|B_1B_2|I)p(B_1B_2|I) = 0. $$ Let's go back to our example. We can use the sum rule to compute the probability of getting a red ball on the second draw independently of what we drew first. This is how it goes: $$ \begin{array}{ccc} p(B|I) &=& p(AB|I) + p(\neg AB|I)\\ &=& p(B|AI)p(A|I) + p(B|\neg AI) p(\neg A|I)\\ &=& \frac{6}{9}\frac{4}{10} + \frac{5}{9}\frac{6}{10}\\ &=& \dots \end{array} $$ ### Example: Medical Diagnosis This example is a modified version of the one found in [Lecture 1](http://www.zabaras.com/Courses/BayesianComputing/IntroToProbabilityAndStatistics.pdf) of the Bayesian Scientific Computing course offered during Spring 2013 by Prof. N. Zabaras at Cornell University. We are going to examine the usefullness of a new tuberculosis test. Let the prior information, I, be: > The percentage of the population infected by tuberculosis is 0.4%. We have run several experiments and determined that: + If a tested patient has the disease, then 80% of the time the test comes out positive. + If a tested patient does not have the disease, then 90% of the time, the test comes out negative. Suppose now that you administer this test to a patient and that the result is positive. How confident are you that the patient does indeed have the disease? Let's use probability theory to answer this question. Let A be the event: > The patient's test is positive. Let B be the event: > The patient has tuberculosis. According to the prior information, we have: $$ p(B|I) = p(\mbox{has tuberculosis}|I) = 0.004, $$ and $$ p(A|B,I) = p(\mbox{test is positive}|\mbox{has tuberculosis},I) = 0.8. $$ Similarly, $$ p(A|\neg B, I) = p(\mbox{test is positive}|\mbox{does not have tuberculosis}, I) = 0.1. $$ We are looking for: $$ \begin{array}{ccc} p(\mbox{has tuberculosis}|\mbox{test is positive},I) &=& P(B|A,I)\\ &=& \frac{p(AB|I)}{p(A|I)} \\ &=& \frac{p(A|B,I)p(B|I)}{p(A|B,I)p(B|I) + p(A|\neg B, I)p(\neg B|I)}\\ &=& \frac{0.8\times 0.004}{0.8\times 0.004 + 0.1 \times 0.996}\\ &\approx& 0.031. \end{array} $$ How much would you pay for such a test? ## Conditional Independence We say that $A$ and $B$ are **independent** (conditional on I), and write, $$ A\perp B|I, $$ if knowledge of one does not yield any information about the other. Mathematically, by $A\perp B|I$, we mean that: $$ p(A|B,I) = p(A|I). $$ Using the product rule, we can easily show that: $$ A\perp B|I \iff p(AB|I) = p(A|I)p(B|I). $$ ### Question + Give an example of $I, A$ and $B$ so that $A\perp B|I$. Now, let $C$ be another event. We say that $A$ and $B$ are **independent** conditional on $C$ (and I), and write: $$ A\perp B|C,I, $$ if knowlege of $C$ makes information about $A$ irrelevant to $B$ (and vice versa). Mathematically, we mean that: $$ p(A|B,C,I) = p(A|C,I). $$ ### Question + Give an example of $I,A,B,C$ so that $A\perp B|C,I$. ## Random Variables The formal mathematical definition of a random variable involves measure theory and is well beyond the scope of this course. Fortunately, we do not have to go through that route to get a theory that is useful in applications. For us, a **random variable** $X$ will just be a variable of our problem whose value is unknown to us. Note that, you should not take the word "random" too literally. If we could, we would change the name to **uncertain** or **unknown** variable. A random variable could correspond to something fixed but unknown, e.g., the number of balls in a box, or it could correspond to something truely random, e.g., the number of particles that hit a [Geiger counter](https://en.wikipedia.org/wiki/Geiger_counter) in a specific time interval. ### Discrete Random Variables We say that a random variable $X$ is discrete, if the possible values it can take are discrete (possibly countably infinite). We write: $$ p(X = x|I) $$ and we read "the probability of $X$ being $x$". If it does not cause any ambiguity, sometimes we will simplify the notation to: $$ p(x) \equiv p(X=x|I). $$ Note that $p(X=x)$ is actually a discrete function of $x$ which depends on our beliefs about $X$. The function $p(x) = p(X=x|I)$ is known as the probability density function of $X$. Now let $Y$ be another random variable. The **sum rule** becomes: $$ p(X=x|I) = \sum_{y}p(X=x,Y=y|I) = \sum_y p(X=x|Y=y,I)p(Y=y|I), $$ or in simpler notation: $$ p(x) = \sum_y p(x,y) = \sum_y p(x|y)p(y). $$ The function $p(X=x, Y=y|I) \equiv p(x, y)$ is known as the joint *probability mass function* of $X$ and $Y$. The **product rule** becomes: $$ p(X=x,Y=y|I) = p(X=x|Y=y,I)p(Y=y|I), $$ or in simpler notation: $$ p(x,y) = p(x|y)p(y). $$ We say that $X$ and $Y$ are **independent** and write: $$ X\perp Y|I, $$ if knowledge of one does not yield any information about the other. Mathematically, $Y$ gives no information about $X$ if: $$ p(x|y) = p(x). $$ From the product rule, however, we get that: $$ p(x) = p(x|y) = \frac{p(x,y)}{p(y)}, $$ from which we see that the joint distribution of $X$ and $Y$ must factorize as: $$ p(x, y) = p(x) p(y). $$ It is trivial to show that if this factorization holds, then $$ p(y|x) = p(y), $$ and thus $X$ yields no information about $Y$ either. ### Continuous Random Variables A random variable $X$ is continuous if the possible values it can take are continuous. The probability of a continuous variable getting a specific value is always zero. Therefore, we cannot work directly with probability mass functions as we did for discrete random variables. We would have to introduce the concepts of the **cumulative distribution function** and the **probability density function**. Fortunately, with the right choice of mathematical symbols, the theory will look exactly the same. Let us start with a real continuous random variable $X$, i.e., a random variable taking values in the real line $\mathbb{R}$. Let $x \in\mathbb{R}$ and consider the probability of $X$ being less than or equal to $x$: $$ F(x) := p(X\le x|I). $$ $F(x)$ is known as the **cumulative distribution function** (CDF). Here are some properties of the CDF whose proof is left as an excersise: + The CDF starts at zero and goes up to one: $$ F(-\infty) = 0\;\mbox{and}\;F(+\infty) = 1. $$ + $F(x)$ is an increasing function of $x$, i.e., $$ x_1 \le x_2 \implies F(x_1)\le F(x_2). $$ + The probability of $X$ being in the interval $[x_1,x_2]$ is: $$ p(x_1 \le X \le x_2|I) = F(x_2) - F(x_1). $$ Now, assume that the derivative of $F(x)$ with respect to $x$ exists. Let us call it $f(x)$: $$ f(x) = \frac{dF(x)}{dx}. $$ Using the fundamental theorem of calculus, it is trivial to show Eq. (\ref{eq:CDF_prob}) implies: \begin{equation} p(x_1 \le X \le x_2|I) = \int_{x_1}^{x_2}f(x)dx. \end{equation} $f(x)$ is known as the **probability density function** (PDF) and it is measured in probability per unit of $X$. To see this note that: $$ p(x \le X \le x + \delta x|I) = \int_{x}^{x+\delta x}f(x')dx' \approx f(x)\delta x, $$ so that: $$ f(x) \approx \frac{p(x \le X \le x + \delta x|I)}{\delta x}. $$ The PDF should satisfy the following properties: + It should be positive $$ f(x) \ge 0, $$ + It should integrate to one: $$ \int_{-\infty}^{\infty} f(x) dx = 1. $$ #### Notation about the PDF of continuous random variables In order to make all the formulas of probability theory the same, we define for a continuous random variable $X$: $$ p(x) := f(x) = \frac{dF(x)}{dx} = \frac{d}{dx}p(X \le x|I). $$ But keep in mind, that if $X$ is continuous $p(x)$ is not a probability but a probability density. That is, it needs a $dx$ to become a probability. Let the PDF $p(x)$ of $X$ and the PDF $p(y)$ of $Y$ ($Y$ is another continuous random variable). We can find the PDF of the random variable $X$ conditioned on $Y$, i.e., the PDF of $X$ if $Y$ is directly observed. This is the **product rule** for continuous random variables: \begin{equation} \label{eq:continuous_bayes} p(y|x) = \frac{p(x, y)}{p(y)}, \end{equation} where $p(x,y)$ is the **joint PDF** of $X$ and $Y$. The **sum rule** for continous random variables is: \begin{equation} \label{eq:continuous_sum} p(x) = \int p(x, y) dy = \int p(x | y) p(y) dy. \end{equation} The similarity between these rules and the discrete ones is obvious. We have prepared a table to help you remember it. | Concept | Discrete Random Variables | Continuous Random Variables | |---|---------------|-----------------| |$p(x)$| in units of robability | in units of probability per unit of $X$| |sum rule| $\sum_y p(x,y) = \sum_y p(x|y)p(y)$ | $\int_y p(x,y) dy = \int_y p(x|y) p(y)$| |product rule| $p(x,y) = p(x|y)p(y)$ | $p(x,y) = p(x|y)p(y)$| ## Expectations Let $X$ be a random variable. The expectation of $X$ is defined to be: $$ \mathbb{E}[X] := \mathbb{E}[X | I] = \int x p(x) dx. $$ Now let $g(x)$ be any function. The expectation of $g(X)$, i.e., the random variable defined after passing $X$ through $g(\cdot)$, is: $$ \mathbb{E}[g(X)] := \mathbb{E}[g(X)|I] = \int g(x)p(x)dx. $$ As usual, calling $\mathbb{E}[\cdot]$ is not a very good name. You may think of $\mathbb{E}[g(X)]$ as the expected value of $g(X)$, but do not take it too far. Can you think of an example in which the expected value is never actually observed? ### Conditional Expectation Let $X$ and $Y$ be two random variables. The conditional expectation of $X$ given $Y=y$ is defined to be: $$ \mathbb{E}[X|Y=y] := \mathbb{E}[X|Y=y,I] = \int xp(x|y)dx. $$ ### Properties of Expectations The following properties of expectations of random variables are extremely helpful. In what follows, $X$ and $Y$ are random variables and $c$ is a constant: + Sum of random variable with a constant: $$ \mathbb{E}[X+c] = \mathbb{E}[X] + c. $$ + Sum of two random variables: $$ \mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]. $$ + Product of random variable with constant: $$ \mathbb{E}[cX] = c\mathbb{E}[X]. $$ + If $X\perp Y$, then: $$ \mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y]. $$ **NOTE**: This property does not hold if $X$ and $Y$ are not independent! + If $f(\cdot)$ is a convex function, then: $$ f(\mathbb{E}[X]) \le \mathbb{E}[f(X)]. $$ **NOTE**: The equality holds only if $f(\cdot)$ is linear! ### Variance of a Random Variable The variance of $X$ is defined to be: $$ \mathbb{V}[X] = \mathbb{E}\left[X - \mathbb{E}[X])^2\right]. $$ It is easy to prove (and a very useful formulat to remember), that: $$ \mathbb{V}[X] = \mathbb{E}[X^2] - \left(\mathbb{E}[X]\right)^2. $$ ### Covariance of Two Random Variables Let $X$ and $Y$ be two random variables. The covariance between $X$ and $Y$ is defined to be: $$ \mathbb{C}[X, Y] = \mathbb{E}\left[\left(X - \mathbb{E}[X]\right) \left(Y-\mathbb{E}[Y]\right)\right] $$ ### Properties of the Variance Let $X$ and $Y$ be random variables and $c$ be a constant. Then: + Sum of random variable with a constant: $$ \mathbb{V}[X + c] = \mathbb{V}[X]. $$ + Product of random variable with a constant: $$ \mathbb{V}[cX] = c^2\mathbb{V}[X]. $$ + Sum of two random variables: $$ \mathbb{V}[X+Y] = \mathbb{V}[X] + \mathbb{V}[Y] + 2\mathbb{C}(X,Y). $$ + Sum of two independent random variables: $$ \mathbb{V}[X+Y] = \mathbb{V}[X] + \mathbb{V}[Y]. $$ # References (<a id="cit-jaynes2003" href="#call-jaynes2003">Jaynes, 2003</a>) E T Jaynes, ``_Probability Theory: The Logic of Science_'', 2003. [online](http://bayes.wustl.edu/etj/prob/book.pdf)

0b9379f5e2cc007f964bf8ad3e9ae1d0d82b2d1e

ipynb

Jupyter Notebook

lectures/lec_02.ipynb

GiveMeData/uq-course

da255bf80ed41cdaca04573ab2dad252cf1ca500

2018-01-14T00:48:35.000Z

2021-01-08T01:30:05.000Z

lectures/lec_02.ipynb

GiveMeData/uq-course

da255bf80ed41cdaca04573ab2dad252cf1ca500

lectures/lec_02.ipynb

GiveMeData/uq-course

da255bf80ed41cdaca04573ab2dad252cf1ca500

2018-01-01T14:24:40.000Z

2021-06-24T22:09:48.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python from scipy import stats from statistics import mean, stdev import pandas as pd import numpy as np import matplotlib.pyplot as plt plt.rcParams["font.family"] = "Times New Roman" import sys import os if "../" not in sys.path: sys.path.append("../") import os os.chdir("..") from envs.data_handler import DataHandler import envs.data_utils as du ``` # Data Shift * Step 1: Calculate the Standard Deviation * Step 2: Create the ordering by the mean value for each <component,failure> group and sort them ascending by mean value, component name, failure name * Step 3: Shift data ```python dh = DataHandler(data_generation='Linear', take_component_id=True, transformation='raw') data_new = du.shift_data(dh.data) data_new.head() ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>Optimal_Affected_Component_Uid</th> <th>Optimal_Failure</th> <th>raw</th> </tr> </thead> <tbody> <tr> <th>0</th> <td>_SEwwu-cdEeet0YmmfbMwkw</td> <td>CF1</td> <td>138.2518</td> </tr> <tr> <th>1</th> <td>_SExXgOcdEeet0YmmfbMwkw</td> <td>CF2</td> <td>25.9559</td> </tr> <tr> <th>2</th> <td>_SEx_HucdEeet0YmmfbMwkw</td> <td>CF3</td> <td>61.6029</td> </tr> <tr> <th>3</th> <td>_SEymDucdEeet0YmmfbMwkw</td> <td>CF3</td> <td>50.0732</td> </tr> <tr> <th>4</th> <td>_SExYKucdEeet0YmmfbMwkw</td> <td>CF3</td> <td>25.9599</td> </tr> </tbody> </table> </div> Perform T-Test: ```python ttest = du.execute_ttest(data_new) num_distinguishable_pairs = len(ttest[ttest['pvalue']<0.05]) total = len(ttest.index) - 2 print('{0} of the {1} <component, failure> combination pairs are statistical significant'.format(num_distinguishable_pairs, total)) ``` 636 of the 912 <component, failure> combination pairs are statistical significant # Parameterization Evaluate different spread multiplication factors for an optimal shifting for each of the transformed datasets (e.g. cube/square root transfromation). We add to the series point of a <component,failure> group a shift value $s$, which is calculated as: \begin{align} s = f * (\sigma(x_{-1}) + \sigma(x)) + tieCount_{t} \end{align} Here, we evaluate different values for $f \in \mathbb{N}^{+}$ as the spread mulitplication factor which is mulitplied with the standard deviation of the previous <component,failure> combination $\sigma(x_{-1})$ and of the current one $\sigma(x)$. Since we shift the groups in the order of their ascending mean value, we add the $tieCount_{t}$ at for the comparison at time point $t$ which is the sum of the standard deviations of all <component,failure> pairs which have the same mean value. ```python def parameterize(start, end, step): result = {} trans = ['raw', 'cube', 'sqt', 'log10', 'ln', 'log2'] factor = range(start, end, step) for t in trans: X = [] Y = [] # initialize the result dictionary for each transformation result[t] = (0,0) # (spread_multiplication_factor, number of statistical significant pairs) # evaluate the different spread mulitplication factors for f in factor: # load the data dh = DataHandler(data_generation='Linear', take_component_id=True, transformation=t) shifted_data = du.shift_data(dh.data, spread_multiplication_factor=f) ttest = du.execute_ttest(shifted_data) num_significant_pairs = len(ttest[ttest['pvalue']<0.05]) # save the results for plotting X.append(f) # factor Y.append(num_significant_pairs) # number of statistical significant pairs if num_significant_pairs > result[t][1]: # we were able to generate at least one more statistical significant pair result[t] = (f, num_significant_pairs) print("{0}: With the spread multiplication factor of {1} we have {2} statistical significant pairs. ".format(t, f, num_significant_pairs), end="\r") plt.plot(X, Y, label=t) print(result) plt.xlabel("Spread Multiplication Factor") plt.ylabel("No. of statistical significant pairs") plt.legend() plt.savefig('data_analysis/03_plots/shifting_parameter_evaluation_' + str(step) + '.pdf') plt.show() ``` ```python parameterize(1, 300, 10) ``` ```python parameterize(1, 50, 1) ``` And then saving optimal shifted data to csv ```python result = {'raw': (151, 667), 'cube': (241, 672), 'sqt': (231, 667), 'log10': (231, 672), 'ln': (231, 671), 'log2': (211, 666)} dh = DataHandler(data_generation='Linear', take_component_id=True, transformation='raw') optimal_shifted_data = dh.data[[dh.data.columns[0], dh.data.columns[1]]] for t, value in result.items(): dh = DataHandler(data_generation='Linear', take_component_id=True, transformation=t) data = du.shift_data(dh.data, spread_multiplication_factor=value[0]) ttest = du.execute_ttest(data) print("Shifting {0} data with a factor of {1} results in {2} distinguishable pairs.".format(t, value[0], len(ttest[ttest['pvalue']<0.05]))) optimal_shifted_data = optimal_shifted_data.merge(data[[data.columns[2]]], how='outer', left_index=True, right_index=True) optimal_shifted_data.to_csv('data/prepared_data/LinearShifted_Id.csv') print("Data saved.") ``` Shifting raw data with a factor of 151 results in 667 distinguishable pairs. Shifting cube data with a factor of 241 results in 672 distinguishable pairs. Shifting sqt data with a factor of 231 results in 667 distinguishable pairs. Shifting log10 data with a factor of 231 results in 672 distinguishable pairs. Shifting ln data with a factor of 231 results in 671 distinguishable pairs. Shifting log2 data with a factor of 211 results in 666 distinguishable pairs. Data saved. # Create datasets with only distinguishable <component,failue> ```python trans = ['raw', 'cube', 'sqt', 'log10', 'ln', 'log2'] for t in trans: dh = DataHandler(data_generation='LinearShifted', take_component_id=True, transformation=t) ttest = du.execute_ttest(dh.data) component_failure_list = du.get_distinguishable_groups(ttest) filtered_data = du.filter_dataset(dh.data, component_failure_list) ttest_2 = du.execute_ttest(filtered_data) num_significant_pairs = len(ttest_2[ttest_2['pvalue']<0.05]) print("{0}: {1}/{2} statistical significant pairs ".format(t, num_significant_pairs, len(ttest_2))) filtered_data.to_csv('data/prepared_data/LinearShifted_Id_' + t + '_dist.csv') print("Data saved.") ``` raw: 728/730 statistical significant pairs cube: 727/737 statistical significant pairs sqt: 726/734 statistical significant pairs log10: 727/736 statistical significant pairs ln: 727/735 statistical significant pairs log2: 727/736 statistical significant pairs Data saved. ```python ```

47972b560c76cbcfc02defa6e55444e50f65a26e

ipynb

Jupyter Notebook

data_analysis/03_shifting.ipynb

hpi-sam/RL_4_Feedback_Control

7e30e660f426f7f62a740e9fd4dafb32e3222690

data_analysis/03_shifting.ipynb

hpi-sam/RL_4_Feedback_Control

7e30e660f426f7f62a740e9fd4dafb32e3222690

2022-01-11T08:06:44.000Z

2022-03-10T10:31:34.000Z

data_analysis/03_shifting.ipynb

hpi-sam/RL_4_Feedback_Control

7e30e660f426f7f62a740e9fd4dafb32e3222690

2022-01-06T08:47:09.000Z

2022-01-06T08:47:09.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Solving Max-Cut Problem with QAOA <em> Copyright (c) 2021 Institute for Quantum Computing, Baidu Inc. All Rights Reserved. </em> ## Overview In the [tutorial on Quantum Approximate Optimization Algorithm](./QAOA_EN.ipynb), we talked about how to encode a classical combinatorial optimization problem into a quantum optimization problem and slove it with Quantum Approximate Optimization Algorithm [1] (QAOA). In this tutorial, we will take the Max-Cut Problem as an example to further elaborate on QAOA. ### Max-Cut Problem The Max-Cut Problem is a common combinatorial optimization problem in graph theory, and it has important applications in statistical physics and circuit design. The maximum cut problem is an NP-hard problem, so there is no efficient algorithm that can solve this problem perfectly. In graph theory, a graph is represented by a pair of sets $G=(V, E)$, where the elements in the set $V$ are the vertices of the graph, and each element in the set $E$ is a pair of vertices, representing an edge connecting these two vertices. For example, the graph in the figure below is represented by $V=\{0,1,2,3\}$ and $E=\{(0,1),(1,2),(2,3),(3, 0)\}$. <div style="text-align:center">Figure 1: A graph with four vertices and four edges </div> A cut on a graph refers to a partition of the graph's vertex set $V$ into two disjoint sets. Each cut corresponds to a set of edges, in which the two vertices of each edge are divided into different sets. So we can define the size of this cut as the size of this set of edges, that is, the number of edges being cut. The Max-Cut Problem is to find a cut that maximizes the number of edges being cut. Figure 2 shows a maximum cut of the graph in Figure 1. The size of the maximum cut is $4$, which means that all edges in the graph are cut. <div style="text-align:center">Figure 2: A maximum cut of the graph in Figure 1 </div> Assuming that the input graph $G=(V, E)$ has $n=|V|$ vertices and $m=|E|$ edges, we can describe the Max-Cut Problem as a combinatorial optimization problem with $n$ bits and $m$ clauses. Each bit corresponds to a vertex $v$ in the graph $G$, and its value $z_v$ is $0$ or $1$, corresponding to the vertex belonging to the set $S_{0}$ or $S_{1}$, respectively. Thus, each value $z$ of these $n$ bits corresponds to a distinct cut. Each clause corresponds to an edge $(u,v)$ in the graph $G$. A clause requires that the two vertices connected by its corresponding edge take different values, namely $z_u\neq z_v$, which means the edge is cut. In other words, when the two vertices connected by the edge are divided into different sets, we say that the clause is satisfied. Therefore, for each edge $(u,v)$ in the graph $G$, we have $$ C_{(u,v)}(z) = z_u+z_v-2z_uz_v, \tag{1} $$ where $C_{(u,v)}(z) = 1$ if and only if the edge is cut. Otherwise, the function is equal to $0$. The objective function of the entire combinatorial optimization problem is $$ C(z) = \sum_{(u,v)\in E}C_{(u,v)}(z) = \sum_{(u,v)\in E}z_u+z_v-2z_uz_v. \tag{2} $$ Therefore, to solve the maximum cut problem is to find a value $z$ that maximizes the objective function in equation (2). ### Encoding Max-Cut Problem Here we take the Max-Cut Problem as an example to further elaborate on QAOA. In order to transform the Max-Cut Problem into a quantum problem, we need to use $n$ qubits, where each qubit corresponds to a vertex in the graph $G$. A qubit being in a quantum state $|0\rangle$ or $|1\rangle$ indicates that its corresponding vertex belongs to the set $S_{0}$ or $S_{1}$, respectively. It is worth noting that $|0\rangle$ and $|1\rangle$ are the two eigenstates of Pauli $Z$ gate, and their eigenvalues are respectively $1$ and $-1$, namely $$ \begin{align} Z|0\rangle&=|0\rangle,\tag{3}\\ Z|1\rangle&=-|1\rangle.\tag{4} \end{align} $$ Therefore, we can use Pauli $Z$ gate to construct the Hamiltonian $H_C$ of the Max-Cut Problem. Because mapping $f(x):x\to(x+1)/2$ maps $-1$ to $0$ and $1$ to $1$, we can replace $z$ in equation (2) with $(Z+I)/2$ ($I$ is the identity matrix) to get the Hamiltonian corresponding to the objective function of the original problem: $$ \begin{align} H_C &= \sum_{(u,v)\in E} \frac{Z_u+I}{2} + \frac{Z_v+I}{2}-2\cdot\frac{Z_u+I}{2} \frac{Z_v+I}{2}\tag{5}\\ &= \sum_{(u,v)\in E} \frac{Z_u+Z_v+2I-(Z_uZ_v+Z_u+Z_v+I)}{2}\tag{6}\\ &= \sum_{(u,v)\in E} \frac{I-Z_uZ_v}{2}.\tag{7} \end{align} $$ The expected value of this Hamiltonian for a quantum state $|\psi\rangle$ is $$ \begin{align} \langle\psi|H_C|\psi\rangle &= \langle\psi|\sum_{(u,v)\in E} \frac{I-Z_uZ_v}{2}|\psi\rangle\tag{8} \\ &= \langle\psi|\sum_{(u,v)\in E} \frac{I}{2}|\psi\rangle-\langle\psi|\sum_{(u,v)\in E} \frac{Z_uZ_v}{2}|\psi\rangle\tag{9}\\ &= \frac{|E|}{2}-\frac{1}{2}\langle\psi|\sum_{(u,v)\in E} Z_uZ_v|\psi\rangle.\tag{10} \end{align} $$ If we define $$ H_D = -\sum_{(u,v)\in E} Z_uZ_v, \tag{11} $$ then finding the quantum state $|\psi\rangle$ that maximizes $\langle\psi|H_C|\psi\rangle$ is equivalent to finding the quantum state $|\psi\rangle$ such that $\langle\psi|H_D|\psi \rangle$ is the largest. ## Paddle Quantum Implementation Now, let's implement QAOA with Paddle Quantum to solve the Max-Cut Problem. There are many ways to find the parameters $\vec{\gamma},\vec{\beta}$. Here we use the gradient descent method in classical machine learning. To implement QAOA with Paddle Quantum, the first thing to do is to import the required packages. Among them, the `networkx` package can help us handle graphs conveniently. ```python from IPython.core.display import HTML display(HTML("<style>pre { white-space: pre !important; }</style>")) ``` <style>pre { white-space: pre !important; }</style> ```python # Import related modules from Paddle Quantum and PaddlePaddle import paddle from paddle_quantum.circuit import UAnsatz from paddle_quantum.utils import pauli_str_to_matrix # Import additional packages needed import numpy as np from numpy import pi as PI import matplotlib.pyplot as plt import networkx as nx ``` Next, we generate the graph $G$ in the Max-Cut Problem. For the convenience of computation, the vertices here are labeled starting from $0$. ```python # n is the number of vertices in the graph G, which is also the number of qubits n = 4 G = nx.Graph() V = range(n) G.add_nodes_from(V) E = [(0, 1), (1, 2), (2, 3), (3, 0)] G.add_edges_from(E) # Print out the generated graph G pos = nx.circular_layout(G) options = { "with_labels": True, "font_size": 20, "font_weight": "bold", "font_color": "white", "node_size": 2000, "width": 2 } nx.draw_networkx(G, pos, **options) ax = plt.gca() ax.margins(0.20) plt.axis("off") plt.show() ``` ### Encoding Hamiltonian In Paddle Quantum, a Hamiltonian can be input in the form of `list`. Here we construct the Hamiltonian $H_D$ in equation (11). ```python # Construct the Hamiltonian H_D in the form of list H_D_list = [] for (u, v) in E: H_D_list.append([-1.0,'z'+str(u) +',z' + str(v)]) print(H_D_list) ``` [[-1.0, 'z0,z1'], [-1.0, 'z1,z2'], [-1.0, 'z2,z3'], [-1.0, 'z3,z0']] As you can see, in this example, the Hamiltonian $H_D$ is $$ H_D = -Z_0Z_1-Z_1Z_2-Z_2Z_3-Z_3Z_0. \tag{12} $$ We can view the matrix form of the Hamiltonian $H_D$ and get information of its eigenvalues: ```python # Convert Hamiltonian H_D from list form to matrix form H_D_matrix = pauli_str_to_matrix(H_D_list, n) # Take out the elements on the diagonal of H_D H_D_diag = np.diag(H_D_matrix).real # Get the maximum eigenvalue of H_D H_max = np.max(H_D_diag) print(H_D_diag) print('H_max:', H_max) ``` [-4. 0. 0. 0. 0. 4. 0. 0. 0. 0. 4. 0. 0. 0. 0. -4.] H_max: 4.0 ### Building the QAOA circuit Earlier we introduced that QAOA needs to apply two unitary transformations $U_C(\gamma)$ and $U_B(\beta)$ alternately on the initial state $|s\rangle = |+\rangle^{\otimes n}$. Here, we use the quantum gates and quantum circuit templates provided in Paddle Quantum to build a quantum circuit to achieve this step. It should be noted that in the Max-Cut Problem, we simplify the problem of maximizing the expected value of the Hamiltonian $H_C$ to the problem of maximizing the expected value of the Hamiltonian $H_D$, so the unitary transformations to be used are $U_D(\gamma)$ and $U_B(\beta)$. By alternately placing two circuit modules with adjustable parameters, we are able to build a QAOA circuit $$ U_B(\beta_p)U_D(\gamma_p)\cdots U_B(\beta_1)U_D(\gamma_1), \tag{13} $$ where $U_D(\gamma) = e^{-i\gamma H_D}$ can be constructed with the circuit in the figure below. Another unitary transformation $U_B(\beta)$ is equivalent to applying a $R_x$ gate to each qubit. <div style="text-align:center">Figure 3: Quantum circuit of unitary transformation $e^{i\gamma Z\otimes Z}$</div> Therefore, the quantum circuit that realizes a layer of unitary transformation $U_B(\beta)U_D(\gamma)$ is shown in Figure 4. <div style="text-align:center">Figure 4: Quantum circuit of unitary transformation $U_B(\beta)U_D(\gamma)$ </div> In Paddle Quantum, the default initial state of each qubit is $|0\rangle$ (the initial state can be customized by input parameters). We can add a layer of Hadamard gates to change the state of each qubit from $|0\rangle$ to $|+\rangle$ so that we get the initial state $|s\rangle = |+\rangle^{\otimes n}$ required by QAOA. In Paddle Quantum, we can add a layer of Hadamard gates to the quantum circuit by calling `superposition_layer()`. ```python def circuit_QAOA(p, gamma, beta): # Initialize the quantum circuit of n qubits cir = UAnsatz(n) # Prepare quantum state |s> cir.superposition_layer() # Build a circuit with p layers for layer in range(p): # Build the circuit of U_D for (u, v) in E: cir.cnot([u, v]) cir.rz(gamma[layer], v) cir.cnot([u, v]) # Build the circuit of U_B, that is, add a layer of R_x gates for v in V: cir.rx(beta[layer], v) return cir ``` After running the constructed QAOA quantum circuit, we obtain the output state $$ |\vec{\gamma},\vec{\beta}\rangle = U_B(\beta_p)U_D(\gamma_p)\cdots U_B(\beta_1)U_D(\gamma_1)|s\rangle. \tag{14} $$ ### Calculating the loss function From the output state of the circuit built in the previous step, we can calculate the objective function of the maximum cut problem $$ F_p(\vec{\gamma},\vec{\beta}) = \langle\vec{\gamma},\vec{\beta}|H_D|\vec{\gamma},\vec{\beta}\rangle. \tag{15} $$ To maximize the objective function is equivalent to minimizing $-F_p$. Therefore, we define $L(\vec{\gamma},\vec{\beta}) = -F_p(\vec{\gamma},\vec{\beta})$ as the loss function, that is, the function to be minimized. Then, we use a classical optimization algorithm to find the optimal parameters $\vec{\gamma},\vec{\beta}$. The following code shows a complete QAOA network built with Paddle Quantum and PaddlePaddle: ```python class Net(paddle.nn.Layer): def __init__(self, p, dtype="float64",): super(Net, self).__init__() self.p = p self.gamma = self.create_parameter(shape=[self.p], default_initializer=paddle.nn.initializer.Uniform(low=0.0, high=2 * PI), dtype=dtype, is_bias=False) self.beta = self.create_parameter(shape=[self.p], default_initializer=paddle.nn.initializer.Uniform(low=0.0, high=2 * PI), dtype=dtype, is_bias=False) def forward(self): # Define QAOA's quantum circuit cir = circuit_QAOA(self.p, self.gamma, self.beta) # Run the quantum circuit cir.run_state_vector() # Calculate the loss function loss = -cir.expecval(H_D_list) return loss, cir ``` ### Training quantum neural network After defining the quantum neural network for QAOA, we use the gradient descent method to update the parameters in the network to maximize the expected value in equation (15). ```python p = 4 # Number of layers in the quantum circuit ITR = 120 # Number of training iterations LR = 0.1 # Learning rate of the optimization method based on gradient descent SEED = 1024 # Set global RNG seed ``` Here, we optimize the network defined above in PaddlePaddle. ```python paddle.seed(SEED) net = Net(p) # Use Adam optimizer opt = paddle.optimizer.Adam(learning_rate=LR, parameters=net.parameters()) # Gradient descent iteration for itr in range(1, ITR + 1): # Run the network defined above loss, cir = net() # Calculate the gradient and optimize loss.backward() opt.minimize(loss) opt.clear_grad() if itr% 10 == 0: print("iter:", itr, "loss:", "%.4f"% loss.numpy()) if itr == ITR: print("\nThe trained circuit:") print(cir) gamma_opt = net.gamma.numpy() print("Optimized parameters gamma:\n", gamma_opt) beta_opt = net.beta.numpy() print("Optimized parameters beta:\n", beta_opt) ``` iter: 10 loss: -3.8886 iter: 20 loss: -3.9134 iter: 30 loss: -3.9659 iter: 40 loss: -3.9906 iter: 50 loss: -3.9979 iter: 60 loss: -3.9993 iter: 70 loss: -3.9998 iter: 80 loss: -3.9999 iter: 90 loss: -4.0000 iter: 100 loss: -4.0000 iter: 110 loss: -4.0000 iter: 120 loss: -4.0000 The trained circuit: --H----*-----------------*--------------------------------------------------X----Rz(3.140)----X----Rx(0.824)----*-----------------*--------------------------------------------------X----Rz(0.737)----X----Rx(2.506)----*-----------------*--------------------------------------------------X----Rz(4.999)----X----Rx(4.854)----*-----------------*--------------------------------------------------X----Rz(0.465)----X----Rx(1.900)-- | | | | | | | | | | | | | | | | --H----X----Rz(3.140)----X----*-----------------*---------------------------|-----------------|----Rx(0.824)----X----Rz(0.737)----X----*-----------------*---------------------------|-----------------|----Rx(2.506)----X----Rz(4.999)----X----*-----------------*---------------------------|-----------------|----Rx(4.854)----X----Rz(0.465)----X----*-----------------*---------------------------|-----------------|----Rx(1.900)-- | | | | | | | | | | | | | | | | --H---------------------------X----Rz(3.140)----X----*-----------------*----|-----------------|----Rx(0.824)---------------------------X----Rz(0.737)----X----*-----------------*----|-----------------|----Rx(2.506)---------------------------X----Rz(4.999)----X----*-----------------*----|-----------------|----Rx(4.854)---------------------------X----Rz(0.465)----X----*-----------------*----|-----------------|----Rx(1.900)-- | | | | | | | | | | | | | | | | --H--------------------------------------------------X----Rz(3.140)----X----*-----------------*----Rx(0.824)--------------------------------------------------X----Rz(0.737)----X----*-----------------*----Rx(2.506)--------------------------------------------------X----Rz(4.999)----X----*-----------------*----Rx(4.854)--------------------------------------------------X----Rz(0.465)----X----*-----------------*----Rx(1.900)-- Optimized parameters gamma: [3.14046713 0.73681226 4.99897226 0.46481489] Optimized parameters beta: [0.82379898 2.50618308 4.85422542 1.90024859] ### Decoding the quantum solution After obtaining the minimum value of the loss function and the corresponding set of parameters $\vec{\gamma}^*,\vec{\beta}^*$, our task has not been completed. In order to obtain an approximate solution to the Max-Cut Problem, it is necessary to decode the solution to the classical optimization problem from the quantum state $|\vec{\gamma}^*,\vec{\beta}^*\rangle$ output by QAOA. Physically, to decode a quantum state, we need to measure it and then calculate the probability distribution of the measurement results: $$ p(z)=|\langle z|\vec{\gamma}^*,\vec{\beta}^*\rangle|^2. \tag{16} $$ Usually, the greater the probability of a certain bit string, the greater the probability that it corresponds to an optimal solution of the Max-Cut problem. Paddle Quantum provides a function to view the probability distribution of the measurement results of the state output by the QAOA quantum circuit: ```python # Repeat the simulated measurement of the circuit output state 1024 times prob_measure = cir.measure(plot=True) ``` After measurement, we can find the bit string with the highest probability of occurrence. Let the vertices whose bit values are $0$ in the bit string belong to the set $S_0$ and the vertices whose bit values are $1$ belong to the set $S_1$. The set of edges between these two vertex sets is a possible maximum cut of the graph. The following code selects the bit string with the greatest chance of appearing in the measurement result, then maps it back to the classic solution, and draws the corresponding maximum cut: - The red vertex belongs to the set $S_0$, - The blue vertex belongs to the set $S_1$, - The dashed line indicates the edge being cut. ```python # Find the most frequent bit string in the measurement results cut_bitstring = max(prob_measure, key=prob_measure.get) print("The bit string form of the cut found:", cut_bitstring) # Draw the cut corresponding to the bit string obtained above on the graph node_cut = ["blue" if cut_bitstring[v] == "1" else "red" for v in V] edge_cut = [ "solid" if cut_bitstring[u] == cut_bitstring[v] else "dashed" for (u, v) in E ] nx.draw( G, pos, node_color=node_cut, style=edge_cut, **options ) ax = plt.gca() ax.margins(0.20) plt.axis("off") plt.show() ``` As you can see, in this example, QAOA has found a maximum cut on the graph. _______ ## References [1] Farhi, E., Goldstone, J. & Gutmann, S. A Quantum Approximate Optimization Algorithm. [arXiv:1411.4028 (2014).](https://arxiv.org/abs/1411.4028)

ddebe3addda1166af869d519ae2c4f871aa10660

ipynb

Jupyter Notebook

tutorial/combinatorial_optimization/MAXCUT_EN.ipynb

gsq7474741/Quantum

16e7d3bf2dba7e94e6faf5c853faf0e913e1f268

2020-07-14T14:10:23.000Z

2020-07-14T14:10:23.000Z

tutorial/combinatorial_optimization/MAXCUT_EN.ipynb

gsq7474741/Quantum

16e7d3bf2dba7e94e6faf5c853faf0e913e1f268

tutorial/combinatorial_optimization/MAXCUT_EN.ipynb

gsq7474741/Quantum

16e7d3bf2dba7e94e6faf5c853faf0e913e1f268

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python import os, sys import h5py import numpy as np from scipy.io import loadmat import cv2 import matplotlib %matplotlib inline import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D from numpy import matrix as mat from sympy import * from numpy import linalg as la ``` ```python def getFx(para, frame): # para中为(13*frame+3)的一套参数，frame传进来是为了确定循环次数 #先写出参数表达式，ABDCDEF六个点的齐次坐标 K = Matrix([[1149.67569986785, 0.0, 508.848621645943], [0.0, 1147.59161666764, 508.064917088557], [0.0, 0.0, 1.0]]) r11, r12, r13, r14, r21, r22, r23, r24, r31, r32, r33 = symbols('r11 r12 r13 r14 r21 r22 r23 r24 r31 r32 r33') Rt = Matrix([[r11, r12, r13, r14], [r21, r22, r23, r24], [r31, r32, r33, 1]]) a, b, c, th, al = symbols('a b c th al') ua, va, wa, ub, vb, wb, uc, vc, wc, ud, vd, wd, ue, ve, we, uf, vf, wf = symbols('ua va wa ub vb wb uc vc wc ud vd wd ue ve we uf vf wf') f = Symbol('f') XA = Matrix([[-a * c * cos(th) * cos(al)], [c-a * c * sin(th)], [-a * c * cos(th) * sin(al)], [1]]) XB = Matrix([[0], [c], [0], [1]]) XC = Matrix([[a * c * cos(th) * cos(al)], [c+a * c * sin(th)], [a * c * cos(th) * sin(al)], [1]]) XD = Matrix([[-b * c], [0], [0], [1]]) XE = Matrix([[0], [0], [0], [1]]) XF = Matrix([[b * c], [0], [0], [1]]) ua, va, wa = K[0,:] * (Rt * XA), K[1,:] * (Rt * XA), K[2,:] * (Rt * XA) ub, vb, wb = K[0,:] * (Rt * XB), K[1,:] * (Rt * XB), K[2,:] * (Rt * XB) uc, vc, wc = K[0,:] * (Rt * XC), K[1,:] * (Rt * XC), K[2,:] * (Rt * XC) ud, vd, wd = K[0,:] * (Rt * XD), K[1,:] * (Rt * XD), K[2,:] * (Rt * XD) ue, ve, we = K[0,:] * (Rt * XE), K[1,:] * (Rt * XE), K[2,:] * (Rt * XE) uf, vf, wf = K[0,:] * (Rt * XF), K[1,:] * (Rt * XF), K[2,:] * (Rt * XF) #根据每一帧的循环，提取出Rt的参数，K是公用的，代入参数写出3D坐标，并计算出u/w，v/w #写成f的形式，即按照六个点分块，每块里面有M帧 getfx = mat(np.zeros((6*frame*2,1))) for i in range(6): for j in range(frame): if i == 0 : f = Matrix([ua/wa, va/wa]) elif i == 1 : f = Matrix([ub/wb, vb/wb]) elif i == 2 : f = Matrix([uc/wc, vc/wc]) elif i == 3 : f = Matrix([ud/wd, vd/wd]) elif i == 4 : f = Matrix([ue/we, ve/we]) else: f = Matrix([uf/wf, vf/wf]) f_value = f.subs({r11:para[13*j], r12:para[13*j+1], r13:para[13*j+2], r14:para[13*j+3], r21:para[13*j+4], r22:para[13*j+5], r23:para[13*j+6], r24:para[13*j+7], r31:para[13*j+8], r32:para[13*j+9], r33:para[13*j+10], th:para[13*j+11], al:para[13*j+12], a:para[-3], b:para[-2], c:para[-1]}) getfx[i*frame*2+j*2] = f_value[0] getfx[i*frame*2+j*2+1] = f_value[1] #返回getfx值，2*frame*6 by 1 return getfx ``` ```python def getJacobian(point, frame, para): # 用参数表示K，R矩阵 focalx, focaly, px, py = symbols('focalx focaly px py') r11, r12, r13, r14, r21, r22, r23, r24, r31, r32, r33 = symbols('r11 r12 r13 r14 r21 r22 r23 r24 r31 r32 r33') Rt = Matrix([[r11, r12, r13, r14], [r21, r22, r23, r24], [r31, r32, r33, 1]]) K = Matrix([[focalx, 0, px], [0, focaly, py], [0, 0, 1]]) # KRt = K * Rt # 用参数表示ABCDEF六个点坐标 a, b, c, th, al = symbols('a b c th al') ua, va, wa, ub, vb, wb, uc, vc, wc, ud, vd, wd, ue, ve, we, uf, vf, wf = symbols('ua va wa ub vb wb uc vc wc ud vd wd ue ve we uf vf wf') f = Symbol('f') if point == 0 : XA = Matrix([[-a * c * cos(th) * cos(al)], [c-a * c * sin(th)], [-a * c * cos(th) * sin(al)], [1]]) ua, va, wa = K[0,:] * (Rt * XA), K[1,:] * (Rt * XA), K[2,:] * (Rt * XA) f = Matrix([ua/wa, va/wa]) elif point == 1 : XB = Matrix([[0], [c], [0], [1]]) ub, vb, wb = K[0,:] * (Rt * XB), K[1,:] * (Rt * XB), K[2,:] * (Rt * XB) f = Matrix([ub/wb, vb/wb]) elif point == 2 : XC = Matrix([[a * c * cos(th) * cos(al)], [c+a * c * sin(th)], [a * c * cos(th) * sin(al)], [1]]) uc, vc, wc = K[0,:] * (Rt * XC), K[1,:] * (Rt * XC), K[2,:] * (Rt * XC) f = Matrix([uc/wc, vc/wc]) elif point == 3 : XD = Matrix([[-b * c], [0], [0], [1]]) ud, vd, wd = K[0,:] * (Rt * XD), K[1,:] * (Rt * XD), K[2,:] * (Rt * XD) f = Matrix([ud/wd, vd/wd]) elif point == 4 : XE = Matrix([[0], [0], [0], [1]]) ue, ve, we = K[0,:] * (Rt * XE), K[1,:] * (Rt * XE), K[2,:] * (Rt * XE) f = Matrix([ue/we, ve/we]) elif point == 5: XF = Matrix([[b * c], [0], [0], [1]]) uf, vf, wf = K[0,:] * (Rt * XF), K[1,:] * (Rt * XF), K[2,:] * (Rt * XF) f = Matrix([uf/wf, vf/wf]) args = Matrix([r11, r12, r13, r14, r21, r22, r23, r24, r31, r32, r33, th, al, a, b, c]) f_X1 = f[0,:].jacobian(args) f_X2 = f[1,:].jacobian(args) JA = Matrix([f_X1, f_X2]) # 2 by 16 matrix JA_value = JA.subs({focalx:1149.676, focaly:1147.592, px:508.849, py:508.065, r11:para[13*frame], r12:para[13*frame+1], r13:para[13*frame+2], r14:para[13*frame+3], r21:para[13*frame+4], r22:para[13*frame+5], r23:para[13*frame+6], r24:para[13*frame+7], r31:para[13*frame+8], r32:para[13*frame+9], r33:para[13*frame+10], th:para[13*frame+11], al:para[13*frame+12], a:para[-3], b:para[-2], c:para[-1]}) #JA_value = JA_value.subs({f:1149.68}) return JA_value ``` ```python def getJ(para, frame): getj = mat(np.zeros((6*frame*2, 13*frame+3))) for m in range(6): for n in range(frame): JA_value = getJacobian(m, n, para) #print(JA_value) getj[2*(m*frame+n):2*(m*frame+n+1), 13*n:13*n+13] = JA_value[:, 0:13] getj[2*(m*frame+n):2*(m*frame+n+1), -3:] = JA_value[:, -3:] return getj ``` ```python def getE(getfx, frame): E = mat(np.zeros((6*frame*2,1))) for i in range(6): for j in range(frame): if i==0 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] elif i==1 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] elif i==2 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] elif i==3 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] elif i==4 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] elif i==5 : E[(i*frame+j)*2] = getfx[i*frame*2+j*2] - x2d[i,0] E[(i*frame+j)*2+1] = getfx[i*frame*2+j*2+1] - x2d[i, 1] return E ``` ```python def LM_opti(frame, x_para, u=1, v=2, step_max=500): J = mat(np.zeros((6*frame*2, 13*frame+3))) E = mat(np.zeros((6*frame*2,1))) # E = f(X) - b ; E_temp = mat(np.zeros((6*frame*2,1))) # E_temp compare with E in L-M x_k = mat(x_para.copy()) #parameter initialization step = 0 # iteration steps mse_last = 0 # mse value after iteration each time step_max = 500 # maximum number of iteration u = 1 v = 2 # u, v initial value # L-M Algorithm obtain optimal parameters while(step < step_max): step += 1 mse, mse_temp = 0, 0 # generate Jacobian Matrix and calculate E getfx = mat(np.zeros((6*frame*2,1))) getfx = getFx(x_k, frame) E = getE(getfx, frame) for i in range(6*frame*2): mse += E[i]**2 mse /= 6*frame*2 # get new J J = mat(np.zeros((6*frame*2, 13*frame+3))) J = getJ(x_k, frame) # delta X = ... #print(J.T * J) dx = mat(np.zeros((13*frame+3,1))) LM = u * mat(np.eye(13*frame+3)) dx = -(J.T * J + LM).I * J.T * E x_k_temp = x_k.copy() x_k_temp += dx #R的更新不能简单赋值 #get R meet R.T*R=I #U * D * V.T = R --> R+ = U * V.T R_old = mat([[x_k_temp[0,0], x_k_temp[1,0], x_k_temp[2,0]], [x_k_temp[4,0], x_k_temp[5,0], x_k_temp[6,0]], [x_k_temp[8,0], x_k_temp[9,0], x_k_temp[10,0]]]) U, sigma, VT = la.svd(R_old) R_new = U * VT x_k_temp[0,0], x_k_temp[1,0], x_k_temp[2,0] = R_new[0,0], R_new[0,1], R_new[0,2] x_k_temp[4,0], x_k_temp[5,0], x_k_temp[6,0] = R_new[1,0], R_new[1,1], R_new[1,2] x_k_temp[8,0], x_k_temp[9,0], x_k_temp[10,0] = R_new[2,0], R_new[2,1], R_new[2,2] ########### # calculate E_temp with x_k_temp # copy from E with x_k getfx_temp = mat(np.zeros((6*frame*2,1))) getfx_temp = getFx(x_k_temp, frame) E_temp = getE(getfx_temp, frame) for i in range(6*frame*2): mse_temp += E_temp[i]**2 mse_temp /= 6*frame*2 # segma value to choose optimization model segma = (mse - mse_temp)/((dx.T * (u * dx - J.T * E))[0,0]) # calculate new u if segma > 0: s = 1.0/3.0 v = 2 x_k = x_k_temp mse = mse_temp u = u * max(s, 1-pow(2*segma,3)) u = u[0,0] else: u = u * v v = v * 2 x_k = x_k_temp print("step = %d, abs(mse-mse_last) = %.8f" %(step, abs(mse-mse_last))) if abs(mse-mse_last)<0.000001: break mse_last = mse print("step = ", step) print("mse = ", mse_last) #print("parameter = ", x_k) return x_k ``` ```python # 数据读取 frame = 1 m = loadmat("valid.mat") # camera intrinsic matrix K = m["annot"][0][0][4] K_cam = K[0][0].tolist() # key point 3D groundtruth gt = m["annot"][0][0][3] img1_gt = gt[135] # array 3 by 17 kp = np.zeros((17,2)) for i in range(17): u = K_cam[0] * mat([img1_gt[0][i], img1_gt[1][i], img1_gt[2][i]]).T v = K_cam[1] * mat([img1_gt[0][i], img1_gt[1][i], img1_gt[2][i]]).T w = K_cam[2] * mat([img1_gt[0][i], img1_gt[1][i], img1_gt[2][i]]).T kp[i][0] = u/w kp[i][1] = v/w # load and show image img = cv2.imread("S9_Posing_1.55011271_000676.jpg") plt.figure("Image") # 图像窗口名称 plt.imshow(img[:,:,[2,1,0]]) plt.axis('on') # 关掉坐标轴为 off plt.title('image1') # 图像题目 plt.show() # visualize key points txt = ['1','2','3','4','5','6','7','8','9','10','11','12','13','14','15','16','17'] img_kp = plt.scatter(kp[:,0], kp[:,1], s = 80, c = 'g', marker = 'X') for i in range(17): plt.annotate(txt[i], xy = (kp[i,0], kp[i,1]), xytext = (kp[i,0]+0.1, kp[i,1]+0.1)) # 这里xy是需要标记的坐标，xytext是对应的标签坐标 plt.axis('on') # 关掉坐标轴为 off plt.title('image_kp') # 图像题目 # visualize ABCDEF plt.figure() img_kp = plt.scatter(kp[0,0], kp[0,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(kp[1,0], kp[1,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(kp[4,0], kp[4,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(kp[8,0], kp[8,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(kp[11,0], kp[11,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(kp[14,0], kp[14,1], s = 80, c = 'g', marker = 'X') plt.axis('on') # 关掉坐标轴为 off plt.title('image_kp_ABCDEF') # 图像题目 plt.show() # save 2D coordinate to list x2d = np.zeros((6 * frame,2)) x2d[0,0] = kp[0,0] for i in range(6): for j in range(frame): if i==0 : x2d[i*frame+j, 0] = kp[14, 0] x2d[i*frame+j, 1] = kp[14, 1] elif i==1 : x2d[i*frame+j, 0] = kp[8, 0] x2d[i*frame+j, 1] = kp[8, 1] elif i==2 : x2d[i*frame+j, 0] = kp[11, 0] x2d[i*frame+j, 1] = kp[11, 1] elif i==3 : x2d[i*frame+j, 0] = kp[4, 0] x2d[i*frame+j, 1] = kp[4, 1] elif i==4 : x2d[i*frame+j, 0] = kp[0, 0] x2d[i*frame+j, 1] = kp[0, 1] elif i==5 : x2d[i*frame+j, 0] = kp[1, 0] x2d[i*frame+j, 1] = kp[1, 1] print(x2d) ``` ```python # parameter initialization for all frame (K_cam, x2d(6*frame by 2)) # x_para(13*frame+3) x_para = np.zeros((13*frame+3,1)) for i in range(frame): x_para[13*i] = -1 # r11 x_para[13*i+5] = 1 # r22 x_para[13*i+10] = -1 # r33 x_para[13*i+3] = 0.0047 x_para[13*i+7] = -0.0997 x_para[13*i+11] = 0 # th x_para[13*i+12] = 0 #al x_para[-3] = 0.35 # a x_para[-2] = 0.25 # b distance = -0.095#0.096 # c x_para[-1] = distance print(mat(x_para.copy())) getfx_ini = getFx(x_para, frame) print(getfx_ini) ``` [[-1. ] [ 0. ] [ 0. ] [ 0.0047] [ 0. ] [ 1. ] [ 0. ] [-0.0997] [ 0. ] [ 0. ] [-1. ] [ 0. ] [ 0. ] [ 0.35 ] [ 0.25 ] [-0.095 ]] [[476.02538041] [284.62882932] [514.25209744] [284.62882932] [552.47881446] [284.62882932] [486.94729956] [393.65003291] [514.25209744] [393.65003291] [541.55689531] [393.65003291]] ```python plt.figure() img_kp = plt.scatter(kp[14,0], kp[14,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[0,0], getfx_ini[1,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[8,0], kp[8,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[2,0], getfx_ini[3,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[11,0], kp[11,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[4,0], getfx_ini[5,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[4,0], kp[4,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[6,0], getfx_ini[7,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[0,0], kp[0,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[8,0], getfx_ini[9,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[1,0], kp[1,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_ini[10,0], getfx_ini[11,0], s = 80, c = 'r', marker = 'X') plt.axis('on') # 关掉坐标轴为 off plt.title('image_kp_ABCDEF_original_opt') # 图像题目 plt.show() ``` ```python x_k = LM_opti(frame, x_para) R_old = mat([[x_k[0,0], x_k[1,0], x_k[2,0]], [x_k[4,0], x_k[5,0], x_k[6,0]], [x_k[8,0], x_k[9,0], x_k[10,0]]]) U, sigma, VT = la.svd(R_old) R_new = U * VT x_k[0,0], x_k[1,0], x_k[2,0] = R_new[0,0], R_new[0,1], R_new[0,2] x_k[4,0], x_k[5,0], x_k[6,0] = R_new[1,0], R_new[1,1], R_new[1,2] x_k[8,0], x_k[9,0], x_k[10,0] = R_new[2,0], R_new[2,1], R_new[2,2] ``` step = 1, abs(mse-mse_last) = 9.83445441 step = 2, abs(mse-mse_last) = 0.07658973 step = 3, abs(mse-mse_last) = 0.03711521 ```python # visualize keypoint after BA # compare with image_kp_ABCDEF print(x_k) getfx_final = getFx(x_k, frame) print(getfx_final) plt.figure() img_kp = plt.scatter(kp[14,0], kp[14,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[0,0], getfx_final[1,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[8,0], kp[8,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[2,0], getfx_final[3,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[11,0], kp[11,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[4,0], getfx_final[5,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[4,0], kp[4,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[6,0], getfx_final[7,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[0,0], kp[0,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[8,0], getfx_final[9,0], s = 80, c = 'r', marker = 'X') img_kp = plt.scatter(kp[1,0], kp[1,1], s = 80, c = 'g', marker = 'X') img_kp = plt.scatter(getfx_final[10,0], getfx_final[11,0], s = 80, c = 'r', marker = 'X') plt.axis('on') # 关掉坐标轴为 off plt.title('image_kp_ABCDEF_original_opt') # 图像题目 plt.show() ``` ```python def getValue(X3D, para): j = 0 X3D_value = X3D.subs({r11:para[13*j], r12:para[13*j+1], r13:para[13*j+2], r14:para[13*j+3], r21:para[13*j+4], r22:para[13*j+5], r23:para[13*j+6], r24:para[13*j+7], r31:para[13*j+8], r32:para[13*j+9], r33:para[13*j+10], th:para[13*j+11], al:para[13*j+12], a:para[-3], b:para[-2], c:para[-1]}) return X3D_value ``` ```python # visualize 3D points and groundtruth # XA,XB,XC,XD,XE,XF with GT[14,8,11,4,0,1] para = x_k.copy() K = Matrix([[1149.67569986785, 0.0, 508.848621645943], [0.0, 1147.59161666764, 508.064917088557], [0.0, 0.0, 1.0]]) r11, r12, r13, r14, r21, r22, r23, r24, r31, r32, r33 = symbols('r11 r12 r13 r14 r21 r22 r23 r24 r31 r32 r33') Rt = Matrix([[r11, r12, r13, r14], [r21, r22, r23, r24], [r31, r32, r33, 1]]) a, b, c, th, al = symbols('a b c th al') ua, va, wa, ub, vb, wb, uc, vc, wc, ud, vd, wd, ue, ve, we, uf, vf, wf = symbols('ua va wa ub vb wb uc vc wc ud vd wd ue ve we uf vf wf') XA = Matrix([[-a * c * cos(th) * cos(al)], [c-a * c * sin(th)], [-a * c * cos(th) * sin(al)], [1]]) XB = Matrix([[0], [c], [0], [1]]) XC = Matrix([[a * c * cos(th) * cos(al)], [c+a * c * sin(th)], [a * c * cos(th) * sin(al)], [1]]) XD = Matrix([[-b * c], [0], [0], [1]]) XE = Matrix([[0], [0], [0], [1]]) XF = Matrix([[b * c], [0], [0], [1]]) A3D = Rt * XA B3D = Rt * XB C3D = Rt * XC D3D = Rt * XD E3D = Rt * XE F3D = Rt * XF j = 0 #A3D_value = A3D.subs({r11:para[13*j], r12:para[13*j+1], r13:para[13*j+2], r14:para[13*j+3], # r21:para[13*j+4], r22:para[13*j+5], r23:para[13*j+6], r24:para[13*j+7], # r31:para[13*j+8], r32:para[13*j+9], r33:para[13*j+10], th:para[13*j+11], # al:para[13*j+12], a:para[-3], b:para[-2], c:para[-1]}) s = 5340.55881868 E3D_value = (getValue(E3D, para) - getValue(E3D, para))*s A3D_value = (getValue(A3D, para) - getValue(E3D, para))*s B3D_value = (getValue(B3D, para) - getValue(E3D, para))*s C3D_value = (getValue(C3D, para) - getValue(E3D, para))*s D3D_value = (getValue(D3D, para) - getValue(E3D, para))*s F3D_value = (getValue(F3D, para) - getValue(E3D, para))*s print(A3D_value) print(B3D_value) print(C3D_value) print(D3D_value) print(E3D_value) print(F3D_value) ``` Matrix([[-143.732977969521], [-459.901052798691], [-36.8592481249503]]) Matrix([[29.9916868960575], [-465.649978271430], [2.31230523863426]]) Matrix([[203.716351761636], [-471.398903744170], [41.4838586022188]]) Matrix([[-121.569097994654], [-7.86570758781875], [-7.18081842657119]]) Matrix([[0], [0], [0]]) Matrix([[121.569097994654], [7.86570758781875], [7.18081842657119]]) ```python Y = mat([img1_gt[0][0], img1_gt[1][0], img1_gt[2][0]]).T #s = 5340.55881868 X1 = (mat([img1_gt[0][14], img1_gt[1][14], img1_gt[2][14]]).T-Y) print(X1) X2 = (mat([img1_gt[0][8], img1_gt[1][8], img1_gt[2][8]]).T-Y) print(X2) X3 = (mat([img1_gt[0][11], img1_gt[1][11], img1_gt[2][11]]).T-Y) print(X3) X4 = (mat([img1_gt[0][4], img1_gt[1][4], img1_gt[2][4]]).T-Y) print(X4) X5 = (mat([img1_gt[0][0], img1_gt[1][0], img1_gt[2][0]]).T-Y) print(X5) X6 = (mat([img1_gt[0][1], img1_gt[1][1], img1_gt[2][1]]).T-Y) print(X6) ``` [[-151.3162711 ] [-440.06270576] [ -93.45364426]] [[ 22.92779888] [-491.06066667] [-118.64034573]] [[ 194.39764354] [-432.74999716] [ -90.46298579]] [[-123.65075656] [ -10.21145788] [ -0.85753061]] [[0.] [0.] [0.]] [[123.63491077] [ 10.39999574] [ 0.92235729]] ```python # 绘制散点图 fig = plt.figure() ax = Axes3D(fig) ax.scatter(X1[0], X1[1], X1[2], s = 80, c = 'g', marker = 'X') ax.scatter(X2[0], X2[1], X2[2], s = 80, c = 'g', marker = 'X') ax.scatter(X3[0], X3[1], X3[2], s = 80, c = 'g', marker = 'X') ax.scatter(X4[0], X4[1], X4[2], s = 80, c = 'g', marker = 'X') ax.scatter(X5[0], X5[1], X5[2], s = 80, c = 'g', marker = 'X') ax.scatter(X6[0], X6[1], X6[2], s = 80, c = 'g', marker = 'X') ax.scatter(A3D_value[0], A3D_value[1], A3D_value[2], s = 80, c = 'r', marker = 'X') ax.scatter(B3D_value[0], B3D_value[1], B3D_value[2], s = 80, c = 'r', marker = 'X') ax.scatter(C3D_value[0], C3D_value[1], C3D_value[2], s = 80, c = 'r', marker = 'X') ax.scatter(D3D_value[0], D3D_value[1], D3D_value[2], s = 80, c = 'r', marker = 'X') ax.scatter(E3D_value[0], E3D_value[1], E3D_value[2], s = 80, c = 'r', marker = 'X') ax.scatter(F3D_value[0], F3D_value[1], F3D_value[2], s = 80, c = 'r', marker = 'X') ax.set_zlabel('Z', fontdict={'size': 15, 'color': 'red'}) ax.set_ylabel('Y', fontdict={'size': 15, 'color': 'red'}) ax.set_xlabel('X', fontdict={'size': 15, 'color': 'red'}) plt.show() ``` ```python ```

419398d133b527499ed6b7355d94028cd05aa0f1

ipynb

Jupyter Notebook

annot/BA and LM --- 1F6P #3.ipynb

XiaotengLu/Human-Torso-Pose-Estimation

997990e74e95832cd377922ea7cc43ec50f82ae0

annot/BA and LM --- 1F6P #3.ipynb

XiaotengLu/Human-Torso-Pose-Estimation

997990e74e95832cd377922ea7cc43ec50f82ae0

annot/BA and LM --- 1F6P #3.ipynb

XiaotengLu/Human-Torso-Pose-Estimation

997990e74e95832cd377922ea7cc43ec50f82ae0

Qwen/Qwen-72B

1. YES 2. YES

__label__kor_Hang

```c++ // Copyright (c) 2020 Patrick Diehl // // SPDX-License-Identifier: BSL-1.0 // Distributed under the Boost Software License, Version 1.0. (See accompanying // file LICENSE_1_0.txt or copy at http://www.boost.org/LICENSE_1_0.txt) ``` # Exercise 1: Classical linear elasticity model Let $\Omega = (0,1) \subset \mathbb R$ and $\overline{\Omega}$ be the closure of $\Omega$, i.e.\ $\Omega=[0,1]$. The continuum local problem consists in finding the displacement $(u\in\overline{\Omega})$ such that: \begin{align} \label{eq:1dlinearelasticity} - E u''(x) = f_b(x), &\quad \forall x \in \Omega, \\ \label{eq:Dirichlet} u(x) = 0, &\quad \text{at}\ x=0,\\ \label{eq:Neumann} Eu'(x) = g, &\quad \text{at}\ x=1, \end{align} where $E$ is the constant modulus of elasticity of the bar, $f_b=f_b(x)$ is a scalar function describing the external body force density (per unit length), and $g \in \mathbb R$ is the traction force applied at end point $x=1$. ```c++ #include <blaze/Math.h> #include <BlazeIterative.hpp> #include <cmath> #include <iostream> #include <vector> #include<run_hpx.cpp> #include<hpx/include/lcos.hpp> #include<hpx/include/parallel_for_loop.hpp> ``` ## Helpfer function for Blaze [Blaze](https://bitbucket.org/blaze-lib/blaze/src/master/) is an open-source, high-performance C++ math library for dense and sparse arithmetic. This linrary has a HPX backend and utilzies the parallel algorithms to accelarted the linear algebara functions. Blaze is not covered in this cours and for more details we refer to * K. Iglberger, G. Hager, J. Treibig, and U. Rüde: High Performance Smart Expression Template Math Libraries . Proceedings of the 2nd International Workshop on New Algorithms and Programming Models for the Manycore Era (APMM 2012) at HPCS 2012 * K. Iglberger, G. Hager, J. Treibig, and U. Rüde: Expression Templates Revisited: A Performance Analysis of Current Methodologies (Download). SIAM Journal on Scientific Computing, 34(2): C42--C69, 2012 ```c++ // Generates a Blaze dynamic matrix of size N times N and fills the matrix with zeros blaze::DynamicMatrix<double> zeroMatrix(unsigned long N) { return blaze::DynamicMatrix<double>( N, N, 0 ); }; ``` ```c++ // Generates a Blaze dynamic vector of size N and fills the vector with zeros blaze::DynamicVector<double> zeroVector(unsigned long N) { return blaze::DynamicVector<double>(N,0); }; ``` ```c++ // Solves the matrix system A \times x = b and returns x blaze::DynamicVector<double> solve(blaze::DynamicMatrix<double> A, blaze::DynamicVector<double> b ) { blaze::iterative::BiCGSTABTag tag; tag.do_log() = true; return solve(A,b,tag); } ``` ## Force function As the external load, a linear $force_b$ function $force : \mathbb{R} \rightarrow \mathbb{R}$ $$ force_b(x) = \begin{cases} 1, if x == 1, \\ 0 , else\end{cases}, x = [0,1]$$ ```c++ double force(double x){ if ( x == 1 ) return 1; return 0; } ``` ## Discretization As the domain $\overline{\Omega}$ we consider the intervall $[0,1]$ and discretize the interval with $n$ elements and using the spacing $h=\frac{1}{n}$ such that $x={0,1\cdot h,2\cdot h,\ldots,n\cdot h}$. ```c++ size_t n = std::pow(2,2); double h= 1./n; n += 1; ``` (unsigned long) 5 ```c++ auto x = zeroVector(n); for(size_t i = 0 ; i < n ; i++) x[i] = i * h; ``` ```c++ // Print the discrete nodes std::cout << x ; ``` ( 0 ) ( 0.25 ) ( 0.5 ) ( 0.75 ) ( 1 ) (std::ostream &) @0x7fce872e3500 <span style="color:blue">Task 1: Replace the for loop in Cell 11 with the hpx::for loop to fill the right-hand side $f$ in parrallel in Cell 12</span>. ## Prepare the external load ```c++ // Get the force vector for the right-hand side auto f = zeroVector(n); ``` ```c++ /* for(size_t i = 0 ; i < n ; i++) { f[i] = force(x[i]); } */ ``` ```c++ run_hpx([](){ }); ``` (void) @0x7fce7831ce48 ```c++ // Print the force vector std::cout << f ; ``` ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 1 ) (std::ostream &) @0x7fce872e3500 ### Assemble the stiffness matrix using finite differences 1. Dirichlet boundary condition at $x=0$: \begin{equation} u_1 = 0. \end{equation} 2. Finite difference schems for In $\overline{\Omega}$: $\forall i=2,\ldots,n-1$: \begin{equation} - E \frac{u_{i-1}-2u_i+u_{i+1}}{h^2} = f_b(x_k). \end{equation} 3. Neumann boundary condition at $x=1$: \begin{equation} E \frac{u_{n-3}-4u_{n-2}+3u_n-1}{2h} = g. \end{equation} For simplicity we assume $E=1$. <span style="color:blue">Task 2: Use aysnchronous programming to assemble the stiffness matrix using hpx::async and hpx::future</span>. 1. Finish the function assemble to fill the matrix from start to end in Cell 16 2. Generate a vector of hpx::futures<void> to collect all future for synchronization in Cell 17 3. Use hpx::async to execute the function assemble asynchronously in Cell 17 4. Use hpx::wait_all to syncrhonize the results in Cell 17 ```c++ // Get the stiffness matrix filled with zeros auto matrix = zeroMatrix(n); ``` ```c++ /* matrix(0,0) = 1; for(size_t i = 1 ; i < n-1 ; i++){ matrix(i,i-1) = -2; matrix(i,i) = 4; matrix(i,i+1) = -2; } matrix(n-1,n-1) = 3*h; matrix(n-1,n-2) = -4*h; matrix(n-1,n-3) = h; matrix *= 1./(2*h*h); */ ``` ```c++ // Assmeble the part of the stiffness matrix where the index goes from start to end void assemble(blaze::DynamicMatrix<double>* matrix, size_t start, size_t end) { #(*matrix)(i,i-1) = -2; #(*matrix)(i,i) = 4; #(*matrix)(i,i+1) = -2; } ``` ```c++ ``` (void) @0x7fce7831ce48 ```c++ std::cout << matrix ; ``` ( 8 0 0 0 0 ) ( -16 32 -16 0 0 ) ( 0 -16 32 -16 0 ) ( 0 0 -16 32 -16 ) ( 0 0 2 -8 6 ) (std::ostream &) @0x7fce872e3500 ```c++ // Solve the matrix system matrix times displacement - f auto displacement = solve(matrix,f); ``` ```c++ std::cout << displacement; ``` ( 0 ) ( 0.25 ) ( 0.5 ) ( 0.75 ) ( 1 ) (std::ostream &) @0x7fce872e3500 # Doing python plots from C++ Doing plots in Python is quite convinent using [matplotlib](https://matplotlib.org/), however, plotting in C++ is a little bit more tricky since we need to write the data as a CSV file and use Python or Matplotlib to plot it. The notebooks have some magic implemented to plot C++ variables directly in Python's matplotlib. Below we use %plot x y to plot a new line and we can repeat this command to add new lines to the same plot. Using %plotc x y will plot all previous added lines and clear the figure. ```c++ %data x displacement ``` x shape is (5,)displacement shape is (5,) ```c++ %%plot plt.xlabel("Position") plt.ylabel("Displacement") plt.plot(x,displacement,label="Simulation") plt.plot(x,x,label="Exact solution") plt.grid() plt.legend() ``` <pre> </pre>

8cd5d2aa1c474aa162c044feca2dc85799442565

ipynb

Jupyter Notebook

exercise/Exercise1.ipynb

STEllAR-GROUP/HPXjupyterTutorial

2deeb2086473a5bf12c6f64e794e0a61fb1cf595

2021-09-30T13:39:19.000Z

2021-09-30T13:39:19.000Z

exercise/Exercise1.ipynb

STEllAR-GROUP/HPXjupyterTutorial

2deeb2086473a5bf12c6f64e794e0a61fb1cf595

exercise/Exercise1.ipynb

STEllAR-GROUP/HPXjupyterTutorial

2deeb2086473a5bf12c6f64e794e0a61fb1cf595

2021-09-30T13:45:40.000Z

2021-09-30T13:45:40.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Numpy (Numeric python) > - 패키지 이름과 같이 **수리적 파이썬 활용**을 위한 파이썬 패키지 > - **선형대수학 구현**과 **과학적 컴퓨팅 연산**을 위한 함수를 제공 > - (key) `nparray` 다차원 배열을 사용하여 **벡터의 산술 연산**이 가능 > - **브로드캐스팅**을 활용하여 shape(형태 혹은 모양)이 다른 데이터의 연산이 가능 >> - 기존 언어에서는 제공 X >> - 굉장히 파워풀한 기능으로서 빅데이터 연산에 굉장히 효율이 좋음 ## Numpy 설치 와 import > - 선행 학습을 통해 클래스와 함수에서 클래스를 불러들여 사용할 수 있다고 배웠습니다. - 다만 직접 작성한 클래스가 아닐경우, 그리고 현재 컴퓨터에 사용해야 할 패키지가 없을경우 간단한 명령어로 설치가능. >> - `pip`, `conda` 명령어 : python 라이브러리 관리 프로그램으로 오픈소스라이브러리를 간편하게 설치 할 수 있도록 하는 명령어 > - 1. 콘솔창에서 실행 시 **`pip` `install` `[패키지명]`** 혹은 **`conda` `install` `[패키지명]`** > - 2. 주피터 노트북으로 실행 시 **`!pip` `install` `[패키지명]`** > - 아나콘다 환경으로 python 환경설정 시 기본적으로 Numpy 설치가 되어있음 ```python # 주피터 노트북에서 Numpy 설치 !pip install numpy ``` DEPRECATION: Configuring installation scheme with distutils config files is deprecated and will no longer work in the near future. If you are using a Homebrew or Linuxbrew Python, please see discussion at https://github.com/Homebrew/homebrew-core/issues/76621 Requirement already satisfied: numpy in /opt/homebrew/lib/python3.9/site-packages (1.21.2) ```python # Numpy 사용을 위해 패키지 불러들이기 import numpy as np # 관례적으로 np라는 약자를 많이 사용하게 됩니다. # 파이썬을 사용하는 대부분의 유저들이 사용하고 있는 닉네임이니 이건 꼭 지켜서 사용해주시는 것을 추천드립니다. ``` ## 데이터분석을 위한 잠깐의 선형대수학 numpy는 기본적으로 수리적 컴퓨팅을 위한 패키지 입니다. 선형대수학을 약간만 이해한다면 데이터를 훨씬 더 깊이있게 다룰 수 있습니다. 출처 : https://art28.github.io/blog/linear-algebra-1/ ### 데이터의 구분에 따른 표현 방법과 예시 #### 스칼라 1, 3.14, 실수 혹은 정수 #### 벡터 [1, 2, 3, 4], "문자열" #### 3 X 4 매트릭스 [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 11, 12]] #### 2 X 3 X 4 텐서 [[[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 11, 12]], [[1, 2, 3, 4], [5, 6, 7, 8], [9, 0, 11, 12]]] ### 데이터로 표현한 선형대수학 출처 : https://m.blog.naver.com/nabilera1/221978354680 ### 데이터 형태에 따른 사칙연산 > 스칼라 +, -, *, / -> 결과도 스칼라 벡터 +, -, 내적 -> +, - 결과는 벡터, 내적 결과는 스칼라 매트릭스 +, -, *, / 텐서 +, -, *, / ### 데이터분석에 자주 사용하는 특수한 연산 벡터와 벡터의 내적 $$\begin{bmatrix}1 & 2 & 3 & 4 \end{bmatrix} \times \begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \end{bmatrix} = 1 * 1 + \ 2 * 2 + 3 * 3 + 4 * 4 = 30$$ # $$ A^TA $$ #### 벡터와 벡터의 내적이 이루어지려면 1. 벡터가 마주하는 shape의 갯수(길이)가 같아야 합니다. 2. 연산 앞에 위치한 벡터는 전치(transpose) 되어야 합니다. 출처 : https://ko.wikipedia.org/wiki/%EC%A0%84%EC%B9%98%ED%96%89%EB%A0%AC #### 벡터 내적으로 방정식 구현 $$y = \begin{bmatrix}1 & 2 & 1 \end{bmatrix} \times \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = 1 * x_1 + \ 2 * x_2 + 1 * x_3 = x_1 + 2x_2 + x_3$$ ## 브로드캐스팅 > 파이썬 넘파이 연산은 브로드캐스팅을 지원합니다. 벡터연산 시 shape이 큰 벡터의 길이만큼 shape이 작은 벡터가 연장되어 연산됩니다. 출처 : http://www.astroml.org/book_figures/appendix/fig_broadcast_visual.html ```python np.arange(3).reshape((3,1))+np.arange(3) ``` array([[0, 1, 2], [1, 2, 3], [2, 3, 4]]) ## Numpy function(유니버셜함수) > `numpy`는 컴퓨팅연산을 위한 다양한 연산함수를 제공합니다. >> 연산함수 기본구조 ex) **`np.sum`**(연산대상, axis=연산방향) **`dtype()`** ### 수리연산 - **`prod()`** - **`dot()`** - **`sum()`** - **`cumprod()`** - **`cumsum()`** - **`abs()`** - **`sqaure()`** - **`sqrt()`** - **`exp()`** - **`log()`** ### 통계연산 - **`mean()`** - **`std()`** - **`var()`** - **`max()`** - **`min()`** - **`argmax()`** - **`argmin()`** ### 로직연산 - **`arange()`** - **`isnan()`** - **`isinf()`** - **`unique()`** ### 기하 - **`shape()`** - **`reshape()`** - **`ndim()`** - **`transpose()`** 각종 연산 함수 참고: https://numpy.org/doc/stable/reference/routines.math.html ### numpy 함수 실습 ```python # 함수 예제를 위한 데이터셋 test_list = [1, 2, 3, 4] test_list2 = [[1, 3], [5, 7]] test_flist = [1, 3.14, -4.5] test_list_2nd = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] test_list_3rd = [[[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]]] test_exp = [0, 1, 10] test_nan = [0, np.nan, np.inf] ``` ```python # 곱연산 np.prod(test_list) ``` 24 ```python # 합연산 np.sum(test_list) ``` 10 ```python # 누적곱연산 벡터형태 np.cumprod(test_list) ``` array([ 1, 2, 6, 24]) ```python # 누적합연산 # 매일/매월 매출에 대한 계산을 할 때 자주 사용 np.cumsum(test_list) ``` array([ 1, 3, 6, 10]) ```python # 절대값 np.abs(test_flist) ``` array([1. , 3.14, 4.5 ]) ```python # 제곱 np.sqrt(test_list) ``` array([1. , 1.41421356, 1.73205081, 2. ]) ```python # 루트 ``` ```python # exp ``` ```python # 로그 ``` ### 통계값 ```python # 평균 np.mean(test_list) ``` 2.5 ```python # 표준편차 np.std(test_list) ``` 1.118033988749895 ```python # 분산 np.var(test_list) ``` 1.25 ```python # 최대값 ``` ```python # 최소값 ``` ```python test_list_2nd ``` [[1, 2, 3], [4, 5, 6], [7, 8, 9]] ```python # 최대값이 존재하고 있는 인덱스 넘버를 리턴 # 출력값이 인덱스 np.argmax(test_list_2nd) ``` 8 ```python # 최소값 인덱스 np.argmin(test_list_2nd) ``` 0 ```python # 범위설정 # range() 함수와 동일하게 작동함 # for i in range(0, 100, 10): # print(i) #(시작포인트, 마지막포인트+1, 스텝수) np.arange(10, 100, 10) ``` array([10, 20, 30, 40, 50, 60, 70, 80, 90]) ```python # 범위설정2 #(시작포인트, 마지막포인트+1, 데이터수) np.linspace(0, 10, 50) # 0~10까지 50개 동일간격으로 ``` array([ 0. , 0.20408163, 0.40816327, 0.6122449 , 0.81632653, 1.02040816, 1.2244898 , 1.42857143, 1.63265306, 1.83673469, 2.04081633, 2.24489796, 2.44897959, 2.65306122, 2.85714286, 3.06122449, 3.26530612, 3.46938776, 3.67346939, 3.87755102, 4.08163265, 4.28571429, 4.48979592, 4.69387755, 4.89795918, 5.10204082, 5.30612245, 5.51020408, 5.71428571, 5.91836735, 6.12244898, 6.32653061, 6.53061224, 6.73469388, 6.93877551, 7.14285714, 7.34693878, 7.55102041, 7.75510204, 7.95918367, 8.16326531, 8.36734694, 8.57142857, 8.7755102 , 8.97959184, 9.18367347, 9.3877551 , 9.59183673, 9.79591837, 10. ]) ```python test_nan ``` [0, nan, inf] ```python # 결측 확인(nan 확인) np.isnan(test_nan) ``` array([False, True, False]) ```python # 발산 확인(무한대 확인) np.isinf(test_nan) ``` array([False, False, True]) ```python test_list_3rd ``` [[[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]]] ```python # 고유값 확인 np.unique(test_list_3rd) len(np.unique(test_list_3rd)) ``` 8 ```python # 데이터 구조(모양)확인(차원 확인) # 안쪽(열)shape부터 확인하자 np.shape(test_list_3rd) ``` (3, 2, 4) ```python # 데이터 shape 변경 # 어떤 조건에서 reshape가능한가? 데이터 내부에 존재하는 속성 갯수가 같아야 함. np.reshape(test_list_3rd, (4,6)) np.reshape(test_list_3rd, (2,2,6)) ``` array([[[1, 2, 3, 4, 5, 6], [7, 8, 1, 2, 3, 4]], [[5, 6, 7, 8, 1, 2], [3, 4, 5, 6, 7, 8]]]) ```python test_list_3rd ``` [[[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]]] ```python # 데이터 차원확인 np.ndim(test_list_3rd) # 기하학적 데이터의 차원수를 이야기하고 데이터분식시 열기준으로 차원을 이야기한다 ``` 3 ```python test_list_2nd ``` [[1, 2, 3], [4, 5, 6], [7, 8, 9]] ```python # 전치행렬 np.transpose(test_list_2nd) ``` array([[1, 4, 7], [2, 5, 8], [3, 6, 9]]) ```python test_list ``` [1, 2, 3, 4] ```python ``` ## Numpy array (배열, 행렬) > - numpy 연산의 기본이 되는 데이터 구조입니다. - 리스트보다 간편하게 만들 수 있으며 **연산이 빠른** 장점이 있습니다. - **브로드캐스팅 연산을 지원**합니다. - 단, **같은 type**의 데이터만 저장 가능합니다. - array 또한 numpy의 기본 함수로서 생성 가능합니다. >> array 함수 호출 기본구조 ex) **`np.array(배열변환 대상 데이터)`** ex) **`np.arange(start, end, step_forward)`** ### numpy array 실습 ```python # 기존 데이터 구조를 array로 변환 test_array = np.array(test_list) test_array2 = np.array(test_list2) test_farray = np.array(test_flist) test_array_2nd = np.array(test_list_2nd) test_array_3rd = np.array(test_list_3rd) ``` ```python # array 생성 확인 test_array ``` array([1, 2, 3, 4]) ```python array_list = [1,2,4.5] ``` ```python # 같은 타입의 데이터만 들어가는지 확인 array_test = np.array(array_list) ``` ```python array_test # 정수, 실수, 문자열 순으로 전체 타입이 설정됨 ``` array([1. , 2. , 4.5]) ```python # 2차원 배열 확인 test_list_2nd # 2차원 배열을 좀 더 편하게 보여줌 test_array_2nd ``` array([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) ```python # 3차원 배열 확인 test_list_3rd test_array_3rd ``` array([[[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]]]) ```python # np.arange 함수로 생성 np.arange(25).reshape(5,5) ``` array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]) ```python np.arange(1,25).reshape(2,12) ``` array([[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]]) ### 특수한 형태의 array를 함수로 생성 함수 호출의 기본구조 > ex) **`np.ones([자료구조 shape])`** >> 자료구조 shape은 정수, **[ ]**리스트, **( )** 튜플 로만 입력가능합니다. - ones() - zeros() - empty() - eye() ```python # 1로 초기화한 array 생성 np.ones([3,3]) ``` array([[1., 1., 1.], [1., 1., 1.], [1., 1., 1.]]) ```python # 0으로 초기화 np.zeros((5,5)) ``` array([[0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.], [0., 0., 0., 0., 0.]]) ```python # 빈 값으로 초기화 np.empty((2,2)) ``` array([[0., 0.], [0., 0.]]) ```python # 항등행렬 초기화 # A X 항등행렬 = A # shape가 안맞는 경우 연산이 가능하도록 만들때 사용 np.eye(4,4) # 예시 # 날짜 매출 # 날짜 매출 # 날짜 매출 # 날짜 매출 # 날짜 매출 # 날짜 날짜 날짜 날짜 날짜 # 매출 없고 없고 없고 없고 # 없고 매출 없고 없고 없고 # 없고 없고 매출 없고 없고 ``` array([[1., 0., 0., 0.], [0., 1., 0., 0.], [0., 0., 1., 0.], [0., 0., 0., 1.]]) ```python np.arange(16).reshape(4,4) @ np.eye(4,4) ``` array([[ 0., 1., 2., 3.], [ 4., 5., 6., 7.], [ 8., 9., 10., 11.], [12., 13., 14., 15.]]) ```python test_array @ np.transpose(test_array) ``` array([1, 2, 3, 4]) ### array 속성 및 내장함수 `np.array` 에는 유용한 수리, 통계 연산을 위한 함수가 갖추어져 있습니다. 다차원 기하학적 연산을 위한 함수도 함께 살펴보겠습니다. > array 내장 속성 호출 기본구조 ex) **`test_array.ndim`** 자주 사용하는 속성 `shape`, `dtype`, `ndim` > array 내장함수 호출 기본구조 ex) **`test_array.prod()`** 위에 학습한 np.sum() 과는 달리 array 변수의 인자를 받아 그대로 사용합니다. #### array 속성 ```python test_array_3rd ``` array([[[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]], [[1, 2, 3, 4], [5, 6, 7, 8]]]) ```python # 데이터 타입확인 test_array_3rd.dtype ``` dtype('int64') ```python # 데이터 타입 확인 test_array_3rd.shape ``` (3, 2, 4) ```python # 데이터 차원 확인 test_array_3rd.ndim ``` 3 ```python # 전치행렬 np.transpose(test_array_3rd) test_array_3rd.T ``` array([[[1, 1, 1], [5, 5, 5]], [[2, 2, 2], [6, 6, 6]], [[3, 3, 3], [7, 7, 7]], [[4, 4, 4], [8, 8, 8]]]) #### array 내장함수 ```python # 내장함수 호출 test_array.mean() np.sqrt(test_array) ``` array([1. , 1.41421356, 1.73205081, 2. ]) ```python # numpy 함수와 키워드가 같습니다. test_array.prod() test_array.sum(axis=) ... ... ``` ### array 연산 컴퓨팅 연산을 위한 패키지인 만큼 편리한 배열 연산 기능을 지원합니다. 여러 array 연산을 통해 다른 자료구조와의 연산 차이를 알아봅시다. ```python test_list = [1,2,3,4,5] test_list2 = [x*2 for x in test_list] test_list2 ``` [2, 4, 6, 8, 10] ```python # array 덧셈, 뺄셈, 곱셈, 나눗셈 test_array = np.array(test_list) test_array > 1 ``` array([False, True, True, True, True]) ```python # 실제 연산속도 차이를 확인하기 위한 큰 데이터 생성 big_list = [x for x in range(400000)] big_array = np.array(big_list) len(big_list), len(big_array) ``` (400000, 400000) ```python # for로 각 리스트에 1씩 값을 더함 big_list2 = [x+1 for x in big_list] # 일반적인 for문보다 빠름(파이썬은 줄단위로 값을 확인) # for index, item in enumerate(big_list): # big_list[index] = item + 1 ``` UsageError: Line magic function `%%time` not found. ```python # array성질을 이용 big_array + 1 ``` array([ 1, 2, 3, ..., 399998, 399999, 400000]) ```python # 행렬내적 first_array = np.arange(15).reshape(5, 3) second_array = np.arange(15).reshape(3, 5) ``` ```python first_array.shape, second_array.shape ``` ((5, 3), (3, 5)) ```python # 행렬내적 연산 first_array @ second_array ``` array([[ 25, 28, 31, 34, 37], [ 70, 82, 94, 106, 118], [115, 136, 157, 178, 199], [160, 190, 220, 250, 280], [205, 244, 283, 322, 361]]) ### 벡터 가중합 벡터의 내적은 가중합을 계산할 때 쓰일 수 있다. **가중합(weighted sum)**이란 복수의 데이터를 단순히 합하는 것이 아니라 각각의 수에 어떤 가중치 값을 곱한 후 이 곱셈 결과들을 다시 합한 것을 말한다. 만약 데이터 벡터가 $x=[x_1, \cdots, x_N]^T$이고 가중치 벡터가 $w=[w_1, \cdots, w_N]^T$이면 데이터 벡터의 가중합은 다음과 같다. $$ \begin{align} w_1 x_1 + \cdots + w_N x_N = \sum_{i=1}^N w_i x_i \end{align} $$ 이 값을 벡터 $x$와 $w$의 곱으로 나타내면 $w^Tx$ 또는 $x^Tw$ 라는 간단한 수식으로 표시할 수 있다. 쇼핑을 할 때 각 물건의 가격은 데이터 벡터, 각 물건의 수량은 가중치로 생각하여 내적을 구하면 총금액을 계산할 수 있다. ```python # 벡터의 가중합 연습문제 # 삼성전자, 셀트리온, 카카오로 포트폴리오를 구성하려한다. # 각 종목의 가격은 80,000원, 270,000원, 160,000원이다. # 삼성전자 100주, 셀트리온 30주, 카카오 50주로 구성하기 위한 매수금액을 구하시오 price = np.array([80000,270000,160000]) item = np.array([100,30,50]) price @ item.T # shape가 맞지 않는데도 결과가 나옴 => @ 연산을하면 item.T로 작동함 price @ item ``` 24100000 ### array 인덱싱, 슬라이싱(매우중요) > 기본적으로 자료구조란 데이터의 묶음, 그 묶음을 관리 할 수 있는 바구니를 이야기 합니다. 데이터 분석을 위해 자료구조를 사용하지만 자료구조안 내용에 접근을 해야 할 경우도 있습니다. >> **인덱싱**이란? 데이터 바구니 안에 든 내용 하나에 접근하는 명령, 인덱스는 내용의 순번 >> **슬라이싱**이란? 데이터 바구니 안에 든 내용 여러가지에 접근 하는 명령 기본적으로 인덱싱과 슬라이싱의 색인은 리스트와 동일합니다. #### 인덱싱, 슬라이싱 실습 ```python # 10부터 19까지 범위를 가지는 array생성 test = np.arange(10,20) ``` ```python # 0부터 3번 인덱스까지 test[:3] ``` array([10, 11, 12]) ```python # 4번 인덱스부터 마지막 인덱스까지 test[4:] ``` array([14, 15, 16, 17, 18, 19]) ```python # 마지막 인덱스부터 뒤에서 3번째 인덱스까지 test[-3:] ``` array([17, 18, 19]) ```python # 0부터 3씩 증가하는 인덱스 test[::3] ``` array([10, 13, 16, 19]) #### 여러가지 인덱싱 및 슬라이싱 방법을 시도해봅시다 ```python index_test2 = np.array(range(25)).reshape([5, 5]) index_test2 ``` array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]) ```python index_test2[2:,1:4] ``` array([[11, 12, 13], [16, 17, 18], [21, 22, 23]]) ```python index_test2[:2,2:] ``` array([[2, 3, 4], [7, 8, 9]]) ```python index_test3 = np.arange(40).reshape(2, 5, 4) index_test3 ``` array([[[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11], [12, 13, 14, 15], [16, 17, 18, 19]], [[20, 21, 22, 23], [24, 25, 26, 27], [28, 29, 30, 31], [32, 33, 34, 35], [36, 37, 38, 39]]]) ```python index_test3[0,3:5,1:3] ``` array([[13, 14], [17, 18]]) ```python index_test3[0:,0:,1] ``` array([[ 1, 5, 9, 13, 17], [21, 25, 29, 33, 37]]) ## 팬시인덱싱 numpy에서 벡터연산을 통해 bool 형태의 벡터를 기준으로 데이터를 선별하는 방법 ```python pet = np.array(['개','고양이','고양이','햄스터','개','햄스터']) num = np.array([1,2,3,4,5,6]) indexing_test = np.random.randn(6,5) # randn : 난수로 행렬생성 ``` ```python indexing_test ``` array([[ 1.21892182, -0.57278769, -2.01837856, 0.37751666, -2.14632145], [ 1.72627741, 0.19978955, -0.45559384, 2.3741275 , 0.61280794], [ 0.38249301, -0.10195438, 1.1064851 , 0.14316647, -0.01103692], [ 0.89879404, -0.08147537, -0.02147208, 0.77555509, -0.45326144], [ 0.12338784, -0.00431352, -1.31633818, 0.85516829, 0.29007829], [-0.17169369, 0.42409219, 1.69908292, 1.43223254, -1.29304307]]) ```python pet == '개' ``` array([ True, False, False, False, True, False]) ```python indexing_test[pet=='개'] ``` array([[ 1.21892182, -0.57278769, -2.01837856, 0.37751666, -2.14632145], [ 0.12338784, -0.00431352, -1.31633818, 0.85516829, 0.29007829]]) ```python indexing_test[~(pet=='개')] ``` array([[ 1.72627741, 0.19978955, -0.45559384, 2.3741275 , 0.61280794], [ 0.38249301, -0.10195438, 1.1064851 , 0.14316647, -0.01103692], [ 0.89879404, -0.08147537, -0.02147208, 0.77555509, -0.45326144], [-0.17169369, 0.42409219, 1.69908292, 1.43223254, -1.29304307]]) ```python (pet=='개') | (pet=='햄스터') ``` array([ True, False, False, True, True, True]) ```python indexing_test[(pet=='개') | (pet=='햄스터')] ``` array([[ 1.21892182, -0.57278769, -2.01837856, 0.37751666, -2.14632145], [ 0.89879404, -0.08147537, -0.02147208, 0.77555509, -0.45326144], [ 0.12338784, -0.00431352, -1.31633818, 0.85516829, 0.29007829], [-0.17169369, 0.42409219, 1.69908292, 1.43223254, -1.29304307]]) ```python num > 3 ``` array([False, False, False, True, True, True]) ```python indexing_test[num>3] ``` array([[ 0.89879404, -0.08147537, -0.02147208, 0.77555509, -0.45326144], [ 0.12338784, -0.00431352, -1.31633818, 0.85516829, 0.29007829], [-0.17169369, 0.42409219, 1.69908292, 1.43223254, -1.29304307]]) ```python # bool 형태는 1,0으로 계산이 된다 (pet =='개').sum() ``` 2 ```python # true가 1개라도 있으면 true (pet =='개').any() ``` True ```python (pet =='개').all() ``` False

37befcb53e5804c7b7f02bac823d4a10b4affb58

ipynb

Jupyter Notebook

python/210910-Python-numpy.ipynb

PpangPpang93/TIL

63e541f223f2317fff06d98cac79f372fd11ed1d

python/210910-Python-numpy.ipynb

PpangPpang93/TIL

63e541f223f2317fff06d98cac79f372fd11ed1d

python/210910-Python-numpy.ipynb

PpangPpang93/TIL

63e541f223f2317fff06d98cac79f372fd11ed1d

Qwen/Qwen-72B

1. YES 2. YES

__label__kor_Hang

*Sebastian Raschka* last modified: 03/31/2014 <hr> I am really looking forward to your comments and suggestions to improve and extend this tutorial! Just send me a quick note via Twitter: [@rasbt](https://twitter.com/rasbt) or Email: [[email protected]](mailto:[email protected]) <hr> ### Problem Category - Statistical Pattern Recognition - Supervised Learning - Parametric Learning - Bayes Decision Theory - Univariate data - 2-class problem - different variances - Gaussian model (2 parameters) - No Risk function <hr> <p><a name="sections"></a> <br></p> # Sections <p>&#8226; <a href="#given">Given information</a><br> &#8226; <a href="#deriving_db">Deriving the decision boundary</a><br> &#8226; <a href="#plotting_db">Plotting the class conditional densities, posterior probabilities, and decision boundary</a><br> &#8226; <a href="#classify_rand">Classifying some random example data</a><br> &#8226; <a href="#emp_err">Calculating the empirical error rate</a><br> <hr> <p><a name="given"></a> <br></p> ## Given information: [<a href="#sections">back to top</a>] <br> ####model: continuous univariate normal (Gaussian) model for the class-conditional densities $ p(x | \omega_j) \sim N(\mu|\sigma^2) $ $ p(x | \omega_j) \sim \frac{1}{\sqrt{2\pi\sigma^2}} \exp{ \bigg[-\frac{1}{2}\bigg( \frac{x-\mu}{\sigma}\bigg)^2 \bigg] } $ ####Prior probabilities: $ P(\omega_1) = P(\omega_2) = 0.5 $ #### Variances of the sample distributions $ \sigma_1^2 = 4, \quad \sigma_2^2 = 1 $ #### Means of the sample distributions $ \mu_1 = 4, \quad \mu_2 = 10 $ <br> <p><a name="deriving_db"></a> <br></p> ## Deriving the decision boundary [<a href="#sections">back to top</a>] <br> ### Bayes' Rule: $ P(\omega_j|x) = \frac{p(x|\omega_j) * P(\omega_j)}{p(x)} $ ###Bayes' Decision Rule: Decide $ \omega_1 $ if $ P(\omega_1|x) > P(\omega_2|x) $ else decide $ \omega_2 $. <br> \begin{equation} \begin{aligned} &\Rightarrow \frac{p(x|\omega_1) * P(\omega_1)}{p(x)} > \frac{p(x|\omega_2) * P(\omega_2)}{p(x)} \end{aligned} \end{equation} ###Bayes' Decision Rule: Decide $ \omega_1 $ if $ P(\omega_1|x) > P(\omega_2|x) $ else decide $ \omega_2 $. <br> $ \Rightarrow \frac{p(x|\omega_1) * P(\omega_1)}{p(x)} > \frac{p(x|\omega_2) * P(\omega_2)}{p(x)} $ We can drop $ p(x) $ since it is just a scale factor. $ \Rightarrow P(x|\omega_1) * P(\omega_1) > p(x|\omega_2) * P(\omega_2) $ $ \Rightarrow \frac{p(x|\omega_1)}{p(x|\omega_2)} > \frac{P(\omega_2)}{P(\omega_1)} $ $ \Rightarrow \frac{p(x|\omega_1)}{p(x|\omega_2)} > \frac{0.5}{0.5} $ $ \Rightarrow \frac{p(x|\omega_1)}{p(x|\omega_2)} > 1 $ $ \Rightarrow \frac{1}{\sqrt{2\pi\sigma_1^2}} \exp{ \bigg[-\frac{1}{2}\bigg( \frac{x-\mu_1}{\sigma_1}\bigg)^2 \bigg] } > \frac{1}{\sqrt{2\pi\sigma_2^2}} \exp{ \bigg[-\frac{1}{2}\bigg( \frac{x-\mu_2}{\sigma_2}\bigg)^2 \bigg] } \quad \bigg| \quad ln $ $ \Rightarrow ln(1) - ln\bigg({\sqrt{2\pi\sigma_1^2}}\bigg) -\frac{1}{2}\bigg( \frac{x-\mu_1}{\sigma_1}\bigg)^2 > ln(1) - ln\bigg({{\sqrt{2\pi\sigma_2^2}}}\bigg) -\frac{1}{2}\bigg( \frac{x-\mu_2}{\sigma_2}\bigg)^2 \quad \bigg| \quad \sigma_1^2 = 4, \quad \sigma_2^2 = 1,\quad \mu_1 = 4, \quad \mu_2 = 10 $ $ \Rightarrow -ln({\sqrt{2\pi4}}) -\frac{1}{2}\bigg( \frac{x-4}{2}\bigg)^2 > -ln({{\sqrt{2\pi}}}) -\frac{1}{2}(x-10)^2 $ $ \Rightarrow -\frac{1}{2} ln({2\pi}) - ln(2) -\frac{1}{8} (x-4)^2 > -\frac{1}{2}ln(2\pi) -\frac{1}{2}(x-10)^2 \quad \bigg| \; \times\; 2 $ $ \Rightarrow -ln({2\pi}) - 2ln(2) - \frac{1}{4}(x-4)^2 > -ln(2\pi) - (x-10)^2 \quad \bigg| \; + ln(2\pi) $ $ \Rightarrow -4ln(4) - (x-4)^2 >- 4(x-10)^2 $ $ \Rightarrow -ln(4) - \frac{1}{4}(x-4)^2 > -(x-10)^2 \quad \big| \; \times \; 4 $ $ \Rightarrow -8ln(2) - x^2 + 8x - 16 > - 4x^2 + 80x - 400 $ $ \Rightarrow 3x^2 - 72x + 384 -8ln(2) > 0 $ $ \Rightarrow x < 7.775 \quad and \quad x > 16.225 $ <p><a name="plotting_db"></a> <br></p> ## Plotting the class conditional densities, posterior probabilities, and decision boundary [<a href="#sections">back to top</a>] <br> ```python %pylab inline import numpy as np from matplotlib import pyplot as plt def pdf(x, mu, sigma): """ Calculates the normal distribution's probability density function (PDF). """ term1 = 1.0 / ( math.sqrt(2*np.pi) * sigma ) term2 = np.exp( -0.5 * ( (x-mu)/sigma )**2 ) return term1 * term2 # generating some sample data x = np.arange(-100, 100, 0.05) # probability density functions pdf1 = pdf(x, mu=4, sigma=4) pdf2 = pdf(x, mu=10, sigma=1) # Class conditional densities (likelihoods) plt.plot(x, pdf1) plt.plot(x, pdf2) plt.title('Class conditional densities (likelihoods)') plt.ylabel('p(x)') plt.xlabel('random variable x') plt.legend(['p(x|w_1) ~ N(4,4)', 'p(x|w_2) ~ N(10,1)'], loc='upper left') plt.ylim([0,0.5]) plt.xlim([-15,20]) plt.show() ``` ```python def posterior(likelihood, prior): """ Calculates the posterior probability (after Bayes Rule) without the scale factor p(x) (=evidence). """ return likelihood * prior # probability density functions posterior1 = posterior(pdf(x, mu=4, sigma=4), 0.5) posterior2 = posterior(pdf(x, mu=10, sigma=1), 0.5) # Class conditional densities (likelihoods) plt.plot(x, posterior1) plt.plot(x, posterior2) plt.title('Posterior Probabilities w. Decision Boundaries') plt.ylabel('P(w)') plt.xlabel('random variable x') plt.legend(['P(w_1|x)', 'p(w_2|X)'], loc='upper left') plt.ylim([0,0.25]) plt.xlim([-15,20]) plt.axvline(7.775, color='r', alpha=0.8, linestyle=':', linewidth=2) plt.axvline(16.225, color='r', alpha=0.8, linestyle=':', linewidth=2) plt.annotate('R2', xy=(10, 0.2), xytext=(10, 0.22)) plt.annotate('R1', xy=(4, 0.2), xytext=(4, 0.22)) plt.annotate('R1', xy=(17, 0.2), xytext=(17.5, 0.22)) plt.show() ``` <p><a name="classify_rand"></a> <br></p> ## Classifying some random example data [<a href="#sections">back to top</a>] <br> ```python # Parameters mu_1 = 4 mu_2 = 10 sigma_1_sqr = 4 sigma_2_sqr = 1 # Generating 10 random samples drawn from a Normal Distribution for class 1 & 2 x1_samples = sigma_1_sqr**0.5 * np.random.randn(20) + mu_1 x2_samples = sigma_1_sqr**0.5 * np.random.randn(20) + mu_2 y = [0 for i in range(20)] # Plotting sample data with a decision boundary plt.scatter(x1_samples, y, marker='o', color='green', s=40, alpha=0.5) plt.scatter(x2_samples, y, marker='^', color='blue', s=40, alpha=0.5) plt.title('Classifying random example data from 2 classes') plt.ylabel('P(x)') plt.xlabel('random variable x') plt.legend(['w_1', 'w_2'], loc='upper right') plt.ylim([-0.1,0.1]) plt.xlim([0,20]) plt.axvline(7.775, color='r', alpha=0.8, linestyle=':', linewidth=2) plt.axvline(16.225, color='r', alpha=0.8, linestyle=':', linewidth=2) plt.annotate('R2', xy=(10, 0.03), xytext=(10, 0.03)) plt.annotate('R1', xy=(4, 0.03), xytext=(4, 0.03)) plt.annotate('R1', xy=(17, 0.03), xytext=(17.5, 0.03)) plt.show() ``` <p><a name="emp_err"></a> <br></p> ## Calculating the empirical error rate [<a href="#sections">back to top</a>] <br> ```python w1_as_w2, w2_as_w1 = 0, 0 for x1,x2 in zip(x1_samples, x2_samples): if x1 > 7.775 and x1 < 16.225: w1_as_w2 += 1 if x2 <= 7.775 and x2 >= 16.225: w2_as_w1 += 1 emp_err = (w1_as_w2 + w2_as_w1) / float(len(x1_samples) + len(x2_samples)) print('Empirical Error: {}%'.format(emp_err * 100)) ``` Empirical Error: 0.0% ```python ``` ```python ``` ```python test complete; Gopal ```

a1319cad159a2934145e5d33d3914c4ebdea8394

ipynb

Jupyter Notebook

tests/others/2_stat_superv_parametric.ipynb

gopala-kr/ds-notebooks

bc35430ecdd851f2ceab8f2437eec4d77cb59423

2019-05-10T09:16:23.000Z

2019-05-10T09:16:23.000Z

tests/others/2_stat_superv_parametric.ipynb

gopala-kr/ds-notebooks

bc35430ecdd851f2ceab8f2437eec4d77cb59423

tests/others/2_stat_superv_parametric.ipynb

gopala-kr/ds-notebooks

bc35430ecdd851f2ceab8f2437eec4d77cb59423

2019-10-14T07:30:18.000Z

2019-10-14T07:30:18.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

## 유클리드 유사도 vs 코사인 유사도 - 유클리드 유사도 구하는 함수 구현하기 - 코사인 유사도 구하는 함수 구현하기 - 코드설명 - 두 결과 비교 - 응용분야 예시 #### Similarity ```The similairt measure is the measure of how much alike two data objects are. Similarity measure in a data mining context is a distance with dimensions representing features of the objects. If this distance is small, it will be the high degree of similarity where large distance will be the low degree of similarity.``` ##### 1. Euclidean distance - literally distance b/w two vectors ```python import sympy sympy.init_printing(use_latex='mathjax') ``` ```python from math import* def e_d(x,y) : "e_d stands for Euclidean distance" c = [(a-b)**2 for a,b in zip(x,y)] return sqrt(sum(c)) print(e_d([1,2], [4,1])) ``` 3.1622776601683795 ```python a = np.array([1,2]) b = np.array([4,1]) c = a-b plt.plot(a[0],a[1], 'ro') plt.plot(b[0],b[1], 'ro') plt.text(0.6, 1.6 , "$a$", fontdict={"size" : 18}) plt.text(3.6, 0.6 , "$b$", fontdict={"size" : 18}) plt.annotate('', xy=a, xytext=(0,0), arrowprops=dict(facecolor='gray', ls="dashed")) plt.annotate('', xy=b, xytext=(0,0), arrowprops=dict(facecolor='gray', ls="dashed")) plt.arrow(4,1,-3,1, width=0.008) plt.xticks(range(6)) plt.yticks(range(6)) plt.show() ``` ```python e_d([1,2], [4,1]), sqrt(10) ``` $$\left ( 3.1622776601683795, \quad 3.1622776601683795\right )$$ ##### 2. Cosine similarity - Cosine similarty metric finds the nomalized dot product of two attributes. ```python def norm(x) : return sqrt(sum([ a**2 for a in x])) def dot_product(x,y): return sum(a*b for a,b in zip(x,y)) def c_s(x,y): """c_s stands for consine similarity""" denominator = norm(x)*norm(y) numerator = dot_product(x,y) return numerator / denominator def c_d(*arg): "c_d stands for consine distance" return 1 - c_s(x,y) print(c_s([1,2],[4,1])) print(c_d([1,2],[4,1])) ``` 0.6507913734559685 0.3492086265440315 ```python ``` #### Example ```python index = ["D1","D2","D3","D4"] columns = ["team","coach","hockey","baseball","soccer","penalty","score","win","loss","season"] datas = { "team" : [5,3,0,0], "coach" : [0,0,7,1], "hockey" : [3,2,0,0], "baseball" : [0,0,2,0], "soccer" : [2,1,1,1], "penalty" : [0,1,0,2], "score" : [0,0,0,2], "win" : [2,1,3,0], "loss" : [0,0,0,3], "season" : [0,1,0,0] } ``` ```python df = pd.DataFrame(index=index, columns=columns, data=datas) df ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>team</th> <th>coach</th> <th>hockey</th> <th>baseball</th> <th>soccer</th> <th>penalty</th> <th>score</th> <th>win</th> <th>loss</th> <th>season</th> </tr> </thead> <tbody> <tr> <th>D1</th> <td>5</td> <td>0</td> <td>3</td> <td>0</td> <td>2</td> <td>0</td> <td>0</td> <td>2</td> <td>0</td> <td>0</td> </tr> <tr> <th>D2</th> <td>3</td> <td>0</td> <td>2</td> <td>0</td> <td>1</td> <td>1</td> <td>0</td> <td>1</td> <td>0</td> <td>1</td> </tr> <tr> <th>D3</th> <td>0</td> <td>7</td> <td>0</td> <td>2</td> <td>1</td> <td>0</td> <td>0</td> <td>3</td> <td>0</td> <td>0</td> </tr> <tr> <th>D4</th> <td>0</td> <td>1</td> <td>0</td> <td>0</td> <td>1</td> <td>2</td> <td>2</td> <td>0</td> <td>3</td> <td>0</td> </tr> </tbody> </table> </div> ```python #cosine similarity ``` ```python def norm(x) : return sqrt(sum([ a**2 for a in x])) def dot_product(x,y): return sum(a*b for a,b in zip(x,y)) def c_s(x,y): """c_s stands for consine similarity""" denominator = norm(x)*norm(y) numerator = dot_product(x,y) return numerator / denominator def c_d(*arg): "c_d stands for consine distance" return 1 - c_s(x,y) print(c_s([1,2],[4,1])) print(c_d([1,2],[4,1])) ``` 0.6507913734559685 0.3492086265440315 ```python C_D1_D2 = c_s(df.loc["D1"],df.loc["D2"]) C_D1_D2 ``` $$0.9356014857063997$$ ```python C_D1_D3 = c_s(df.loc["D1"],df.loc["D3"]) C_D1_D3 ``` $$0.1555231582719478$$ ```python C_D1_D4 = c_s(df.loc["D1"],df.loc["D4"]) C_D1_D4 ``` $$0.07079923254047886$$ ```python C_D2_D3 = c_s(df.loc["D2"],df.loc["D3"]) C_D2_D3 ``` $$0.12222646627042817$$ ```python C_D2_D4 = c_s(df.loc["D2"],df.loc["D4"]) C_D2_D4 ``` $$0.16692446522239712$$ ```python C_D3_D4 = c_s(df.loc["D3"],df.loc["D4"]) C_D3_D4 ``` $$0.23122932520643197$$ ```python #Euclidean similarty ``` ```python from math import* def e_d(x,y) : "e_d stands for Euclidean distance" c = [(a-b)**2 for a,b in zip(x,y)] return sqrt(sum(c)) print(e_d([1,2], [4,1])) ``` 3.1622776601683795 ```python E_D1_D2 = e_d(df.loc["D1"],df.loc["D2"]) E_D1_D2 ``` $$3.0$$ ```python E_D1_D3 = e_d(df.loc["D1"],df.loc["D3"]) E_D1_D3 ``` $$9.433981132056603$$ ```python E_D1_D4 = e_d(df.loc["D1"],df.loc["D4"]) E_D1_D4 E_D1_D4 ``` $$7.54983443527075$$ ```python E_D2_D3 = e_d(df.loc["D2"],df.loc["D3"]) E_D2_D3 ``` $$8.48528137423857$$ ```python E_D2_D4 = e_d(df.loc["D2"],df.loc["D4"]) E_D2_D4 ``` $$5.477225575051661$$ ```python E_D3_D4 = e_d(df.loc["D3"],df.loc["D4"]) E_D3_D4 ``` $$8.12403840463596$$ ```python data1 = {"D1_D2" : [C_D1_D2, E_D1_D2] , "D1_D3" : [C_D1_D3, E_D1_D3], "D1_D4" : [C_D1_D4, E_D1_D4], "D2_D3" : [C_D2_D3, E_D2_D3], "D2_D4" : [C_D2_D4, E_D2_D4], "D3_D4" : [C_D3_D4, E_D3_D4] } ``` ```python df2 = pd.DataFrame(columns=["D1_D2","D1_D3","D1_D4","D2_D3","D2_D4","D3_D4"], index=["Cosine_similarity","Euclidean_similarty"], data=data1) df2 ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>D1_D2</th> <th>D1_D3</th> <th>D1_D4</th> <th>D2_D3</th> <th>D2_D4</th> <th>D3_D4</th> </tr> </thead> <tbody> <tr> <th>Cosine_similarity</th> <td>0.935601</td> <td>0.155523</td> <td>0.070799</td> <td>0.122226</td> <td>0.166924</td> <td>0.231229</td> </tr> <tr> <th>Euclidean_similarty</th> <td>3.000000</td> <td>9.433981</td> <td>7.549834</td> <td>8.485281</td> <td>5.477226</td> <td>8.124038</td> </tr> </tbody> </table> </div> ```python ``` ```python ``` ```python data1 = {"D1_D2" : [C_D1_D2, E_D1_D2**2] , "D1_D3" : [C_D1_D3, E_D1_D3**2], "D1_D4" : [C_D1_D4, E_D1_D4**2], "D2_D3" : [C_D2_D3, E_D2_D3**2], "D2_D4" : [C_D2_D4, E_D2_D4**2], "D3_D4" : [C_D3_D4, E_D3_D4**2] } ``` ```python df2 = pd.DataFrame(columns=["D1_D2","D1_D3","D1_D4","D2_D3","D2_D4","D3_D4"], index=["Cosine_similarity","Euclidean_similarty"], data=data1) df2 ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>D1_D2</th> <th>D1_D3</th> <th>D1_D4</th> <th>D2_D3</th> <th>D2_D4</th> <th>D3_D4</th> </tr> </thead> <tbody> <tr> <th>Cosine_similarity</th> <td>0.935601</td> <td>0.155523</td> <td>0.070799</td> <td>0.122226</td> <td>0.166924</td> <td>0.231229</td> </tr> <tr> <th>Euclidean_similarty</th> <td>9.000000</td> <td>89.000000</td> <td>57.000000</td> <td>72.000000</td> <td>30.000000</td> <td>66.000000</td> </tr> </tbody> </table> </div> ```python ``` ```python ``` ```python ``` ```python from sklearn.feature_extraction.text import CountVectorizer from sklearn.metrics.pairwise import euclidean_distances ``` ```python corpus =[ "All my cats in a row", "When my cat sits down, she looks like a Furby toy!", "The cat from outer space", "Sunshine loves to sit like this for some reason." ] ``` ```python vectorizer = CountVectorizer() features = vectorizer.fit_transform(corpus).todense() print(vectorizer.vocabulary_ ) ``` {'all': 0, 'my': 11, 'cats': 2, 'in': 7, 'row': 14, 'when': 25, 'cat': 1, 'sits': 17, 'down': 3, 'she': 15, 'looks': 9, 'like': 8, 'furby': 6, 'toy': 24, 'the': 21, 'from': 5, 'outer': 12, 'space': 19, 'sunshine': 20, 'loves': 10, 'to': 23, 'sit': 16, 'this': 22, 'for': 4, 'some': 18, 'reason': 13} ```python for i in features: print(i) ``` [[1 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0]] [[0 1 0 1 0 0 1 0 1 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 1 1]] [[0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0]] [[0 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 0]] ```python for i in features : print(euclidean_distances(features[0], i)) ``` [[0.]] [[3.60555128]] [[3.16227766]] [[3.74165739]] ```python for i in features: print(cosine_similarity(features[0], i)) ```

df647fd8a7e8e13ed94480c57a98fba34ff8b6d1

ipynb

Jupyter Notebook

01_linearalgebra/01_Euclidean Distance & Cosine distance.ipynb

seokyeongheo/study-math-with-python

18266dc137e46ea299cbd89241e474d7fd610122

01_linearalgebra/01_Euclidean Distance & Cosine distance.ipynb

seokyeongheo/study-math-with-python

18266dc137e46ea299cbd89241e474d7fd610122

01_linearalgebra/01_Euclidean Distance & Cosine distance.ipynb

seokyeongheo/study-math-with-python

18266dc137e46ea299cbd89241e474d7fd610122

2018-06-07T05:57:02.000Z

2018-06-07T05:57:02.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Laboratório 5: Câncer e Tratamento Químico ### Referente ao capítulo 10 Queremos minimizar a densidade de um tumor em um organismo e os efeitos colaterais das drogas para o tratamento de câncer por quimioterapia em um período de tempo fixo. É assumido que o tumor tenha um crescimento Gompertzian. A hipótese *log-kill* de Skipper é utilizada para modelar a ocisão de células cancerosas. Ela diz que a morte de células devido aos químicos é proporcional à população de tumor. Considere $N(t)$ a densidade normalizada do tumor no tempo $t$. Assim: $$N'(t) = rN(t)\ln\left(\frac{1}{N(t)}\right) - u(t)\delta N(t)$$ onde $r$ é a taxa de crescimento do tumor, $\delta$ a magnitude da dose e $u(t)$ descreve a força do efeito da droga (famacocinética). Escolhemos o funcional da seguinte maneira para minimizar tanto a densidade de tumor quanto os efeitos da droga. Supomos aqui que quanto maior a força do efeito da droga, maior seu efeito negativo, o que é razoável. $$\min_u \int_0^T aN(t)^2 + u(t)^2 dt$$ Além disso, $u(t) \geq 0$ e $N(0) = N_0$. ## Condições Necessárias ### Hamiltoniano $$ H = aN^2 + u^2 + \lambda\left[rN\ln\left(\frac{1}{N}\right) - u\delta N\right] $$ ### Equação adjunta $$ \lambda '(t) = - H_N = -2aN - \lambda\left(r\ln\left(\frac{1}{N}\right) - rN\frac{1}{1/N}\frac{1}{N^2} - u\delta\right) = -2aN + \lambda\left[r-r\ln\left(\frac{1}{N}\right) + u\delta\right] $$ ### Condição de transversalidade $$ \lambda(T) = 0 $$ ### Condição de otimalidade $$ H_u = 2u - \lambda \delta N $$ Nesse caso temos apenas uma desigualdade, a inferior e, portanto, $H_u < 0$ não é possível. Temos que $\delta > 0$ dada sua interpretação e $N(t) > 0$, afinal sua derivada não é nem definida caso contrário. Como o problema é de minimização, temos que lembrar que as desigualdades ficam invertidas em relação a $H_u$. $$ H_u > 0 \implies u^*(t) = 0 \implies \lambda \delta N < 0 \implies \lambda(t) < 0 $$ $$ H_u = 0 \implies 0 \le u^*(t) = \frac{\delta}{2}\lambda(t)N(t) $$ Assim $u^*(t) = \max\left\{0, \frac{\delta}{2}\lambda(t)N(t)\right\}$ ### Importanto as bibliotecas ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import solve_ivp import sympy as sp import sys sys.path.insert(0, '../pyscripts/') from optimal_control_class import OptimalControl ``` ### Usando a biblitoca sympy ```python x_sp,u_sp,lambda_sp, r_sp, a_sp, delta_sp = sp.symbols('N u lambda r a delta') H = a_sp*x_sp**2 + u_sp**2 + lambda_sp*(r_sp*x_sp*sp.log(1/x_sp) - u_sp*delta_sp*x_sp) H ``` $\displaystyle N^{2} a + \lambda \left(- N \delta u + N r \log{\left(\frac{1}{N} \right)}\right) + u^{2}$ ```python print('H_x = {}'.format(sp.diff(H,x_sp))) print('H_u = {}'.format(sp.diff(H,u_sp))) print('H_lambda = {}'.format(sp.diff(H,lambda_sp))) ``` H_x = 2*N*a + lambda*(-delta*u + r*log(1/N) - r) H_u = -N*delta*lambda + 2*u H_lambda = -N*delta*u + N*r*log(1/N) Resolvendo para $H_u = 0$ ```python eq = sp.Eq(sp.diff(H,u_sp), 0) sp.solve(eq,u_sp) ``` [N*delta*lambda/2] Aqui podemos descrever as funções necessárias para a classe. ```python parameters = {'a': None, 'delta': None, 'r': None} diff_state = lambda t, N, u, par: -N*par['delta']*u + N*par['r']*np.log(1/N) diff_lambda = lambda t, N, u, lambda_, par: -2*N*par['a'] + lambda_*(par['delta']*u + par['r']*(1 - np.log(1/N))) update_u = lambda t, N, lambda_, par: np.maximum(0, 0.5*par['delta']*lambda_*N) ``` ## Aplicando a classe ao exemplo Vamos fazer algumas exeperimentações. Sinta-se livre para variar os parâmetros. Nesse caso passaremos o limite inferior como padrão inicial. Note que `np.inf` deve ser escrito quando não há limite. ```python problem = OptimalControl(diff_state, diff_lambda, update_u, bounds = [(0,np.inf)]) ``` ```python N0 = 0.975 T = 20 parameters['a'] = 3 parameters['delta'] = 0.45 parameters['r'] = 0.3 ``` ```python t,x,u,lambda_ = problem.solve(N0, T, parameters) ax = problem.plotting(t,x,u,lambda_) ``` Note o formato do controle: Inicia com valor alto e depois reduz conforme se aproxima do dia 20. Isto é consistente com as práticas médicas atuais. Todavia, também observamos que o tumor tende a crescer após o dia 12. Por isso, vamos aumentar a importância de reduzir as células cancerosas, em troca de mais danos pelos remédios. Vamos notar que a força da droga é muito maior quando reduzimos a importância de seu efeito negativo. O formato das curvas é, entretanto, muito similar. ```python parameters['a'] = 10 t,x,u,lambda_ = problem.solve(N0, T, parameters) ax = problem.plotting(t,x,u,lambda_) ``` Vamos comparar o controle escolhido conforme diferentes valor iniciais da densidade $N_0$. ```python parameters['a'] = 3 N0_values = [0.1, 0.5, 0.7, 0.9] u_values = [] for N0 in N0_values: _,_,u,_ = problem.solve(N0, T, parameters) u_values.append(u) fig = plt.figure(figsize = (10,5)) plt.xlabel("Tempo") plt.ylabel("Químico") plt.title("Quimioterapia utilizada") for i, N0 in enumerate(N0_values): plt.plot(t, u_values[i],label = r'$N_0$ = {}'.format(N0)) plt.legend() plt.grid(alpha = 0.5) ``` Podemos observar acima que apenas o estágio inicial do controle é afetado pela densidade inicial, até que atinge uma espécie de equilíbrio. Outro ponto importante é o quanto a magnitude da dose influencia na dosagem do químico ao longo do tempo. Observe que quanto maior a magnitude, o valor inicial é mais alto, porém o decaimento é mais rápido, usando menos químico no total, como era esperado. ```python N0 = 0.8 delta_values = [0.25, 0.5, 0.75] u_values = [] for delta in delta_values: parameters['delta'] = delta _,_,u,_ = problem.solve(N0, T, parameters) u_values.append(u) fig = plt.figure(figsize = (10,5)) plt.xlabel("Tempo") plt.ylabel("Químico") plt.title("Quimioterapia utilizada") for i, delta in enumerate(delta_values): plt.plot(t, u_values[i],label = r'$\delta$ = {}'.format(delta)) plt.legend() plt.grid(alpha = 0.5) ``` Podemos ver também diferentes valores temporais $T$ de duração do tratamento alteram o comportamento da quantidade de células cancerosas. Na verdade, o formato fica muito similar em todos os casos, variando o tempo de estabilidade no meio do caminho. ```python parameters['delta'] = 0.4 T_values = [10, 20, 40, 80] x_values = [] u_values = [] t_values = [] for T in T_values: t,x,u,_ = problem.solve(N0, T, parameters) t_values.append(t) x_values.append(x) u_values.append(u) fig, ax = plt.subplots(2,2,figsize = (10,8)) fig.suptitle(r'Comparação temporal de $N(t)$') for k in range(4): i = k//2 j = k%2 ax[i][j].set_ylabel(r"$N(t)$") ax[i][j].set_title("Densisade de células cancerosas") ax[i][j].plot(t_values[k], x_values[k]) ax[i][j].grid(alpha = 0.5) fig, ax = plt.subplots(2,2,figsize = (10,8)) fig.suptitle(r'Comparação temporal de $u(t)$') for k in range(4): i = k//2 j = k%2 ax[i][j].set_ylabel(r"$u(t)$") ax[i][j].set_title("Força do efeito da quimioterapia") ax[i][j].plot(t_values[k], u_values[k], color = 'green') ax[i][j].grid(alpha = 0.5) ``` Você deve ter observado em todos os exemplos que $u(t)$ atinge 0 apenas pontualmente. É necessária a restrição $u(t) \ge 0$? ## Experimentação ```python #N0 = 1 #T = 5 #parameters['r'] = 0.3 #parameters['a'] = 10 #parameters['delta'] = 0.4 # #t,x,u,lambda_ = problem.solve(N0, T, parameters) #roblem.plotting(t,x,u,lambda_) ``` ### Este é o final do notebook

5a776214ccedf4b2ee94d7f932cfc25f8dbdb0bc

ipynb

Jupyter Notebook

notebooks/.ipynb_checkpoints/Laboratory5-checkpoint.ipynb

lucasmoschen/optimal-control-biological

642a12b6a3cb351429018120e564b31c320c44c5

2021-11-03T16:27:39.000Z

2021-11-03T16:27:39.000Z

notebooks/.ipynb_checkpoints/Laboratory5-checkpoint.ipynb

lucasmoschen/optimal-control-biological

642a12b6a3cb351429018120e564b31c320c44c5

notebooks/.ipynb_checkpoints/Laboratory5-checkpoint.ipynb

lucasmoschen/optimal-control-biological

642a12b6a3cb351429018120e564b31c320c44c5

Qwen/Qwen-72B

1. YES 2. YES

__label__por_Latn

``` import numpy as np import matplotlib.pyplot as plt from scipy.optimize import curve_fit import sympy #from astropy.visualization import astropy_mpl_style, quantity_support #from google.colab import drive #drive.mount('/content/drive') c=299792458 ``` Una matriz en python y algunas operaciones. Con esto uds van a construir un cuaderno para calcular eventos espacio-tiempo en 2 sistema de referencia. **En la siguiente celda** ``` vx = input('Ingrese la velocidad para el boost en x') vy = input('Ingrese la velocidad para el boost en y') vz = input('Ingrese la velocidad para el boost en z') vx = float(vx) betax=vx/c vy = float(vy) betay=vy/c vz = float(vz) betaz=vz/c v2=vx**2+vy**2+vz**2 beta2=betax**2+betay**2+betaz**2 if v2>c**2: print('Este valor de velocidad no es posible, está por encima de c, los cálculos NO SON CORRECTOS') ``` Ingrese la velocidad para el boost en x5000 Ingrese la velocidad para el boost en y4000 Ingrese la velocidad para el boost en z3000 ``` #Ahora se calcula directamente la T. de Lorentz con la forma general Mgeneral=[[gamma, -gamma*betax, -gamma*betay, -gamma*betaz],[-gamma*betax, 1+(gamma-1)*betax**2/beta2, (gamma-1)*betax*betay/beta2,(gamma-1)*betax*betaz/beta2],[-gamma*betay, (gamma-1)*betay*betax/beta2 ,1+(gamma-1)*betay**2/beta2, (gamma-1)*betay*betaz/beta2],[-gamma*betaz, (gamma-1)*betaz*betax/beta2, (gamma-1)*betay*betaz/beta2, 1+(gamma-1)*betaz**2/beta2]] print(Mgeneral) ``` [[1.1547005383792517, -1.9258332015464707e-05, -1.5406665612371767e-05, -1.1554999209278823e-05], [-1.9258332015464707e-05, 1.0773502691896257, 0.06188021535170067, 0.04641016151377551], [-1.5406665612371767e-05, 0.06188021535170068, 1.0495041722813605, 0.037128129211020405], [-1.1554999209278823e-05, 0.04641016151377551, 0.037128129211020405, 1.0278460969082652]]

c9bd3220e1cc5badfc41eb492edd196101ab4c43

ipynb

Jupyter Notebook

datos/colabs/Transformaciones de Lorentz.ipynb

Sekilloda/sekilloda.github.io

1a272eb607400a71a2971569e6ac2426f81661f7

datos/colabs/Transformaciones de Lorentz.ipynb

Sekilloda/sekilloda.github.io

1a272eb607400a71a2971569e6ac2426f81661f7

datos/colabs/Transformaciones de Lorentz.ipynb

Sekilloda/sekilloda.github.io

1a272eb607400a71a2971569e6ac2426f81661f7

Qwen/Qwen-72B

1. YES 2. YES

__label__spa_Latn

# Linear Regression with Regularization Regularization is a way to prevent overfitting and allows the model to generalize better. We'll cover the *Ridge* and *Lasso* regression here. ## The Need for Regularization Unlike polynomial fitting, it's hard to imagine how linear regression can overfit the data, since it's just a single line (or a hyperplane). One situation is that features are **correlated** or redundant. Suppose there are two features, both are exactly the same, our predicted hyperplane will be in this format: $$ \hat{y} = w_0 + w_1x_1 + w_2x_2 $$ and the true values of $x_2$ is almost the same as $x_1$ (or with some multiplicative factor and noise). Then, it's best to just drop $w_2x_2$ term and use: $$ \hat{y} = w_0 + w_1x_1 $$ to fit the data. This is a simpler model. But we don't know whether $x_1$ and $x_2$ is **actually** redundant or not, at least with bare eyes, and we don't want to manually drop a parameter just because we feel like it. We want to model to learn to do this itself, that is, to *prefer a simpler model that fits the data well enough*. To do this, we add a *penalty term* to our loss function. Two common penalty terms are L2 and L1 norm of $w$. ## L2 and L1 Penalty ### 0. No Penalty (or Linear) This is linear regression without any regularization (from [previous article](/blog_content/linear_regression/linear_regression_tutorial.html#writing-sse-loss-in-matrix-notation)): $$ L(w) = \sum_{i=1}^{n} \left( y^i - wx^i \right)^2 $$ ### 1. L2 Penalty (or Ridge) We can add the **L2 penalty term** to it, and this is called **L2 regularization**.: $$ L(w) = \sum_{i=1}^{n} \left( y^i - wx^i \right)^2 + \lambda\sum_{j=0}^{d}w_j^2 $$ This is called L2 penalty just because it's a L2-norm of $w$. In fancy term, this whole loss function is also known as **Ridge regression**. Let's see what's going on. Loss function is something we **minimize**. Any terms that we add to it, we also want it to be minimized (that's why it's called *penalty term*). The above means we want $w$ that fits the data well (first term), but we also want the values of $w$ to be small as possible (second term). The lambda ($\lambda$) is there to adjust how much to penalize $w$. Note that `sklearn` refers to this as alpha ($\alpha$) instead, but whatever. It's tricky to know the appropriate value for lambda. You just have to try them out, in exponential range (0.01, 0.1, 1, 10, etc), then select the one that has the lowest loss on validation set, or doing k-fold cross validation. Setting $\lambda$ to be very low means we don't penalize the complex model much. Setting it to $0$ is the original linear regression. Setting it high means we strongly prefer simpler model, at the cost of how well it fits the data. #### Closed-form solution of Ridge It's not hard to find a closed-form solution for Ridge, first write the loss function in matrix notation: $$ L(w) = {\left\lVert y - Xw \right\rVert}^2 + \lambda{\left\lVert w \right\rVert}_2^2 $$ Then the gradient is: $$ \nabla L_w = -2X^T(y-Xw) + 2\lambda w $$ Setting to zero and solve: $$ \begin{align} 0 &= -2X^T(y-Xw) + 2\lambda w \\ &= X^T(y-Xw) - \lambda w \\ &= X^Ty - X^TXw - \lambda w \\ &= X^Ty - (X^TX + \lambda I_d) w \end{align} $$ Move that to other side and we get a closed-form solution: $$ \begin{align} (X^TX + \lambda I_d) w &= X^Ty \\ w &= (X^TX + \lambda I_d)^{-1}X^Ty \end{align} $$ which is almost the same as linear regression without regularization. ### 2. L1 Penalty (or Lasso) As you might guess, you can also use L1-norm for **L1 regularization**: $$ L(w) = \sum_{i=1}^{n} \left( y^i - wx^i \right)^2 + \lambda\sum_{j=0}^{d}\left|w_j\right| $$ Again, in fancy term, this loss function is also known as **Lasso regression**. Using matrix notation: $$ L(w) = {\left\lVert y - Xw \right\rVert}^2 + \lambda{\left\lVert w \right\rVert}_1 $$ It's more complex to get a closed-form solution for this, so we'll leave it here. ## Visualizing the Loss Surface with Regularization Let's see what these penalty terms mean geometrically. ### L2 loss surface <center> \ </center> This simply follows the 3D equation: $$ L(w) = {\left\lVert w \right\rVert}_2^2 = w_0^2 + w_1^2 $$ The center of the bowl is lowest, since `w = [0,0]`, but that is not even a line and it won't predict anything useful. #### L2 loss surface under different lambdas When you multiply the L2 norm function with lambda, $L(w) = \lambda(w_0^2 + w_1^2)$, the width of the bowl changes. The lowest (and flattest) one has lambda of 0.25, which you can see it penalizes The two subsequent ones has lambdas of 0.5 and 1.0. <center> \ </center> ### L1 loss surface Below is the loss surface of L1 penalty: <center> \ </center> Similarly the equation is $L(w) = \lambda(\left| w_0 \right| + \left| w_1 \right|)$. ### Contour of different penalty terms If the L2 norm is 1, you get a unit circle ($w_0^2 + w_1^2 = 1$). In the same manner, you get "unit" shapes in other norms: <center> \ </center> **When you walk along these lines, you get the same loss, which is 1** These shapes can hint us different behaviors of each norm, which brings us to the next question. ## Which one to use, L1 or L2? What's the point of using different penalty terms, as it seems like both try to push down the size of $w$. **Turns out L1 penalty tends to produce sparse solutions**. This means many entries in $w$ are zeros. This is good if you want the model to be simple and compact. Why is that? ### Geometrical Explanation *Note: these figures are generated with unusually high lambda to exaggerate the plot* First let's bring both linear regression and penalty loss surface together (left), and recall that we want to find the **minimum loss when both surfaces are summed up** (right): <center> \ </center> Ridge regression is like finding the middle point where the loss of a sum between linear regression and L2 penalty loss is lowest: <center> \ </center> You can imagine starting with the linear regression solution (red point) where the loss is the lowest, then you move towards the origin (blue point), where the penalty loss is lowest. **The more lambda you set, the more you'll be drawn towards the origin, since you penalize the values of $w_i$ more** so it wants to get to where they're all zeros: <center> \ </center> Since the loss surfaces of linear regression and L2 norm are both ellipsoid, the solution found for Ridge regression **tends to be directly between both solutions**. Notice how the summed ellipsoid is still right in the middle. --- For Lasso: <center> \ </center> And this is the Lasso solution for lambda = 30 and 60: <center> \ \ </center> Notice that the ellipsoid of linear regression **approaches, and finally hits a corner of L1 loss**, and will always stay at that corner. What does a corner of L1 norm means in this situation? It means $w_1 = 0$. Again, this is because the contour lines **at the same loss value** of L2 norm reaches out much farther than L1 norm: <center> \ </center> If the linear regression finds an optimal contact point along the L2 circle, then it will stop since there's no use to move sideways where the loss is usually higher. However, with L1 penalty, it can drift toward a corner, because it's **the same loss along the line** anyway (I mean, why not?) and thus is exploited, if the opportunity arises. ```python ```

ae3c1d979dcc6c2f9f9c9b741f3938a5b50781ab

ipynb

Jupyter Notebook

blog_content_source/linear_regression/linear_regression_regularized.ipynb

aunnnn/ml-tutorial

b40a6fb04dd4dc560f87486f464b292d84f02fdf

blog_content_source/linear_regression/linear_regression_regularized.ipynb

aunnnn/ml-tutorial

b40a6fb04dd4dc560f87486f464b292d84f02fdf

blog_content_source/linear_regression/linear_regression_regularized.ipynb

aunnnn/ml-tutorial

b40a6fb04dd4dc560f87486f464b292d84f02fdf

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Variablen Wenn Sie ein neues Jupyter Notebook erstellen, wählen Sie `Python 3.6` als Typ des Notebooks aus. Innerhalb des Notebooks arbeiten Sie dann mit Python in der Version 3.6. Um zu verstehen, welche Bedeutung Variablen haben, müssen Sie also Variablen in Python 3.6 verstehen. In Python ist eine Variable ein Kurzname für einen Speicherbereich, in den Daten abgelegt werden können. Auf diese Daten kann dann unter dem Namen der Variablen zugegriffen werden. Dies haben wir prinzipiell bereits in den ersten Übungen durchgeführt. Beispiele: `d_i = 20 # Innendurchmesser eines Rohres` Den Name einer Variablen muss eindeutig sein. Er darf nicht mit einer Ziffer beginnen und darf keine Operationszeichen wie `'+', '-', '*', ':', '^'` oder `'#'` enthalten. Soll einer Variablen ein Wert zugewiesen werden, so geschieht das mittels des Gleichheitszeichens, siehe oben. Das `'#'`-Zeichen leitet einen Kommentar ein. Alles, was nach diesem Zeichen folgt, hat einen rein informativen Charakter aber wird von Python ignoriert. Sobald eine Variable angelegt ist, kann diese in Berechnungen verwendet werden, z.B. ```python import math ``` ```python d_i = 20e-3 # Innendurchmesser eines Rohres in m A_i = math.pi*d_i**2/4 # Lichter Querschnitt in m**2 ``` In einer Zelle dürfen mehrere Operationen stehen, siehe oben. Eine Zelle kann eine Ausgabe haben. Das ist dann immer das Ergebnis der Letzten Operation. In der oben stehenden Zelle steht als letztes eine Wertzuweisung `A_i = ...`, die keine Ausgabe erzeugt. Um interaktiv arbeiten zu können, sollte man nicht zu lange Berechnungen in einzelnen Zellen durchführen. Statt dessen sollte man nach wenigen sinnvollen Schritten Zwischenergebnisse anzeigen, um den Gang der Arbeit nachvollziehen zu können. Stellt man einen Fehler fest, so können Werte geändert und die Zelle erneut ausgeführt werden. Um das Ergebnis der oben durchgeführten Berechnung anzuzeigen, kann man die zuletzt erzeugte Variable aufrufen. Die Zelle sähe dann so aus: ```python d_i = 20e-3 # Innendurchmesser eines Rohres in m A_i = math.pi*d_i**2/4 # Lichter Querschnitt in m**2 A_i ``` 0.0003141592653589793 Variablen haben einen Typ, den man sich durch die Funktion `type(variable)` anzeigen kann, z.B.: ```python type(A_i) ``` float # Aufgabe Untersuchen Sie, welchen Typ die Variablen `a=1` `x=5.0` `Name = 'Bernd'` und `Punkt = (3,4)` haben. ```python a = 1 type(a) ``` int ```python x=5.0 type(x) ``` float ```python name='Bernd' type(name) ``` str ```python Punkt=(3,4) type(Punkt) ``` tuple Untersuchen Sie, welche Auswirkung der Befehl `2*name` ```python 2*name ``` 'BerndBernd' mit dem oben definierten Namen hat. Welche Auswirkung hat analog `2*Punkt`? ```python 2*Punkt ``` (3, 4, 3, 4) Stellen Sie eine Vermutung auf, welchen Typ das Produkt `a*x` und die Summe `a+x` mit den oben festgelegten Werten von `a` und `x` haben und überprüfen Sie diese. ```python type(a+x) ``` float ```python type(a*x) ``` float Um die Anwendungen in der Mathematik nicht aus den Augen zu verlieren, bearbeiten Sie die folgende Aufgabe: # Aufgabe Berechnen Sie das Gewicht von 250m Kupferrohr CU15$\times$1 in kg. Entnehmen Sie dafür die Dichte $\varrho$ von Kupfer ihrem Tabellenbuch. Die Zusammenhänge sind durch die folgenden Formeln gegeben: \begin{align} A &= \dfrac{\pi\,(d_a^2 - d_i^2)}{4} \\[2ex] V &= A\, l \\[2ex] m &= \varrho\, V \end{align} ```python import math ``` ```python # Ihre Lösung beginnt hier d_a = 15e-2 # d_a in dm d_i = 13e-2 # d_i in dm l = 250e1 # l in dm A = math.pi*(d_a**2 - d_i**2)/4 # Querschnitt in dm**2 V = A*l # Volumen in dm**3 rho = 8.96 # kg/dm**3 m = rho*V m # Masse in kg ``` 98.52034561657587

36e883b6d78e79d9e4d5e5a6226591b05d222628

ipynb

Jupyter Notebook

src/03-Variablen_lsg.ipynb

w-meiners/anb-first-steps

6cb3583f77ae853922acd86fa9e48e9cf5188596

src/03-Variablen_lsg.ipynb

w-meiners/anb-first-steps

6cb3583f77ae853922acd86fa9e48e9cf5188596

src/03-Variablen_lsg.ipynb

w-meiners/anb-first-steps

6cb3583f77ae853922acd86fa9e48e9cf5188596

Qwen/Qwen-72B

1. YES 2. YES

__label__deu_Latn

$$ \def\abs#1{\left\lvert #1 \right\rvert} \def\Set#1{\left\{ #1 \right\}} \def\mc#1{\mathcal{#1}} \def\M#1{\boldsymbol{#1}} \def\R#1{\mathsf{#1}} \def\RM#1{\boldsymbol{\mathsf{#1}}} \def\op#1{\operatorname{#1}} \def\E{\op{E}} \def\d{\mathrm{\mathstrut d}} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator*{\argmin}{arg\,min} \def\norm#1{\left\lVert #1 \right\rVert} $$ # What is adversarial attack and why we care about it? Despite the effectiveness on a variety of tasks, deep neural networks can be very vulnerable to adversarial attacks. In security-critical domains like facial recognition authorization and autonomous vehicles, such vulnerability against adversarial attacks makes the model highly unreliable. An adversarial attack tries to add an imperceptible perturbation to the sample so that a trained neural network would classify it incorrectly. The following is an example of adversarial attack. ## Adversarial Attacks ### Categories of Attacks #### Poisoning Attack and Evasion Attack Poisoning attacks involve manipulating the training process. The manipulation can happen on the training set (by replacing original samples or inserting fake samples) or the training algorithm itself (by changing the logic of the training algorithm). Such attacks either directly cause poor performance of the trained model or make the model fail on certain samples so as to construct a backdoor for future use. Evasion attacks aim to manipulate the benign samples $\M x$ by a small perturbation $\M \delta$ so that a trained model can no longer classify it correctly. Usually, such perturbations are so small that a human observer cannot notice them. In other words, the perturbed sample "evades" from the classification of the model. #### Targeted Attack and Non-Targeted Attack A targeted attack tries to perturb the benign samples $\M x$ so that the trained model classifies it as a given certain class $t\in \mc Y$. A non-targeted attack only tries to perturb the benign samples $\M x$ so that the trained model classifies them incorrectly. #### White-Box Attack and Black-Box Attack For white-box attacks, the attacker has access to all the knowledge of the targeted model. For neural networks, the attacker knows all the information about the network structure, parameters, gradients, etc. For black-box attacks, the attacker only knows the outputs of the model when feeding inputs into it. In practice, black-box attacks usually rely on generating adversarial perturbations from another model that the attacker has full access to. Black-box attacks are more usual in applications, but the robustness against white-box attacks is the ultimate goal of a robust model because it reveals the fundamental weakness of neuron networks. Thus, most of the study of adversarial robustness focuses on white-box, non-targeted, evasion attacks. ### Examples of White-box Attacking Algorithms Most white-box attacking algorithms are based on using the gradient calculated by the model to perturb the samples. Two typical examples are Fast Gradient Sign Method (FGSM) attack and its multi-step variant, Projected Gradient Descent (PGD) attack. FGSM attack generates the perturbation as $$\begin{align} \M \delta = \epsilon\text{sgn}\nabla_{\M x}L(\M x, y), \end{align}$$ where $\epsilon$ controls the perturbation size. The adversarial sample is $$\begin{align} \M x' = \M x + \M \delta. \end{align}$$ FGSM attack can be seen as trying to maximizing (single step) the loss of the model. PGD attack tries perform the same task, but in a iterative way (at the cost of higher computational time): $$\begin{align} \M x^{t+1} = \Pi_{\M x+\epsilon}\left(\M x^t + \alpha\text{sgn}\nabla_{\M x}L(\M x, y)\right), \end{align}$$ where $\alpha$ is the step size. $\M x^t$ denote the generated adversarial sample in step $t$, with $\M x^0$ being the original sample. $\Pi$ refers to a projection operation that clips the generated adversarial sample into the valid region: the $\epsilon$-ball around $\M x$, which is $\{\M x':\norm{\M x'-\M x}\leq \epsilon \}$. In practice, a PGD attack with a relatively small adversarial power $\epsilon$ (small enough to be neglected by human observers) is able to reduce the accuracy of a well-trained model to nearly zero. Because of such effectiveness, researchers often use PGD attacks as a basic check of the adversarial robustness of their models. #### Implementation of FGSM attack ```python import torch import torch.nn as nn import torch.nn.functional as F import torch.optim as optim from torchvision import datasets, transforms import numpy as np import matplotlib.pyplot as plt # NOTE: This is a hack to get around "User-agent" limitations when downloading MNIST datasets # see, https://github.com/pytorch/vision/issues/3497 for more information from six.moves import urllib opener = urllib.request.build_opener() opener.addheaders = [('User-agent', 'Mozilla/5.0')] urllib.request.install_opener(opener) ``` ##### Inputs - **epsilons** - List of epsilon values to use for the run. It is important to keep 0 in the list because it represents the model performance on the original test set. Also, intuitively we would expect the larger the epsilon, the more noticeable the perturbations but the more effective the attack in terms of degrading model accuracy. Since the data range here is $[0,1]$, no epsilon value should exceed 1. - **pretrained_model** - path to the pretrained MNIST model which was trained with [pytorch/examples/mnist ](https://github.com/pytorch/examples/tree/master/mnist). For simplicity, download the pretrained model [here](https://drive.google.com/drive/folders/1fn83DF14tWmit0RTKWRhPq5uVXt73e0h?usp=sharing). - **use_cuda** - boolean flag to use CUDA if desired and available. Note, a GPU with CUDA is not critical for this tutorial as a CPU will not take much time. ```python epsilons = [0, .05, .1, .15, .2, .25, .3] pretrained_model = "lenet_mnist_model.pth" use_cuda=True ``` ##### Model Under Attack As mentioned, the model under attack is the same MNIST model from [pytorch/examples/mnist](https://github.com/pytorch/examples/tree/master/mnist). You may train and save your own MNIST model or you can download and use the provided model. The *Net* definition and test dataloader here have been copied from the MNIST example. The purpose of this section is to define the model and dataloader, then initialize the model and load the pretrained weights. ```python # LeNet Model definition class Net(nn.Module): def __init__(self): super(Net, self).__init__() self.conv1 = nn.Conv2d(1, 10, kernel_size=5) self.conv2 = nn.Conv2d(10, 20, kernel_size=5) self.conv2_drop = nn.Dropout2d() self.fc1 = nn.Linear(320, 50) self.fc2 = nn.Linear(50, 10) def forward(self, x): x = F.relu(F.max_pool2d(self.conv1(x), 2)) x = F.relu(F.max_pool2d(self.conv2_drop(self.conv2(x)), 2)) x = x.view(-1, 320) x = F.relu(self.fc1(x)) x = F.dropout(x, training=self.training) x = self.fc2(x) return F.log_softmax(x, dim=1) # MNIST Test dataset and dataloader declaration test_loader = torch.utils.data.DataLoader( datasets.MNIST('../data', train=False, download=True, transform=transforms.Compose([ transforms.ToTensor(), ])), batch_size=1, shuffle=True) # Define what device we are using print("CUDA Available: ",torch.cuda.is_available()) device = torch.device("cuda" if (use_cuda and torch.cuda.is_available()) else "cpu") # Initialize the network model = Net().to(device) # Load the pretrained model model.load_state_dict(torch.load(pretrained_model, map_location='cpu')) # Set the model in evaluation mode. In this case this is for the Dropout layers model.eval() ``` CUDA Available: True Net( (conv1): Conv2d(1, 10, kernel_size=(5, 5), stride=(1, 1)) (conv2): Conv2d(10, 20, kernel_size=(5, 5), stride=(1, 1)) (conv2_drop): Dropout2d(p=0.5, inplace=False) (fc1): Linear(in_features=320, out_features=50, bias=True) (fc2): Linear(in_features=50, out_features=10, bias=True) ) ##### FGSM Attack Now, we can define the function that creates the adversarial examples by perturbing the original inputs. The ``fgsm_attack`` function takes three inputs, *image* is the original clean image ($x$), *epsilon* is the pixel-wise perturbation amount ($\epsilon$), and *data_grad* is gradient of the loss w.r.t the input image ($\nabla_{x} L(\mathbf{\theta}, \mathbf{x}, y)$). The function then creates perturbed image as \begin{align}perturbed\_image = image + epsilon*sign(data\_grad) = x + \epsilon * sign(\nabla_{x} L(\mathbf{\theta}, \mathbf{x}, y))\end{align} Finally, in order to maintain the original range of the data, the perturbed image is clipped to range $[0,1]$. ```python # FGSM attack code def fgsm_attack(image, epsilon, data_grad): # Collect the element-wise sign of the data gradient sign_data_grad = data_grad.sign() # Create the perturbed image by adjusting each pixel of the input image perturbed_image = image + epsilon*sign_data_grad # Adding clipping to maintain [0,1] range perturbed_image = torch.clamp(perturbed_image, 0, 1) # Return the perturbed image return perturbed_image ``` ##### Testing Function Finally, the central result of this tutorial comes from the ``test`` function. Each call to this test function performs a full test step on the MNIST test set and reports a final accuracy. However, notice that this function also takes an *epsilon* input. This is because the ``test`` function reports the accuracy of a model that is under attack from an adversary with strength $\epsilon$. More specifically, for each sample in the test set, the function computes the gradient of the loss w.r.t the input data ($data\_grad$), creates a perturbed image with ``fgsm_attack`` ($perturbed\_data$), then checks to see if the perturbed example is adversarial. In addition to testing the accuracy of the model, the function also saves and returns some successful adversarial examples to be visualized later. ```python def test( model, device, test_loader, epsilon ): # Accuracy counter correct = 0 adv_examples = [] # Loop over all examples in test set for data, target in test_loader: # Send the data and label to the device data, target = data.to(device), target.to(device) # Set requires_grad attribute of tensor. Important for Attack data.requires_grad = True # Forward pass the data through the model output = model(data) init_pred = output.max(1, keepdim=True)[1] # get the index of the max log-probability # If the initial prediction is wrong, dont bother attacking, just move on if init_pred.item() != target.item(): continue # Calculate the loss loss = F.nll_loss(output, target) # Zero all existing gradients model.zero_grad() # Calculate gradients of model in backward pass loss.backward() # Collect datagrad data_grad = data.grad.data # Call FGSM Attack perturbed_data = fgsm_attack(data, epsilon, data_grad) # Re-classify the perturbed image output = model(perturbed_data) # Check for success final_pred = output.max(1, keepdim=True)[1] # get the index of the max log-probability if final_pred.item() == target.item(): correct += 1 # Special case for saving 0 epsilon examples if (epsilon == 0) and (len(adv_examples) < 5): adv_ex = perturbed_data.squeeze().detach().cpu().numpy() adv_examples.append( (init_pred.item(), final_pred.item(), adv_ex) ) else: # Save some adv examples for visualization later if len(adv_examples) < 5: adv_ex = perturbed_data.squeeze().detach().cpu().numpy() adv_examples.append( (init_pred.item(), final_pred.item(), adv_ex) ) # Calculate final accuracy for this epsilon final_acc = correct/float(len(test_loader)) print("Epsilon: {}\tTest Accuracy = {} / {} = {}".format(epsilon, correct, len(test_loader), final_acc)) # Return the accuracy and an adversarial example return final_acc, adv_examples ``` ##### Run Attack The last part of the implementation is to actually run the attack. Here, we run a full test step for each epsilon value in the *epsilons* input. For each epsilon we also save the final accuracy and some successful adversarial examples to be plotted in the coming sections. Notice how the printed accuracies decrease as the epsilon value increases. Also, note the $\epsilon=0$ case represents the original test accuracy, with no attack. ```python accuracies = [] examples = [] # Run test for each epsilon for eps in epsilons: acc, ex = test(model, device, test_loader, eps) accuracies.append(acc) examples.append(ex) ``` Epsilon: 0 Test Accuracy = 9810 / 10000 = 0.981 Epsilon: 0.05 Test Accuracy = 9426 / 10000 = 0.9426 Epsilon: 0.1 Test Accuracy = 8510 / 10000 = 0.851 Epsilon: 0.15 Test Accuracy = 6826 / 10000 = 0.6826 Epsilon: 0.2 Test Accuracy = 4303 / 10000 = 0.4303 Epsilon: 0.25 Test Accuracy = 2087 / 10000 = 0.2087 Epsilon: 0.3 Test Accuracy = 871 / 10000 = 0.0871 ##### Results Accuracy vs Epsilon The first result is the accuracy versus epsilon plot. As alluded to earlier, as epsilon increases we expect the test accuracy to decrease. This is because larger epsilons mean we take a larger step in the direction that will maximize the loss. Notice the trend in the curve is not linear even though the epsilon values are linearly spaced. For example, the accuracy at $\epsilon=0.05$ is only about 4% lower than $\epsilon=0$, but the accuracy at $\epsilon=0.2$ is 25% lower than $\epsilon=0.15$. Also, notice the accuracy of the model hits random accuracy for a 10-class classifier between $\epsilon=0.25$ and $\epsilon=0.3$. ```python plt.figure(figsize=(5,5)) plt.plot(epsilons, accuracies, "*-") plt.yticks(np.arange(0, 1.1, step=0.1)) plt.xticks(np.arange(0, .35, step=0.05)) plt.title("Accuracy vs Epsilon") plt.xlabel("Epsilon") plt.ylabel("Accuracy") plt.show() ``` ##### Sample Adversarial Examples Remember the idea of no free lunch? In this case, as epsilon increases the test accuracy decreases **BUT** the perturbations become more easily perceptible. In reality, there is a tradeoff between accuracy degredation and perceptibility that an attacker must consider. Here, we show some examples of successful adversarial examples at each epsilon value. Each row of the plot shows a different epsilon value. The first row is the $\epsilon=0$ examples which represent the original “clean” images with no perturbation. The title of each image shows the “original classification -> adversarial classification.” Notice, the perturbations start to become evident at $\epsilon=0.15$ and are quite evident at $\epsilon=0.3$. However, in all cases humans are still capable of identifying the correct class despite the added noise. ```python # Plot several examples of adversarial samples at each epsilon cnt = 0 plt.figure(figsize=(8,10)) for i in range(len(epsilons)): for j in range(len(examples[i])): cnt += 1 plt.subplot(len(epsilons),len(examples[0]),cnt) plt.xticks([], []) plt.yticks([], []) if j == 0: plt.ylabel("Eps: {}".format(epsilons[i]), fontsize=14) orig,adv,ex = examples[i][j] plt.title("{} -> {}".format(orig, adv)) plt.imshow(ex, cmap="gray") plt.tight_layout() plt.show() ``` ## Adversarial Training with Perturbed Examples To defend adversarial attacks, a direct idea is to add perturbations during the training process. Mardy et al. \cite{madry2017} proposed to formulate a robust optimization problem to minimize the adversarial risk instead of the usual empirical risk: $$\begin{align} \min_{\theta} \E_{(\M x, y)\in \mc D} \left[\max_{\norm{\M\delta}_p<\epsilon}L(\M x+\M\delta, y)\right]. \end{align}$$ The inner maximization tries to find perturbed samples that produce a high loss, which is also the goal of PGD attacks. The outer minimization problem tries to find the model parameters that minimize the adversarial loss given by the inner adversaries. This robust optimization approach effectively trains more robust models. It has been a benchmark for evaluating the adversarial robustness of models and is often seen as the standard way of adversarial training. Based on this method, many variants were proposed in the following years. They involve using a more sophisticated regularizer or adaptively adjusting the adversarial power. Those methods that add perturbations during training share a common disadvantage of high computational cost. ## Adversarial Training with Stochastic Networks Stochastic networks refer to the neuron networks that involve random noise layers. Liu et al. proposed Random Self Ensemble (RSE). Their method injects spherical Gaussian noise into different layers of a network and uses the ensemble of multiple forward pass as the final output. The variance of their added noise is treated as a hyper-parameter to be tuned. RSE shows good robustness against PGD attack and C\&W attack. Similar to RSE, He et al. proposed Parametric Noise Injection (PNI). Rather than fixed variance, they applied an additional intensity parameter to control the variance of the noise. This intensity parameter is trained together with the model parameters. Inspired by the idea of trainable noise, Eustratiadis et al. proposed Weight-Covariance Alignment (WCA) method \cite{wca}. This method adds trainable Gaussian noise to the activation of the penultimate layer of the network. Let $\M g_{\theta}:\mathcal X\rightarrow \mathbb R^D$ be the neural network parameterized by $\theta$ except the final layer and $f_{\M W, \M b}:\mathbb R^D \rightarrow \mathbb R^K$ be the final linear layer parameterized by weight matrix $\M W^{K\times D}$ and bias vector $\M b^{K\times 1}$, where $K=\abs{\mc Y}$ is the number of classes . This WCA method adds a Gaussian noise $\M u \sim \mathcal N_{0, \M\Sigma}$ to the output of penultimate layer $\M g_{\theta}(x)$, where $\M\Sigma^{D\times D}$ is the covariance matrix. Thus, the final output becomes $$\begin{align} f_{\M W, \M b}\left(\M g_\theta(\M x)\right) = \M W\left(\M g_\theta (\M x)+\M u\right) + \M b. \end{align}$$ The loss function is defined as $$\begin{align} L=L_{\text{CE}} + L_{\text{WCA}}+ \lambda \sum_{y\in \mathcal{Y}}\M W_y^{\intercal} \M W_y, \end{align}$$ where $L_{\text{CE}}$ is the usual cross-entropy loss, and $L_{\text{WCA}}$ is a term that encourage the noise and the weight of last layer to be aligned with each other. The third term is gives $l^2$ penalty to $\M W_y$ with large magnitude. The WCA regularizer is defined as $$\begin{align} L_{\text{WCA}} = -\log\sum_{y\in \mathcal{Y}}\M W_y \M\Sigma\M W_y^\intercal . \end{align}$$ where $\M W_y$ is the weight vector of the last layer that is associated with class $y$. The WCA regularizer encourages the weights associated with the last layer to be well aligned with the covariance matrix of the noise. Larger trained variance corresponding to one feature means this feature is harder to perturb, so putting more weight on such features will force the final layer to focus more on these robust features. Models trained with WCA show better performance against PGD attacks on various datasets comparing with the aforementioned approaches. In addition, because of the fact that WCA does not involve generating adversarial samples, the computational time is significantly lower than adversarial training with perturbations. The method we propose is inspired by this WCA method. But instead of adding noise to the penultimate layer, we directly add noise to the output of the final layer. ## Training a neural network with noisy logits We consider training a model with a noisy representation $\R{Z}$ satisfying the Markov chain: $$\R{X}\to \R{Z} \to \hat{\R{Y}}$$ Hence, the estimate for $P_{\R{Y}|\R{X}}$ is given by $P_{\R{Y}|\R{Z}}$ and $P_{\R{Z}|\R{X}}$ as $$P_{\hat{\R{Y}}|\R{X}} (y|x) = E\left[\left.P_{\hat{\R{Y}}|\R{Z}}(y|\R{Z}) \right|\R{X}=x\right].$$ In particular, we propose to set $\R{Z}$ to be the noisy logits of $\hat{\R{Y}}$, i.e., $P_{\hat{Y}|\R{Z}}$ is defined by the pmf obtained with the usual softmax function $$ p_{\hat{\R{Y}}|\RM{z}} (y|\M{z}) := \frac{\exp(z_y)}{\sum_{y'\in \mathcal{Y}} \exp(z_{y'})}, $$ so $z_y$ is the logit for class $y$. The noisy logit is defined as $$ \R{Z} = \RM{z}:=[g(y|\R{X})+\R{u}_y]_{y\in \mathcal{Y}} $$ where $$g(y|x)\in \mathbb{R}$$ for $(x,y)\in \mathcal{X}\times \mathcal{Y}$ is computed by a neural network to be trained, and $\R{u}_y\sim \mathcal{N}_{0,\sigma_y^2}$ for $y\in \mathcal{Y}$ are independent gaussian random variables with variance $\sigma_y^2>0$. For simplicity, $$ $$\begin{align} \M{g}(x)&:= [g(y|x)]_{y\in \mathcal{Y}}\\ \RM{u}&:=[\R{u}_y]_{y\in \mathcal{Y}}\\ \M{\Sigma}&:=\M{\sigma} \M{I} \M{\sigma}^\intercal \quad \text{with }\M{\sigma}:=[ \sigma_y]_{y\in \mathcal{Y}}, \end{align}$$ $$ which are referred to as the (noiseless) logits, additive noise (vector) and its (diagonal) covariance matrix respectively. Hence, $P_{\R{Z}|\R{X}}$ is defined by the multivariate gaussian density function $$ p_{\RM{z}|\R{X}}(\M{z}|x) = \mathcal{N}_{\M{g}(x),\Sigma}(\M{z}) $$ for $x\in \mathcal{X}$ and $\M{z}\in \mathbb{R}^{\abs{\mathcal{Y}}}$. The loss function used for training the neural network is derived from $$ L := E\left[-\log p_{\hat{\R{Y}}|\R{X}}(\R{Y}|\R{X})\right] - \log \sum_{y\in \mathcal{Y}}\sigma_y^2 + \lambda\left(\sum_{y\in \mathcal{Y}}\sigma_y^2\right) $$ ```python ```

ece557c6386cb4ddf4efe5dcfa5fe5fb26961420

ipynb

Jupyter Notebook

part3/Noisy_logits.ipynb

ccha23/miml

6a41de1c0bb41d38e3cdc6e9c27363215b7729b9

2021-08-17T15:16:11.000Z

2021-08-17T15:16:11.000Z

part3/Noisy_logits.ipynb

ccha23/miml

6a41de1c0bb41d38e3cdc6e9c27363215b7729b9

part3/Noisy_logits.ipynb

ccha23/miml

6a41de1c0bb41d38e3cdc6e9c27363215b7729b9

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Reduced Helmholtz equation of state: carbon dioxide **Water equation of state:** You can see the full, state-of-the-art equation of state for water, which also uses a reduced Helmholtz approach: the IAPWS 1995 formulation (Wagner 2002). This equation is state is available using CoolProp with the `Water` fluid. One modern approach for calculating thermodynamic properties of real fluids uses a reduced Helmholtz equation of state, using the reduced Helmholtz free energy function $\alpha$: \begin{equation} \alpha (\tau, \delta) = \frac{a}{RT} = \frac{u - Ts}{RT} \end{equation} which is a function of reduced density $\delta$ and reduced temperature $\tau$: \begin{equation} \delta = \frac{\rho}{\rho_{\text{crit}}} \quad \text{and} \quad \tau = \frac{T_{\text{crit}}}{T} \end{equation} The reduced Helmhotz free energy function, $\alpha(\tau, \delta)$, is given as the sum of ideal gas and residual components: \begin{equation} \alpha(\tau, \delta) = \alpha_{IG} (\tau, \delta) + \alpha_{\text{res}} (\tau, \delta) \;, \end{equation} which are both given as high-order fits using many coefficients: ```python import matplotlib.pyplot as plt %matplotlib inline # these are mostly for making the saved figures nicer from IPython.display import set_matplotlib_formats set_matplotlib_formats('pdf', 'png') plt.rcParams['figure.dpi']= 150 plt.rcParams['savefig.dpi'] = 150 import numpy as np import cantera as ct from scipy import integrate, optimize from pint import UnitRegistry ureg = UnitRegistry() Q_ = ureg.Quantity ``` ```python import sympy sympy.init_printing(use_latex='mathjax') T, R, tau, delta = sympy.symbols('T, R, tau, delta', real=True) a_vars = sympy.symbols('a0, a1, a2, a3, a4, a5, a6, a7', real=True) theta_vars = sympy.symbols('theta3, theta4, theta5, theta6, theta7', real=True) n_vars = sympy.symbols('n0, n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11', real=True) alpha_ideal = sympy.log(delta) + a_vars[0] + a_vars[1]*tau + a_vars[2]*sympy.log(tau) for i in range(3, 8): alpha_ideal += a_vars[i] * sympy.log(1.0 - sympy.exp(-tau * theta_vars[i-3])) display(sympy.Eq(sympy.symbols('alpha_IG'), alpha_ideal)) alpha_res = ( n_vars[0] * delta * tau**0.25 + n_vars[1] * delta * tau**1.25 + n_vars[2] * delta * tau**1.50 + n_vars[3] * delta**3 * tau**0.25 + n_vars[4] * delta**7 * tau**0.875 + n_vars[5] * delta * tau**2.375 * sympy.exp(-delta) + n_vars[6] * delta**2 * tau**2 * sympy.exp(-delta) + n_vars[7] * delta**5 * tau**2.125 * sympy.exp(-delta) + n_vars[8] * delta * tau**3.5 * sympy.exp(-delta**2) + n_vars[9] * delta * tau**6.5 * sympy.exp(-delta**2) + n_vars[10] * delta**4 * tau**4.75 * sympy.exp(-delta**2) + n_vars[11] * delta**2 * tau**12.5 * sympy.exp(-delta**3) ) display(sympy.Eq(sympy.symbols('alpha_res'), alpha_res)) ``` $\displaystyle \alpha_{IG} = a_{0} + a_{1} \tau + a_{2} \log{\left(\tau \right)} + a_{3} \log{\left(1.0 - e^{- \tau \theta_{3}} \right)} + a_{4} \log{\left(1.0 - e^{- \tau \theta_{4}} \right)} + a_{5} \log{\left(1.0 - e^{- \tau \theta_{5}} \right)} + a_{6} \log{\left(1.0 - e^{- \tau \theta_{6}} \right)} + a_{7} \log{\left(1.0 - e^{- \tau \theta_{7}} \right)} + \log{\left(\delta \right)}$ $\displaystyle \alpha_{res} = \delta^{7} n_{4} \tau^{0.875} + \delta^{5} n_{7} \tau^{2.125} e^{- \delta} + \delta^{4} n_{10} \tau^{4.75} e^{- \delta^{2}} + \delta^{3} n_{3} \tau^{0.25} + \delta^{2} n_{11} \tau^{12.5} e^{- \delta^{3}} + \delta^{2} n_{6} \tau^{2} e^{- \delta} + \delta n_{0} \tau^{0.25} + \delta n_{1} \tau^{1.25} + \delta n_{2} \tau^{1.5} + \delta n_{5} \tau^{2.375} e^{- \delta} + \delta n_{8} \tau^{3.5} e^{- \delta^{2}} + \delta n_{9} \tau^{6.5} e^{- \delta^{2}}$ ## Carbon dioxide equation of state The coefficients $a_i$, $\theta_i$, and $n_i$ are given for carbon dioxide: ```python # actual coefficients coeffs_a = [ 8.37304456, -3.70454304, 2.500000, 1.99427042, 0.62105248, 0.41195293, 1.04028922, 8.327678e-2 ] coeffs_theta = [ 3.151630, 6.111900, 6.777080, 11.32384, 27.08792 ] coeffs_n = [ 0.89875108, -0.21281985e1, -0.68190320e-1, 0.76355306e-1, 0.22053253e-3, 0.41541823, 0.71335657, 0.30354234e-3, -0.36643143, -0.14407781e-2, -0.89166707e-1, -0.23699887e-1 ] ``` Through some math, we can find an expression for pressure: \begin{equation} P = R T \rho \left[ 1 + \delta \left(\frac{\partial \alpha_{\text{res}}}{\partial \delta} \right)_{\tau} \right] \end{equation} Use this expression to estimate the pressure at $T$ = 350 K and $v$ = 0.01 m$^3$/kg, and compare against that obtained from Cantera. We can use our symbolic expression for $\alpha_{\text{res}} (\tau, \delta)$ and take the partial derivative: ```python # use Cantera fluid to get specific gas constant and critical properties f = ct.CarbonDioxide() gas_constant = ct.gas_constant / f.mean_molecular_weight temp_crit = f.critical_temperature density_crit = f.critical_density # conditions of interest temp = 350 specific_volume = 0.01 density = 1.0 / specific_volume # take the partial derivative of alpha_res with respect to delta derivative_alpha_delta = sympy.diff(alpha_res, delta) # substitute all coefficients derivative_alpha_delta = derivative_alpha_delta.subs( [(n, n_val) for n, n_val in zip(n_vars, coeffs_n)] ) def get_pressure( temp, specific_vol, fluid, derivative_alpha_delta, tau, delta ): '''Calculates pressure for reduced Helmholtz equation of state''' red_density = (1.0 / specific_vol) / fluid.critical_density red_temp_inv = fluid.critical_temperature / temp gas_constant = ct.gas_constant / fluid.mean_molecular_weight dalpha_ddelta = derivative_alpha_delta.subs( [(delta, red_density), (tau, red_temp_inv)] ) pres = ( gas_constant * temp * (1.0 / specific_vol) * (1.0 + red_density * dalpha_ddelta) ) return pres pres = get_pressure(temp, specific_volume, f, derivative_alpha_delta, tau, delta) print(f'Pressure: {pres / 1e6: .3f} MPa') f.TV = temp, specific_volume print(f'Cantera pressure: {f.P / 1e6: .3f} MPa') ``` Pressure: 5.464 MPa Cantera pressure: 5.475 MPa Our calculation and that from Cantera agree fairly well! They are not exactly the same because Cantera uses a slightly different formulation for the equation of state. Let's compare the calculations now for a range of specific volumes and multiple temperatures: ```python fig, ax = plt.subplots(figsize=(8, 4)) specific_volumes = np.geomspace(0.001, 0.01, num=20) temperatures = [300, 400, 500] for temp in temperatures: pressures = np.zeros(len(specific_volumes)) pressures_cantera = np.zeros(len(specific_volumes)) for idx, spec_vol in enumerate(specific_volumes): pressures[idx] = get_pressure( temp, spec_vol, f, derivative_alpha_delta, tau, delta ) f.TV = temp, spec_vol pressures_cantera[idx] = f.P ax.loglog(specific_volumes, pressures/1000., 'o', color='blue') ax.loglog(specific_volumes, pressures_cantera/1000., color='blue') bbox_props = dict(boxstyle='round', fc='w', ec='0.3', alpha=0.9) ax.text(2e-3, 4e3, '300 K', ha='center', va='center', bbox=bbox_props) ax.text(2e-3, 1.6e4, '400 K', ha='center', va='center', bbox=bbox_props) ax.text(2e-3, 7e4, '500 K', ha='center', va='center', bbox=bbox_props) ax.legend(['Reduced Helmholtz', 'Cantera']) plt.grid(True, which='both') plt.xlabel('Specific volume (m^3/kg)') plt.ylabel('Pressure (kPa)') fig.tight_layout() plt.show() ``` We can see that the pressure calculated using the reduced Helmholtz equation of state matches closely with that from Cantera, which uses a different but similarly advanced equation of state. ## Bibliography Wagner, W., & Pruß, A. (2002). The IAPWS Formulation 1995 for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use. Journal of Physical and Chemical Reference Data, 31(2), 387–535. https://doi.org/10.1063/1.1461829

6f06dd9bbc1022316a219d9c7ad0d2c55634f68c

ipynb

Jupyter Notebook

content/properties-pure/reduced-helmholtz.ipynb

msb002/computational-thermo

9302288217a36e0ce29e320688a3f574921909a5

content/properties-pure/reduced-helmholtz.ipynb

msb002/computational-thermo

9302288217a36e0ce29e320688a3f574921909a5

content/properties-pure/reduced-helmholtz.ipynb

msb002/computational-thermo

9302288217a36e0ce29e320688a3f574921909a5

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

**Notas para contenedor de docker:** Comando de docker para ejecución de la nota de forma local: nota: cambiar `dir_montar` por la ruta de directorio que se desea mapear a `/datos` dentro del contenedor de docker. ``` dir_montar=<ruta completa de mi máquina a mi directorio>#aquí colocar la ruta al directorio a montar, por ejemplo: #dir_montar=/Users/erick/midirectorio. ``` Ejecutar: ``` $docker run --rm -v $dir_montar:/datos --name jupyterlab_prope_r_kernel_tidyverse -p 8888:8888 -d palmoreck/jupyterlab_prope_r_kernel_tidyverse:3.0.16 ``` Ir a `localhost:8888` y escribir el password para jupyterlab: `qwerty` Detener el contenedor de docker: ``` docker stop jupyterlab_prope_r_kernel_tidyverse ``` Documentación de la imagen de docker `palmoreck/jupyterlab_prope_r_kernel_tidyverse:3.0.16` en [liga](https://github.com/palmoreck/dockerfiles/tree/master/jupyterlab/prope_r_kernel_tidyverse). --- Para ejecución de la nota usar: [docker](https://www.docker.com/) (instalación de forma **local** con [Get docker](https://docs.docker.com/install/)) y ejecutar comandos que están al inicio de la nota de forma **local**. O bien dar click en alguno de los botones siguientes: [](https://mybinder.org/v2/gh/palmoreck/dockerfiles-for-binder/jupyterlab_prope_r_kernel_tidyerse?urlpath=lab/tree/Propedeutico/Python/clases/2_calculo_DeI/1_aproximacion_a_derivadas_e_integrales.ipynb) esta opción crea una máquina individual en un servidor de Google, clona el repositorio y permite la ejecución de los notebooks de jupyter. [](https://repl.it/languages/python3) esta opción no clona el repositorio, no ejecuta los notebooks de jupyter pero permite ejecución de instrucciones de Python de forma colaborativa con [repl.it](https://repl.it/). Al dar click se crearán nuevos ***repl*** debajo de sus users de ***repl.it***. En Python podemos utilizar el paquete *SymPy* para realizar cómputo algebraico o simbólico, ver [computer algebra](https://en.wikipedia.org/wiki/Computer_algebra). ```python import sympy #usamos import para importar módulos o paquetes de python ``` # Representación de símbolos matemáticos como objetos de Python Para usar el contenido al que podemos acceder dentro de un paquete de Python utilizamos `sympy.<aquí escribir el contenido a usar>` ## Clase `Symbol` Usamos la clase `Symbol` para crear un objeto `Symbol` de Python: ```python x = sympy.Symbol("x") #el nombre del símbolo es x y se asigna a la variable x ``` ```python x ``` $\displaystyle x$ Y tenemos funciones ya en las librerías incluidas en Python cuando se instala en nuestras máquinas como `type`: ```python type(x) #podemos revisar qué tipo de objeto es con la función type ``` sympy.core.symbol.Symbol ```python y = sympy.Symbol("y") ``` ```python y ``` $\displaystyle y$ ```python type(y) ``` sympy.core.symbol.Symbol Podemos pasar argumentos a la clase `Symbol` para identificar el tipo del objeto. ```python x = sympy.Symbol("x") y = sympy.Symbol("y", positive=True) #argumento positive igual a True z = sympy.Symbol("z", negative=True) ``` Una vez que hemos creado un objeto tipo `Symbol` podemos usar funciones como `sqrt`: ```python sympy.sqrt(x**2) ``` $\displaystyle \sqrt{x^{2}}$ ```python sympy.sqrt(y**2) ``` $\displaystyle y$ ```python sympy.sqrt(z**2) ``` $\displaystyle - z$ *SymPy* nos devuelve simplificaciones útiles si identificamos el tipo del objeto: ```python n1 = sympy.Symbol("n1") n2 = sympy.Symbol("n2", integer=True) n3 = sympy.Symbol("n3", odd=True) n4 = sympy.Symbol("n4", even=True) ``` ```python sympy.cos(n1*sympy.pi) ``` $\displaystyle \cos{\left(\pi n_{1} \right)}$ ```python sympy.cos(n2*sympy.pi) ``` $\displaystyle \left(-1\right)^{n_{2}}$ ```python sympy.cos(n3*sympy.pi) ``` $\displaystyle -1$ ```python sympy.cos(n4*sympy.pi) ``` $\displaystyle 1$ Podemos definir símbolos en una sola línea con la función de `symbols` como sigue: ```python a, b, c = sympy.symbols("a, b, c") #obsérvese el uso de tuplas del lado izquierdo de la igualdad ``` ```python a ``` $\displaystyle a$ ```python b ``` $\displaystyle b$ ## Expresiones Para representar en *SymPy* la expresión algebraica $1 + 2x^2 + 3x^3 - x^2 + 5$ creamos al símbolo $x$: ```python x = sympy.Symbol("x") ``` ```python expr = 1 + 2*x**2 + 3*x**3 - x**2 + 5 ``` ```python expr ``` $\displaystyle 3 x^{3} + x^{2} + 6$ **obsérvese que se ha simplificado la expresión.** ## Subs **Ejemplo evaluar la expresión $3x^3 + x^2 + 6$ en $x=1, 2$** ```python x = sympy.Symbol("x") ``` ```python expr = 3*x**3 + x**2 + 6 ``` Podemos usar el método `subs` del objeto `expr` para substituir el valor de $x$ en $1$: ```python expr.subs(x,1) ``` $\displaystyle 10$ ```python expr.subs(x,2) ``` $\displaystyle 34$ **Ejemplo: evaluar $xy + z^2x$ en $x = 1.25$, $y=0.4$, $z=3.2$** ```python x, y, z = sympy.symbols("x,y,z") ``` ```python expr = x*y + z**2*x ``` ```python vals = {x: 1.25, y: 0.4, z: 3.2} #obsérvese el uso de diccionarios ``` ```python expr.subs(vals) ``` $\displaystyle 13.3$ ## Simplify **Ejemplo: $2(x^2 -x) -x(x+1)$** ```python x = sympy.Symbol("x") ``` ```python expr2 = 2*(x**2 - x) - x*(x+1) ``` ```python expr2 ``` $\displaystyle 2 x^{2} - x \left(x + 1\right) - 2 x$ Usamos la función `simplify` de *SymPy*: ```python sympy.simplify(expr2) ``` $\displaystyle x \left(x - 3\right)$ Y es equivalente a usar el método *simplify* del objeto `expr2`: ```python expr2.simplify() ``` $\displaystyle x \left(x - 3\right)$ **Ejemplo: $2\sin(x)\cos(x)$** ```python x = sympy.Symbol("x") ``` ```python expr3 = 2*sympy.sin(x)*sympy.cos(x) ``` ```python expr3 ``` $\displaystyle 2 \sin{\left(x \right)} \cos{\left(x \right)}$ ```python expr3.simplify() ``` $\displaystyle \sin{\left(2 x \right)}$ ## Expand **Ejemplo:** $(x+1)*(x+2)$ ```python x = sympy.Symbol("x") ``` ```python expr = (x+1)*(x+2) ``` ```python sympy.expand(expr) ``` $\displaystyle x^{2} + 3 x + 2$ ```python expr.expand() ``` $\displaystyle x^{2} + 3 x + 2$ ## Factor **Ejemplo:** $x^2 -1$ ```python x = sympy.Symbol("x") ``` ```python expr = x**2 -1 ``` ```python expr.factor() ``` $\displaystyle \left(x - 1\right) \left(x + 1\right)$ ```python sympy.factor(expr) ``` $\displaystyle \left(x - 1\right) \left(x + 1\right)$ ## Ecuaciones **Ejemplo:** $x^2+2x-3=0$ ```python x = sympy.Symbol("x") ``` ```python sympy.solve(x**2 + 2*x -3) ``` [-3, 1] **Ejemplo:** $ax^2+bx+c=0$ para la variable $x$ ```python x = sympy.Symbol("x") ``` ```python a,b,c = sympy.symbols("a, b, c") ``` ```python sympy.solve(a*x**2 + b*x + c, x) #aquí especificamos la variable que es incógnita ``` [(-b + sqrt(-4*a*c + b**2))/(2*a), -(b + sqrt(-4*a*c + b**2))/(2*a)] Y hay ejemplos de ecuaciones que no pueden resolverse de forma cerrada (en términos de sus coeficientes y operaciones) como $x^5 - x^2 +1 = 0$ ```python x = sympy.Symbol("x") ``` ```python sympy.solve(x**5 - x**2 +1) ``` [CRootOf(x**5 - x**2 + 1, 0), CRootOf(x**5 - x**2 + 1, 1), CRootOf(x**5 - x**2 + 1, 2), CRootOf(x**5 - x**2 + 1, 3), CRootOf(x**5 - x**2 + 1, 4)] ## Referencias * [SymPy](https://www.sympy.org/en/index.html) y [Numerical Python by Robert Johansson, Apress](https://www.apress.com/gp/book/9781484242452)

7254c7209156004f5a3b0f2899d5ddbcbf9e4ff3

ipynb

Jupyter Notebook

Python/clases/2_calculo_DeI/0_modulo_sympy.ipynb

CarlosJChV/Propedeutico

d903192ffa64a7576faace68c2256e69bc11087c

2019-07-07T07:51:19.000Z

2022-03-04T18:17:36.000Z

Python/clases/2_calculo_DeI/0_modulo_sympy.ipynb

CarlosJChV/Propedeutico

d903192ffa64a7576faace68c2256e69bc11087c

2019-06-12T01:15:41.000Z

2021-08-01T18:20:04.000Z

Python/clases/2_calculo_DeI/0_modulo_sympy.ipynb

CarlosJChV/Propedeutico

d903192ffa64a7576faace68c2256e69bc11087c

2019-06-07T15:24:05.000Z

2021-07-27T03:05:41.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__spa_Latn

```python from sympy import * init_printing() ''' r_GEO = 36000 + 6371 KM r_LEO = 2000 + 6371 KM G = 6.674e-11 Me = 5.972e24 ''' M, E = symbols("M E", Functions = True) e_c, a, G, M_e, r, mu = symbols("e_c a G M_e r mu", Contstants = True) T_circular, T_elliptical, T_GEO, T_GTO, T_LEO, r_LEO, r_GEO, T_tot = symbols("T_circular T_elliptical T_GEO T_GTO T_LEO r_LEO r_GEO T_tot", Constants = True) t, x, y, Y = symbols("t x y Y", Variables = True) mu_calculated = (6.674e-11 * 5.972e24) ``` The orbital period of a circular Orbit: ```python Eq(T_circular, 2*pi*sqrt(r**3 / mu)) ``` Where mu is: ```python Eq(mu, G*M_e) ``` Then, the GEO's orbital period in hours is: ```python r_GEO_Calculated = (36000 + 6371)*1000 T_GEO_Calculated = 2*pi*sqrt(r_GEO_Calculated**3 / mu_calculated) Eq(T_GEO, T_GEO_Calculated.evalf()/60/60) ``` And the LEO's orbital period in hours is: ```python r_LEO_Calculated = (2000 + 6371)*1000 T_LEO_Calculated = 2*pi*sqrt(r_LEO_Calculated**3 / mu_calculated) Eq(T_LEO, T_LEO_Calculated.evalf()/60/60) ``` _____________________________________________ # Finding the GTO. The goal is to get both 'e' (the eccentricity of our GTO) and 'a' (its semi-major axis). So, we need 2 eqns. The equation of the GEO (A circle equation): ```python geo = Eq(y**2, r**2-x**2) geo ``` The equation of the GTO (An ellipse equation): ```python gto = Eq(((x+a*e_c)**2/a**2)+(y**2/a**2*(1-e_c**2)), 1) gto ``` We Wanna solve these two eqns to get the semi-major axis of our GTO and the raduis of our LEO. first, substitute the GEO's eqn in the GTO's eqn. ```python toSolve = gto.subs({y**2:geo.rhs}) toSolve ``` Now we can solve for x. ```python solX = solveset(toSolve, x) solX ``` Now we can calculate the y coordinate for each x. ```python solY1 = solveset(Eq(geo.lhs, geo.rhs.subs({x:list(solX)[0]})), y) solY1 ``` ```python solY2 = solveset(Eq(geo.lhs, geo.rhs.subs({x:list(solX)[1]})), y) solY2 ``` We have 4 different possible points for the intersection between a circle and an ellipse, but the intersection between the GEO and the GTO is going to be at only one point with an x coordinate of '-r_GEO' (the radius of the GEO). Now, we can get the first eqn. ```python geoAndGtoIntersection = solveset(Eq(list(solX)[0], -r_GEO).subs({r:r_GEO}), a) geoAndGtoIntersection ``` Surbrisingly, there are 2 possible values for a. But we're not interrested in the negative value. So our first eqn is: ```python eqn1 = Eq(a, list(list(geoAndGtoIntersection.args)[2])[1]) eqn1 ``` To get another eqn, we can do the same but this time with the LOE. The intersection between our LEO and GTO is exactly at the x coordinate of 'r_LEO'. ```python gtoAndLeoIntersection = solveset(Eq(list(solX)[1], r).subs({r:r_LEO}), a) gtoAndLeoIntersection ``` Again, there are 2 possible values for 'r_LEO' but we need the positive one. ```python eqn2 = Eq(a, list(list(gtoAndLeoIntersection.args)[2])[0]) eqn2 ``` This is the positive because 0 < e_c < 1. Now, we have 2 eqns and 2 variables. And we're ready to get 'a' and 'e_c' ```python e_c_Exp = Eq(e_c, solveset(eqn1.subs({a:eqn2.rhs, r_GEO:r_GEO_Calculated, r_LEO:r_LEO_Calculated}), e_c).args[0]) e_c_Calculated = e_c_Exp.rhs e_c_Exp ``` ```python s = solveset(eqn2.subs({r_LEO:r_LEO_Calculated, e_c:e_c_Calculated})).args[0] a_Exp = Eq(a, s) a_Calculated = a_Exp.rhs a_Exp ``` There's another way for finding 'a'. ```python p1 = plot(sqrt(r_GEO_Calculated**2-x**2), -sqrt(r_GEO_Calculated**2-x**2), sqrt(r_LEO_Calculated**2-x**2), -sqrt(r_LEO_Calculated**2-x**2), sqrt(a_Calculated**2*(1-e_c_Calculated**2)*(1-((x+a_Calculated*e_c_Calculated)**2/a_Calculated**2))), -sqrt(a_Calculated**2*(1-e_c_Calculated**2)*(1-((x+a_Calculated*e_c_Calculated)**2/a_Calculated**2))),(x, -5*10**7, 5*10**7),xlim = (-7.5*10**7, 7.5*10**7), ylim=((-5*10**7, 5*10**7))) ``` From the geometry we can say that: ```python Eq(a, (r_LEO + r_GEO)/2) ``` This could've saved us a lot of math work :) __________________________________ # Now let's calculate the periods. The orbital period of an elliptical orbit is: ```python Eq(T_elliptical, 2*pi*sqrt(a**3 / mu)) ``` ```python T_GTO_Calculated = 2*pi*sqrt(a_Calculated**3/mu_calculated) Eq(T_GTO, T_GTO_Calculated.evalf()/60/60) ``` So, the total time required to put our satellite in a GEO using Hohmann transfer is: ```python Eq(T_tot, T_GTO / 2 + T_LEO / 2) ``` The total time required to put our satellite in a GEO of a 36,000 Kilometers above sea level in hours is: ```python Eq(T_tot, (T_GTO_Calculated / 2 + T_LEO_Calculated / 2).evalf()/60/60) ```

1e53111d5c0a13baee4c6e7b73cb80cf171ffb77

ipynb

Jupyter Notebook

Python Notebooks/Hohmann Transfer.ipynb

Yaamani/Satellite-Simulation

f9b3363e79b62a30724c53c99fdb097a68ff324d

Python Notebooks/Hohmann Transfer.ipynb

Yaamani/Satellite-Simulation

f9b3363e79b62a30724c53c99fdb097a68ff324d

Python Notebooks/Hohmann Transfer.ipynb

Yaamani/Satellite-Simulation

f9b3363e79b62a30724c53c99fdb097a68ff324d

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

**This notebook is an exercise in the [Computer Vision](https://www.kaggle.com/learn/computer-vision) course. You can reference the tutorial at [this link](https://www.kaggle.com/ryanholbrook/convolution-and-relu).** --- # Introduction # In this exercise, you'll work on building some intuition around feature extraction. First, we'll walk through the example we did in the tutorial again, but this time, with a kernel you choose yourself. We've mostly been working with images in this course, but what's behind all of the operations we're learning about is mathematics. So, we'll also take a look at how these feature maps can be represented instead as arrays of numbers and what effect convolution with a kernel will have on them. Run the cell below to get started! ```python # Setup feedback system from learntools.core import binder binder.bind(globals()) from learntools.computer_vision.ex2 import * ``` # Apply Transformations # The next few exercises walk through feature extraction just like the example in the tutorial. Run the following cell to load an image we'll use for the next few exercises. ```python import numpy as np import tensorflow as tf import matplotlib.pyplot as plt plt.rc('figure', autolayout=True) plt.rc('axes', labelweight='bold', labelsize='large', titleweight='bold', titlesize=18, titlepad=10) plt.rc('image', cmap='magma') image_path = '../input/computer-vision-resources/car_illus.jpg' image = tf.io.read_file(image_path) image = tf.io.decode_jpeg(image, channels=1) image = tf.image.resize(image, size=[400, 400]) plt.figure(figsize=(6, 6)) plt.imshow(tf.squeeze(image), cmap='gray') plt.axis('off') plt.show(); ``` You can run this cell to see some standard kernels used in image processing. ```python import learntools.computer_vision.visiontools as visiontools from learntools.computer_vision.visiontools import edge, bottom_sobel, emboss, sharpen kernels = [edge, bottom_sobel, emboss, sharpen] names = ["Edge Detect", "Bottom Sobel", "Emboss", "Sharpen"] plt.figure(figsize=(12, 12)) for i, (kernel, name) in enumerate(zip(kernels, names)): plt.subplot(1, 4, i+1) visiontools.show_kernel(kernel) plt.title(name) plt.tight_layout() ``` # 1) Define Kernel # Use the next code cell to define a kernel. You have your choice of what kind of kernel to apply. One thing to keep in mind is that the *sum* of the numbers in the kernel determines how bright the final image is. Generally, you should try to keep the sum of the numbers between 0 and 1 (though that's not required for a correct answer). In general, a kernel can have any number of rows and columns. For this exercise, let's use a $3 \times 3$ kernel, which often gives the best results. Define a kernel with `tf.constant`. ```python # YOUR CODE HERE: Define a kernel with 3 rows and 3 columns. kernel = tf.constant([ [-1, 1, -1], [1, 8, 1], [-1, 1, -1], ]) # Uncomment to view kernel visiontools.show_kernel(kernel) # Check your answer q_1.check() ``` ```python # Lines below will give you a hint or solution code #q_1.hint() #q_1.solution() ``` Now we'll do the first step of feature extraction, the filtering step. First run this cell to do some reformatting for TensorFlow. ```python # Reformat for batch compatibility. image = tf.image.convert_image_dtype(image, dtype=tf.float32) image = tf.expand_dims(image, axis=0) kernel = tf.reshape(kernel, [*kernel.shape, 1, 1]) kernel = tf.cast(kernel, dtype=tf.float32) ``` # 2) Apply Convolution # Now we'll apply the kernel to the image by a convolution. The *layer* in Keras that does this is `layers.Conv2D`. What is the *backend function* in TensorFlow that performs the same operation? ```python # YOUR CODE HERE: Give the TensorFlow convolution function (without arguments) conv_fn = tf.nn.conv2d # Check your answer q_2.check() ``` <IPython.core.display.Javascript object> <span style="color:#33cc33">Correct</span> ```python # Lines below will give you a hint or solution code #q_2.hint() #q_2.solution() ``` Once you've got the correct answer, run this next cell to execute the convolution and see the result! ```python image_filter = conv_fn( input=image, filters=kernel, strides=1, # or (1, 1) padding='SAME', ) plt.imshow( # Reformat for plotting tf.squeeze(image_filter) ) plt.axis('off') plt.show(); ``` Can you see how the kernel you chose relates to the feature map it produced? # 3) Apply ReLU # Now detect the feature with the ReLU function. In Keras, you'll usually use this as the activation function in a `Conv2D` layer. What is the *backend function* in TensorFlow that does the same thing? ```python # YOUR CODE HERE: Give the TensorFlow ReLU function (without arguments) relu_fn = tf.nn.relu # Check your answer q_3.check() ``` ```python # Lines below will give you a hint or solution code #q_3.hint() #q_3.solution() ``` Once you've got the solution, run this cell to detect the feature with ReLU and see the result! The image you see below is the feature map produced by the kernel you chose. If you like, experiment with some of the other suggested kernels above, or, try to invent one that will extract a certain kind of feature. ```python image_detect = relu_fn(image_filter) plt.imshow( # Reformat for plotting tf.squeeze(image_detect) ) plt.axis('off') plt.show(); ``` In the tutorial, our discussion of kernels and feature maps was mainly visual. We saw the effect of `Conv2D` and `ReLU` by observing how they transformed some example images. But the operations in a convolutional network (like in all neural networks) are usually defined through mathematical functions, through a computation on numbers. In the next exercise, we'll take a moment to explore this point of view. Let's start by defining a simple array to act as an image, and another array to act as the kernel. Run the following cell to see these arrays. ```python # Sympy is a python library for symbolic mathematics. It has a nice # pretty printer for matrices, which is all we'll use it for. import sympy sympy.init_printing() from IPython.display import display image = np.array([ [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 1, 0, 1, 1, 1], [0, 1, 0, 0, 0, 0], ]) kernel = np.array([ [1, -1], [1, -1], ]) display(sympy.Matrix(image)) display(sympy.Matrix(kernel)) # Reformat for Tensorflow image = tf.cast(image, dtype=tf.float32) image = tf.reshape(image, [1, *image.shape, 1]) kernel = tf.reshape(kernel, [*kernel.shape, 1, 1]) kernel = tf.cast(kernel, dtype=tf.float32) ``` # 4) Observe Convolution on a Numerical Matrix # What do you see? The image is simply a long vertical line on the left and a short horizontal line on the lower right. What about the kernel? What effect do you think it will have on this image? After you've thought about it, run the next cell for the answer. ```python # View the solution (Run this code cell to receive credit!) q_4.check() ``` Now let's try it out. Run the next cell to apply convolution and ReLU to the image and display the result. ```python image_filter = tf.nn.conv2d( input=image, filters=kernel, strides=1, padding='VALID', ) image_detect = tf.nn.relu(image_filter) # The first matrix is the image after convolution, and the second is # the image after ReLU. display(sympy.Matrix(tf.squeeze(image_filter).numpy())) display(sympy.Matrix(tf.squeeze(image_detect).numpy())) ``` Is the result what you expected? # Conclusion # In this lesson, you learned about the first two operations a convolutional classifier uses for feature extraction: **filtering** an image with a **convolution** and **detecting** the feature with the **rectified linear unit**. # Keep Going # Move on to [**Lesson 3**](https://www.kaggle.com/ryanholbrook/maximum-pooling) to learn the final operation: **condensing** the feature map with **maximum pooling**! --- *Have questions or comments? Visit the [Learn Discussion forum](https://www.kaggle.com/learn-forum) to chat with other Learners.*

adee5c7500430b75308db90d56f0d9efe32fcce4

ipynb

Jupyter Notebook

Computer_Vision/exercise-convolution-and-relu.ipynb

olonok69/kaggle

bf61ba510c83fd55262939ac6c5a62b7c855ba53

Computer_Vision/exercise-convolution-and-relu.ipynb

olonok69/kaggle

bf61ba510c83fd55262939ac6c5a62b7c855ba53

Computer_Vision/exercise-convolution-and-relu.ipynb

olonok69/kaggle

bf61ba510c83fd55262939ac6c5a62b7c855ba53

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Semantics: PrefScLTL. In this notebook, we ensure that semantics of our proposed preference logic are sound. Proposed semantics: * $(w_1, w_2) \models \alpha_1~\trianglerighteq~\alpha_2$ iff $w_1 \models \alpha_1$ and $w_2 \models \alpha_2 \land \neg \alpha_1$ We expect the remaining operator semantics to follow from this definition. * $(w_1, w_2) \models \alpha_1~\triangleright~\alpha_2$ iff $(w_1, w_2) \models \alpha_1~\trianglerighteq~\alpha_2$ and $(w_1, w_2) \not\models \alpha_2~\trianglerighteq~\alpha_1$ * $(w_1, w_2) \models \alpha_1~\sim~\alpha_2$ iff $(w_1, w_2) \models \alpha_1~\trianglerighteq~\alpha_2$ and $(w_1, w_2) \models \alpha_2~\trianglerighteq~\alpha_1$ In what follows, we derive these semantics to ensure soundness of our definitions. ## Indifference. Every atomic preference formula induces four partitions of $\Sigma^\omega$. Correspondingly, we define 4 propositions each for $w_1, w_2$ to denote which scLTL formulas the words satisfy. * `w1_00` means $w1$ satisfies $\not \alpha_1$, $\not \alpha_2$. * `w1_01` means $w1$ satisfies $\not \alpha_1$, $\alpha_2$. * `w1_10` means $w1$ satisfies $\alpha_1$, $\not \alpha_2$. * `w1_11` means $w1$ satisfies $\alpha_1$, $\alpha_2$. * `w2_00` means $w2$ satisfies $\not \alpha_1$, $\not \alpha_2$. * `w2_01` means $w2$ satisfies $\not \alpha_1$, $\alpha_2$. * `w2_10` means $w2$ satisfies $\alpha_1$, $\not \alpha_2$. * `w2_11` means $w2$ satisfies $\alpha_1$, $\alpha_2$. ```python from sympy import * from sympy.logic import simplify_logic ``` ```python w1_00, w1_01, w1_10, w1_11, w2_00, w2_01, w2_10, w2_11 = symbols('w1_00 w1_01 w1_10 w1_11 w2_00 w2_01 w2_10 w2_11') ``` ```python # Constraint 1: w1 must be in one of the classes. w1_constraint1 = w1_00 | w1_01 | w1_10 | w1_11 w2_constraint1 = w2_00 | w2_01 | w2_10 | w2_11 w1_constraint1, w2_constraint1 ``` (w1_00 | w1_01 | w1_10 | w1_11, w2_00 | w2_01 | w2_10 | w2_11) ```python # Constraint 2: w1 \models \alpha_1 and w_2 \models \neg \alpha_1 \land \alpha_2 w1_constraint2 = w1_10 | w1_11 w2_constraint2 = w2_01 w1_constraint2, w2_constraint2 ``` (w1_10 | w1_11, w2_01) ```python # Constraint 3: w1 \not\models \alpha_2 and w_2 \not\models \neg \alpha_2 \land \alpha_1 w1_constraint3 = w1_10 | w1_10 | w1_00 w2_constraint3 = w2_01 | w2_11 | w2_00 w1_constraint3, w2_constraint3 ``` (w1_00 | w1_10, w2_00 | w2_01 | w2_11) ```python # Semantics for strict preference. # w1 must satisfy constraints 1, 2, 3. w1_satisfy = w1_constraint1 & w1_constraint2 & w1_constraint3 w2_satisfy = w2_constraint1 & w2_constraint2 & w2_constraint3 w1_satisfy = simplify_logic(w1_satisfy) w2_satisfy = simplify_logic(w2_satisfy) print("w1_satisfy", w1_satisfy) print("w2_satisfy", w2_satisfy) ``` w1_satisfy w1_10 | (w1_00 & w1_11) w2_satisfy w2_01 In `w1_satisfy`, note that the clause `(w1_00 & w1_11)` is trivially false because w1 cannot simultaneously satisfy and violate both $\alpha_1, \alpha_2$. Therefore, `w1_satisfy = w1_10`. And, `w2_satisfy = w2_01`. We recognize this to be the semantics we have defined for our logic. Q.E.D.

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ipynb

Jupyter Notebook

jp-notebooks/pref-scltl-semantics.ipynb

abhibp1993/preference-planning

a6384457debee65735eb24eed678f8f98f69d113

jp-notebooks/pref-scltl-semantics.ipynb

abhibp1993/preference-planning

a6384457debee65735eb24eed678f8f98f69d113

jp-notebooks/pref-scltl-semantics.ipynb

abhibp1993/preference-planning

a6384457debee65735eb24eed678f8f98f69d113

Qwen/Qwen-72B

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__label__eng_Latn

# CS-109A Introduction to Data Science ## Lab 11: Neural Network Basics - Introduction to `tf.keras` **Harvard University**<br> **Fall 2019**<br> **Instructors:** Pavlos Protopapas, Kevin Rader, Chris Tanner<br> **Lab Instructors:** Chris Tanner and Eleni Kaxiras. <br> **Authors:** Eleni Kaxiras, David Sondak, and Pavlos Protopapas. ```python ## RUN THIS CELL TO PROPERLY HIGHLIGHT THE EXERCISES import requests from IPython.core.display import HTML styles = requests.get("https://raw.githubusercontent.com/Harvard-IACS/2018-CS109A/master/content/styles/cs109.css").text HTML(styles) ``` <style> blockquote { background: #AEDE94; } h1 { padding-top: 25px; padding-bottom: 25px; text-align: left; padding-left: 10px; background-color: #DDDDDD; color: black; } h2 { padding-top: 10px; padding-bottom: 10px; text-align: left; padding-left: 5px; background-color: #EEEEEE; color: black; } div.exercise { background-color: #ffcccc; border-color: #E9967A; border-left: 5px solid #800080; padding: 0.5em; } span.sub-q { font-weight: bold; } div.theme { background-color: #DDDDDD; border-color: #E9967A; border-left: 5px solid #800080; padding: 0.5em; font-size: 18pt; } div.gc { background-color: #AEDE94; border-color: #E9967A; border-left: 5px solid #800080; padding: 0.5em; font-size: 12pt; } p.q1 { padding-top: 5px; padding-bottom: 5px; text-align: left; padding-left: 5px; background-color: #EEEEEE; color: black; } header { padding-top: 35px; padding-bottom: 35px; text-align: left; padding-left: 10px; background-color: #DDDDDD; color: black; } </style> ```python import numpy as np import matplotlib.pyplot as plt import matplotlib.image as mpimg import numpy as np import pandas as pd %matplotlib inline from PIL import Image ``` ```python from __future__ import absolute_import, division, print_function, unicode_literals # TensorFlow and tf.keras import tensorflow as tf tf.keras.backend.clear_session() # For easy reset of notebook state. print(tf.__version__) # You should see a 2.0.0 here! ``` 2.0.0 #### Instructions for running `tf.keras` with Tensorflow 2.0: 1. Create a `conda` virtual environment by cloning an existing one that you know works ``` conda create --name myclone --clone myenv ``` 2. Go to [https://www.tensorflow.org/install/pip](https://www.tensorflow.org/install/pip) and follow instructions for your machine. 3. In a nutshell: ``` pip install --upgrade pip pip install tensorflow==2.0.0 ``` All references to Keras should be written as `tf.keras`. For example: ``` model = tf.keras.models.Sequential([ tf.keras.layers.Flatten(input_shape=(28, 28)), tf.keras.layers.Dense(128, activation='relu'), tf.keras.layers.Dropout(0.2), tf.keras.layers.Dense(10, activation='softmax') ]) model.compile(optimizer='adam', loss='sparse_categorical_crossentropy', metrics=['accuracy']) tf.keras.models.Sequential tf.keras.layers.Dense, tf.keras.layers.Activation, tf.keras.layers.Dropout, tf.keras.layers.Flatten, tf.keras.layers.Reshape tf.keras.optimizers.SGD tf.keras.preprocessing.image.ImageDataGenerator tf.keras.regularizers tf.keras.datasets.mnist ``` You could avoid the long names by using ``` from tensorflow import keras from tensorflow.keras import layers ``` These imports do not work on some systems, however, because they pick up previous versions of `keras` and `tensorflow`. That is why I avoid them in this lab. ## Learning Goals In this lab we will understand the basics of neural networks and how to start using a deep learning library called `keras`. By the end of this lab, you should: - Understand how a simple neural network works and code some of its functionality from scratch. - Be able to think and do calculations in matrix notation. Also think of vectors and arrays as tensors. - Know how to install and run `tf.keras`. - Implement a simple real world example using a neural network. ## Part 1: Neural Networks 101 Suppose we have an input vector $X=${$x_1, x_2, ... x_L$} to a $k$-layered network. <BR><BR> Each layer has its own number of nodes. For the first layer in our drawing that number is $J$. We can store the weights for each node in a vector $\mathbf{W} \in \mathbb{R}^{JxL+1}$ (accounting for bias). Similarly, we can store the biases from each node in a vector $\mathbf{b} \in \mathbb{R}^{I}$. The affine transformation is then written as $$\mathbf{a} = \mathbf{W^T}X + \mathbf{b}$$ <BR> What we then do is "absorb" $\mathbf{b}$ into $X$ by adding a column of ones to $X$. Our $X$ matrix than becomes $\mathbf{X} \in \mathbb{R}^{JxL+1}$ and our equation: <BR><BR>$$\mathbf{a} = \mathbf{W^T}_{plusones}X$$ <br>We have that $\mathbf{a} \in \mathbb{R}^{J}$ as well. Next we evaluate the output from each node. We write $$\mathbf{u} = \sigma\left(\mathbf{a}\right)$$ where $\mathbf{u}\in\mathbb{R}^{J}$. We can think of $\sigma$ operating on each individual element of $\mathbf{a}$ separately or in matrix notation. If we denote each component of $\mathbf{a}$ by $a_{j}$ then we can write $$u_{j} = \sigma\left(a_{j}\right), \quad j = 1, ... J.$$<br> In our code we will implement all these equations in matrix notation. `tf.keras` (Tensorflow) and `numpy` perform the calculations in matrix format. <br><br> Image source: *"Modern Mathematical Methods for Computational Science and Engineering"* Efthimios Kaxiras and Athanassios Fokas. Let's assume that we have 3 input points (L = 3), two hidden layers ($k=2$), and 2 nodes in each layer ($J=2$)<br> ### Input Layer $𝑋$={$𝑥_1,𝑥_2,x_3$} ### First Hidden Layer \begin{equation} \begin{aligned} a^{(1)}_1 = w^{(1)}_{10} + w^{(1)}_{11}x_1 + w^{(1)}_{12}x_2 + w^{(1)}_{13}x_3 \\ a^{(1)}_2 = w^{(1)}_{20} + w^{(1)}_{21}x_1 + w^{(1)}_{22}x_2 + w^{(1)}_{23}x_3 \\ \end{aligned} \end{equation} <br> All this in matrix notation: $$\mathbf{a} = \mathbf{W^T}X$$ <br> NOTE: in $X$ we have added a column of ones to account for the bias<BR><BR> **Then the sigmoid is applied**: \begin{equation} \begin{aligned} u^{(1)}_1 = \sigma(a^{(1)}_1) \\ u^{(1)}_2 = \sigma(a^{(1)}_2) \\ \end{aligned} \end{equation} or in matrix notation: $$\mathbf{u} = \sigma\left(\mathbf{a}\right)$$ ### Second Hidden Layer \begin{equation} \begin{aligned} a^{(2)}_1 = w^{(2)}_{10} + w^{(2)}_{11}u^{(1)}_1 + w^{(2)}_{12}u^{(1)}_2 + w^{(2)}_{13}u^{(1)}_3 \\ a^{(2)}_2 = w^{(2)}_{20} + w^{(2)}_{21}u^{(1)}_1 + w^{(2)}_{22}u^{(1)}_2 + w^{(2)}_{23}u^{(1)}_3 \\ \end{aligned} \end{equation} <br> **Then the sigmoid is applied**: \begin{equation} \begin{aligned} u^{(2)}_1 = \sigma(a^{(2)}_1) \\ u^{(2)}_2 = \sigma(a^{(2)}_2) \\ \end{aligned} \end{equation} ### Output Layer #### If the output is categorical: For example with four classes ($M=4$): $Y$={$y_1, y_2, y_3, y_4$}, we have the affine and then the sigmoid is lastly applied: \begin{equation} \begin{aligned} a^{(3)}_1 = w^{(3)}_{10} + w^{(3)}_{11}u^{(2)}_1 + w^{(3)}_{12}u^{(2)}_2 \\ a^{(3)}_2 = w^{(3)}_{20} + w^{(3)}_{21}u^{(2)}_1 + w^{(3)}_{22}u^{(2)}_2 \\ a^{(3)}_3 = w^{(3)}_{30} + w^{(3)}_{31}u^{(2)}_1 + w^{(3)}_{32}u^{(2)}_2 \\ a^{(3)}_4 = w^{(3)}_{40} + w^{(3)}_{41}u^{(2)}_1 + w^{(3)}_{42}u^{(2)}_2 \\ \end{aligned} \end{equation} <br> \begin{equation} \begin{aligned} y_1 = \sigma(a^{(3)}_1) \\ y_2 = \sigma(a^{(3)}_2) \\ y_3 = \sigma(a^{(3)}_3) \\ y_3 = \sigma(a^{(3)}_4) \\ \end{aligned} \end{equation} $\sigma$ will be softmax in the case of multiple classes and sigmoid for binary. <BR> #### If the output is a number (regression): We have a single y as output: \begin{equation} \begin{aligned} y = w^{(3)}_{10}+ w^{(3)}_{11}u^{(2)}_1 + w^{(3)}_{12}u^{(2)}_2 + w^{(3)}_{13}u^{(2)}_3 \\ \end{aligned} \end{equation} #### Matrix Multiplication and constant addition ```python a = np.array([[1, 0], [0, 1], [2, 3]]) b = np.array([[4, 1, 1], [2, 2, 1]]) print(np.matrix(a)) print('------') print(np.matrix(b)) ``` [[1 0] [0 1] [2 3]] ------ [[4 1 1] [2 2 1]] ```python # both Tensorflow and numpy take care of transposing. c = tf.matmul(a, b) # the tensorflow way print(c) d = np.dot(a, b) # the numpy way print(d) ``` tf.Tensor( [[ 4 1 1] [ 2 2 1] [14 8 5]], shape=(3, 3), dtype=int64) [[ 4 1 1] [ 2 2 1] [14 8 5]] ```python # how do we add the constant in the matrix a = [[1, 0], [0, 1]] ones = np.ones((len(a),1)) a = np.append(a, ones, axis=1) a ``` array([[1., 0., 1.], [0., 1., 1.]]) <div class="exercise"><b>1. In class exercise : Plot the sigmoid</b></div> Define the `sigmoid` and the `tanh`. For `tanh` you may use `np.tanh` and for the `sigmoid` use the general equation: \begin{align} \sigma = \dfrac{1}{1+e^{-2(x-c)/a}} \qquad\text{(1.1)} \textrm{} \end{align} Generate a list of 500 $x$ points from -5 to 5 and plot both functions. What do you observe? What do variables $c$ and $a$ do? ```python # your code here ``` ```python # %load solutions/sigmoid.py # The smaller the `a`, the sharper the function is. # Variable `c` moves the function along the x axis def sigmoid(x,c,a): z = ((x-c)/a) return 1.0 / (1.0 + np.exp(-z)) x = np.linspace(-5.0, 5.0, 500) # input points c = 1. a = 0.5 plt.plot(x, sigmoid(x, c, a), label='sigmoid') plt.plot(x, np.tanh(x), label='tanh') plt.grid(); plt.legend(); ``` <div class="exercise"><b>2. In class exercise: Approximate a Gaussian function using a node and manually adjusting the weights. Start with one layer with one node and move to two nodes.</b></div> The task is to approximate (learn) a function $f\left(x\right)$ given some input $x$. For demonstration purposes, the function we will try to learn is a Gaussian function: \begin{align} f\left(x\right) = e^{-x^{2}} \textrm{} \end{align} Even though we represent the input $x$ as a vector on the computer, you should think of it as a single input. #### 2.1 Start by plotting the above function using the $x$ dataset you created earlier ```python x = np.linspace(-5.0, 5.0, 500) # input points def gaussian(x): return np.exp(-x*x) f = gaussian(x) plt.plot(x, f, label='gaussian') plt.legend() ``` ```python f.shape ``` (500,) #### 2.2 Now, let's code the single node as per the image above. Write a function named `affine` that does the transformation. The definition is provided below. Then create a simpler sigmoid with just one variable. We choose a **sigmoid** activation function and specifically the **logistic** function. Sigmoids are a family of functions and the logistic function is just one member in that family. $$\sigma\left(z\right) = \dfrac{1}{1 + e^{-z}}.$$ <br> Define both functions in code. ```python def affine(x, w, b): """Return affine transformation of x INPUTS ====== x: A numpy array of points in x w: An array representing the weight of the perceptron b: An array representing the biases of the perceptron RETURN ====== z: A numpy array of points after the affine transformation z = wx + b """ # Code goes here return z ``` ```python # your code here ``` ```python # %load solutions/affine-sigmoid.py def affine(x, w, b): return w * x + b def sigmoid(z): return 1.0 / (1.0 + np.exp(-z)) ``` And now we plot the activation function and the true function. What do you think will happen if you change $w$ and $b$? ```python w = [-5.0, 0.1, 5.0] # Create a list of weights b = [0.0, -1.0, 1.0] # Create a list of biases fig, ax = plt.subplots(1,1, figsize=(9,5)) SIZE = 16 # plot our true function, the gaussian ax.plot(x, f, lw=4, ls='-.', label='True function') # plot 3 "networks" for wi, bi in zip(w, b): h = sigmoid(affine(x, wi, bi)) ax.plot(x, h, lw=4, label=r'$w = {0}$, $b = {1}$'.format(wi,bi)) ax.set_title('Single neuron network', fontsize=SIZE) # Create labels (very important!) ax.set_xlabel('$x$', fontsize=SIZE) # Notice we make the labels big enough to read ax.set_ylabel('$y$', fontsize=SIZE) ax.tick_params(labelsize=SIZE) # Make the tick labels big enough to read ax.legend(fontsize=SIZE, loc='best') # Create a legend and make it big enough to read ``` We didn't do an exhaustive search of the weights and biases, but it sure looks like this single perceptron is never going to match the actual function. Again, we shouldn't be suprised about this. The output layer of the network is simple the logistic function, which can only have so much flexibility. Let's try to make our network more flexible by using **more nodes**! ### Multiple Perceptrons in a Single Layer It appears that a single neuron is somewhat limited in what it can accomplish. What if we expand the number of nodes/neurons in our network? We have two obvious choices here. One option is to add depth to the network by putting layers next to each other. The other option is to stack neurons on top of each other in the same layer. Now the network has some width, but is still only one layer deep. ```python x = np.linspace(-5.0, 5.0, 500) # input points f = np.exp(-x*x) # data w = np.array([3.5, -3.5]) b = np.array([3.5, 3.5]) # Affine transformations z1 = w[0] * x + b[0] z2 = w[1] * x + b[1] # Node outputs h1 = sigmoid(z1) h2 = sigmoid(z2) ``` Now let's plot things and see what they look like. ```python fig, ax = plt.subplots(1,1, figsize=(9,5)) ax.plot(x, f, lw=4, ls = '-.', label='True function') ax.plot(x, h1, lw=4, label='First neuron') ax.plot(x, h2, lw=4, label='Second neuron') # Set title ax.set_title('Comparison of Neuron Outputs', fontsize=SIZE) # Create labels (very important!) ax.set_xlabel('$x$', fontsize=SIZE) # Notice we make the labels big enough to read ax.set_ylabel('$y$', fontsize=SIZE) ax.tick_params(labelsize=SIZE) # Make the tick labels big enough to read ax.legend(fontsize=SIZE, loc='best') # Create a legend and make it big enough to read ``` Just as we expected. Some sigmoids. Of course, to get the network prediction we must combine these two sigmoid curves somehow. First we'll just add $h_{1}$ and $h_{2}$ without any weights to see what happens. #### Note We are **not** doing classification here. We are trying to predict an actual function. The sigmoid activation is convenient when doing classification because you need to go from $0$ to $1$. However, when learning a function, we don't have as good of a reason to choose a sigmoid. ```python # Network output wout = np.ones(2) # Set the output weights to unity to begin bout = -1 # bias yout = wout[0] * h1 + wout[1] * h2 + bout ``` And plot. ```python fig, ax = plt.subplots(1,1, figsize=(9,5)) ax.plot(x, f, ls='-.', lw=4, label=r'True function') ax.plot(x, yout, lw=4, label=r'$y_{out} = h_{1} + h_{2}$') # Create labels (very important!) ax.set_xlabel('$x$', fontsize=SIZE) # Notice we make the labels big enough to read ax.set_ylabel('$y$', fontsize=SIZE) ax.tick_params(labelsize=SIZE) # Make the tick labels big enough to read ax.legend(fontsize=SIZE, loc='best') # Create a legend and make it big enough to read ``` Very cool! The two nodes interact with each other to produce a pretty complicated-looking function. It still doesn't match the true function, but now we have some hope. In fact, it's starting to look a little bit like a Gaussian! We can do better. There are three obvious options at this point: 1. Change the number of nodes 2. Change the activation functions 3. Change the weights #### We will leave this simple example for some other time! Let's move on to fashion items! ## Part 2: Tensors, Fashion, and Reese Witherspoon We can think of tensors as multidimensional arrays of real numerical values; their job is to generalize matrices to multiple dimensions. While tensors first emerged in the 20th century, they have since been applied to numerous other disciplines, including machine learning. Tensor decomposition/factorization can solve, among other, problems in unsupervised learning settings, temporal and multirelational data. For those of you that will get to handle images for Convolutional Neural Networks, it's a good idea to have the understanding of tensors of rank 3. We will use the following naming conventions: - scalar = just a number = rank 0 tensor ($a$ ∈ $F$,) <BR><BR> - vector = 1D array = rank 1 tensor ( $x = (\;x_1,...,x_i\;)⊤$ ∈ $F^n$ ) <BR><BR> - matrix = 2D array = rank 2 tensor ( $\textbf{X} = [a_{ij}] ∈ F^{m×n}$ ) <BR><BR> - 3D array = rank 3 tensor ( $\mathscr{X} =[t_{i,j,k}]∈F^{m×n×l}$ ) <BR><BR> - $\mathscr{N}$D array = rank $\mathscr{N}$ tensor ( $\mathscr{T} =[t_{i1},...,t_{i\mathscr{N}}]∈F^{n_1×...×n_\mathscr{N}}$ ) <-- Things start to get complicated here... #### Tensor indexing We can create subarrays by fixing some of the given tensor’s indices. We can create a vector by fixing all but one index. A 2D matrix is created when fixing all but two indices. For example, for a third order tensor the vectors are <br><BR> $\mathscr{X}[:,j,k]$ = $\mathscr{X}[j,k]$ (column), <br> $\mathscr{X}[i,:,k]$ = $\mathscr{X}[i,k]$ (row), and <BR> $\mathscr{X}[i,j,:]$ = $\mathscr{X}[i,j]$ (tube) <BR> #### Tensor multiplication We can multiply one matrix with another as long as the sizes are compatible ((n × m) × (m × p) = n × p), and also multiply an entire matrix by a constant. Numpy `numpy.dot` performs a matrix multiplication which is straightforward when we have 2D or 1D arrays. But what about > 3D arrays? The function will choose according to the matching dimentions but if we want to choose we should use `tensordot`, but, again, we **do not need tensordot** for this class. ### Reese Witherspoon This image is from the dataset [Labeled Faces in the Wild](http://vis-www.cs.umass.edu/lfw/person/Reese_Witherspoon.html) used for machine learning training. Images are 24-bit RGB images (height, width, channels) with 8 bits for each of R, G, B channel. Explore and print the array. ```python # load and show the image FILE = '../fig/Reese_Witherspoon.jpg' img = mpimg.imread(FILE) imgplot = plt.imshow(img) ``` ```python print(f'The image is a: {type(img)} of shape {img.shape}') img[3:5, 3:5, :] ``` The image is a: <class 'numpy.ndarray'> of shape (150, 150, 3) array([[[241, 241, 241], [242, 242, 242]], [[241, 241, 241], [242, 242, 242]]], dtype=uint8) #### Slicing tensors: slice along each axis ```python # we want to show each color channel fig, axes = plt.subplots(1, 3, figsize=(10,10)) for i, subplot in zip(range(3), axes): temp = np.zeros(img.shape, dtype='uint8') temp[:,:,i] = img[:,:,i] subplot.imshow(temp) subplot.set_axis_off() plt.show() ``` #### Multiplying Images with a scalar (just for fun, does not really help us in any way) ```python temp = img temp = temp * 2 plt.imshow(temp) ``` For more on image manipulation by `matplotlib` see: [matplotlib-images](https://matplotlib.org/3.1.1/tutorials/introductory/images.html) ### Anatomy of an Artificial Neural Network In Part 1 we hand-made a neural network by writing some simple python functions. We focused on a regression problem where we tried to learn a function. We practiced using the logistic activation function in a network with multiple nodes, but a single or two hidden layers. Some of the key observations were: * Increasing the number of nodes allows us to represent more complicated functions * The weights and biases have a very big impact on the solution * Finding the "correct" weights and biases is really hard to do manually * There must be a better method for determining the weights and biases automatically We also didn't assess the effects of different activation functions or different network depths. ### https://www.tensorflow.org/guide/keras `tf.keras` is TensorFlow's high-level API for building and training deep learning models. It's used for fast prototyping, state-of-the-art research, and production. `Keras` is a library created by François Chollet. After Google released Tensorflow 2.0, the creators of `keras` recommend that "Keras users who use multi-backend Keras with the TensorFlow backend switch to `tf.keras` in TensorFlow 2.0. `tf.keras` is better maintained and has better integration with TensorFlow features". #### IMPORTANT: In `Keras` everything starts with a Tensor of N samples as input and ends with a Tensor of N samples as output. ### The 3 parts of an ANN - **Part 1: the input layer** (our dataset) - **Part 2: the internal architecture or hidden layers** (the number of layers, the activation functions, the learnable parameters and other hyperparameters) - **Part 3: the output layer** (what we want from the network) In the rest of the lab we will practice with end-to-end neural network training 1. Load the data 2. Define the layers of the model. 3. Compile the model. 4. Fit the model to the train set (also using a validation set). 5. Evaluate the model on the test set. 6. Plot metrics such as accuracy. 7. Predict on random images from test set. 8. Predict on a random image from the web! ```python seed = 7 np.random.seed(seed) ``` ### Fashion MNIST MNIST, the set of handwritten digits is considered the Drosophila of Machine Learning. It has been overused, though, so we will try a slight modification to it. **Fashion-MNIST** is a dataset of clothing article images (created by [Zalando](https://github.com/zalandoresearch/fashion-mnist)), consisting of a training set of 60,000 examples and a test set of 10,000 examples. Each example is a **28 x 28** grayscale image, associated with a label from **10 classes**. The creators intend Fashion-MNIST to serve as a direct drop-in replacement for the original MNIST dataset for benchmarking machine learning algorithms. It shares the same image size and structure of training and testing splits. Each pixel is 8 bits so its value ranges from 0 to 255. Let's load and look at it! #### 1. Load the data ```python x_train.shape ``` (60000, 28, 28) ```python %%time # get the data from keras fashion_mnist = tf.keras.datasets.fashion_mnist # load the data splitted in train and test! how nice! (x_train, y_train),(x_test, y_test) = fashion_mnist.load_data() # normalize the data by dividing with pixel intensity # (each pixel is 8 bits so its value ranges from 0 to 255) x_train, x_test = x_train / 255.0, x_test / 255.0 # classes are named 0-9 so define names for plotting clarity class_names = ['T-shirt/top', 'Trouser', 'Pullover', 'Dress', 'Coat', 'Sandal', 'Shirt', 'Sneaker', 'Bag', 'Ankle boot'] # plot plt.figure(figsize=(10,10)) for i in range(25): plt.subplot(5,5,i+1) plt.xticks([]) plt.yticks([]) plt.grid(False) plt.imshow(x_train[i], cmap=plt.cm.binary) plt.xlabel(class_names[y_train[i]]) plt.show() ``` ```python plt.imshow(x_train[3], cmap=plt.cm.binary) ``` ```python x_train.shape, x_test.shape ``` ((60000, 28, 28), (10000, 28, 28)) ```python y_train.shape ``` (60000,) #### 2. Define the layers of the model. ```python # type together model = tf.keras.models.Sequential([ tf.keras.layers.Flatten(input_shape=(28, 28)), tf.keras.layers.Dense(154, activation='relu'), tf.keras.layers.Dense(64, activation='relu'), #tf.keras.layers.Dropout(0.2), tf.keras.layers.Dense(10, activation='softmax') ]) ``` #### 3. Compile the model ```python loss_fn = tf.keras.losses.SparseCategoricalCrossentropy() optimizer = tf.keras.optimizers.Adam() model.compile(optimizer=optimizer, loss=loss_fn, metrics=['accuracy']) ``` ```python model.summary() ``` Model: "sequential" _________________________________________________________________ Layer (type) Output Shape Param # ================================================================= flatten (Flatten) (None, 784) 0 _________________________________________________________________ dense (Dense) (None, 154) 120890 _________________________________________________________________ dense_1 (Dense) (None, 64) 9920 _________________________________________________________________ dense_2 (Dense) (None, 10) 650 ================================================================= Total params: 131,460 Trainable params: 131,460 Non-trainable params: 0 _________________________________________________________________ ```python tf.keras.utils.plot_model( model, #to_file='model.png', # if you want to save the image show_shapes=True, # True for more details than you need show_layer_names=True, rankdir='TB', expand_nested=False, dpi=96 ) ``` [Everything you wanted to know about a Keras Model and were afraid to ask](https://www.tensorflow.org/api_docs/python/tf/keras/Model) #### 4. Fit the model to the train set (also using a validation set) This is the part that takes the longest. ----------------------------------------------------------- **ep·och** <BR> noun: epoch; plural noun: epochs. A period of time in history or a person's life, typically one marked by notable events or particular characteristics. Examples: "the Victorian epoch", "my Neural Netwok's epochs". <BR> ----------------------------------------------------------- ```python %%time # the core of the network training history = model.fit(x_train, y_train, validation_split=0.33, epochs=50, verbose=2) ``` Train on 40199 samples, validate on 19801 samples Epoch 1/50 40199/40199 - 3s - loss: 0.5265 - accuracy: 0.8121 - val_loss: 0.4014 - val_accuracy: 0.8543 Epoch 2/50 40199/40199 - 3s - loss: 0.3875 - accuracy: 0.8587 - val_loss: 0.3988 - val_accuracy: 0.8521 Epoch 3/50 40199/40199 - 3s - loss: 0.3502 - accuracy: 0.8711 - val_loss: 0.3711 - val_accuracy: 0.8645 Epoch 4/50 40199/40199 - 3s - loss: 0.3199 - accuracy: 0.8818 - val_loss: 0.3884 - val_accuracy: 0.8573 Epoch 5/50 40199/40199 - 3s - loss: 0.3066 - accuracy: 0.8870 - val_loss: 0.3463 - val_accuracy: 0.8763 Epoch 6/50 40199/40199 - 3s - loss: 0.2861 - accuracy: 0.8931 - val_loss: 0.3719 - val_accuracy: 0.8655 Epoch 7/50 40199/40199 - 3s - loss: 0.2775 - accuracy: 0.8964 - val_loss: 0.3487 - val_accuracy: 0.8752 Epoch 8/50 40199/40199 - 3s - loss: 0.2641 - accuracy: 0.9008 - val_loss: 0.3437 - val_accuracy: 0.8779 Epoch 9/50 40199/40199 - 3s - loss: 0.2528 - accuracy: 0.9043 - val_loss: 0.3454 - val_accuracy: 0.8809 Epoch 10/50 40199/40199 - 3s - loss: 0.2430 - accuracy: 0.9093 - val_loss: 0.3273 - val_accuracy: 0.8857 Epoch 11/50 40199/40199 - 3s - loss: 0.2335 - accuracy: 0.9127 - val_loss: 0.3207 - val_accuracy: 0.8866 Epoch 12/50 40199/40199 - 3s - loss: 0.2259 - accuracy: 0.9141 - val_loss: 0.3352 - val_accuracy: 0.8876 Epoch 13/50 40199/40199 - 3s - loss: 0.2189 - accuracy: 0.9175 - val_loss: 0.3281 - val_accuracy: 0.8915 Epoch 14/50 40199/40199 - 3s - loss: 0.2128 - accuracy: 0.9184 - val_loss: 0.3394 - val_accuracy: 0.8850 Epoch 15/50 40199/40199 - 3s - loss: 0.2038 - accuracy: 0.9215 - val_loss: 0.3257 - val_accuracy: 0.8934 Epoch 16/50 40199/40199 - 3s - loss: 0.1982 - accuracy: 0.9249 - val_loss: 0.3501 - val_accuracy: 0.8895 Epoch 17/50 40199/40199 - 3s - loss: 0.1912 - accuracy: 0.9277 - val_loss: 0.3758 - val_accuracy: 0.8849 Epoch 18/50 40199/40199 - 3s - loss: 0.1828 - accuracy: 0.9295 - val_loss: 0.3625 - val_accuracy: 0.8889 Epoch 19/50 40199/40199 - 3s - loss: 0.1817 - accuracy: 0.9315 - val_loss: 0.3517 - val_accuracy: 0.8911 Epoch 20/50 40199/40199 - 3s - loss: 0.1747 - accuracy: 0.9332 - val_loss: 0.3463 - val_accuracy: 0.8929 Epoch 21/50 40199/40199 - 3s - loss: 0.1707 - accuracy: 0.9349 - val_loss: 0.3691 - val_accuracy: 0.8892 Epoch 22/50 40199/40199 - 3s - loss: 0.1651 - accuracy: 0.9372 - val_loss: 0.3567 - val_accuracy: 0.8932 Epoch 23/50 40199/40199 - 3s - loss: 0.1611 - accuracy: 0.9374 - val_loss: 0.3737 - val_accuracy: 0.8914 Epoch 24/50 40199/40199 - 3s - loss: 0.1582 - accuracy: 0.9397 - val_loss: 0.3833 - val_accuracy: 0.8906 Epoch 25/50 40199/40199 - 3s - loss: 0.1518 - accuracy: 0.9416 - val_loss: 0.3730 - val_accuracy: 0.8921 Epoch 26/50 40199/40199 - 3s - loss: 0.1485 - accuracy: 0.9429 - val_loss: 0.4085 - val_accuracy: 0.8897 Epoch 27/50 40199/40199 - 3s - loss: 0.1415 - accuracy: 0.9462 - val_loss: 0.3862 - val_accuracy: 0.8909 Epoch 28/50 40199/40199 - 3s - loss: 0.1433 - accuracy: 0.9451 - val_loss: 0.4204 - val_accuracy: 0.8907 Epoch 29/50 40199/40199 - 3s - loss: 0.1374 - accuracy: 0.9483 - val_loss: 0.3937 - val_accuracy: 0.8920 Epoch 30/50 40199/40199 - 3s - loss: 0.1330 - accuracy: 0.9498 - val_loss: 0.4073 - val_accuracy: 0.8887 Epoch 31/50 40199/40199 - 3s - loss: 0.1339 - accuracy: 0.9493 - val_loss: 0.4131 - val_accuracy: 0.8891 Epoch 32/50 40199/40199 - 3s - loss: 0.1288 - accuracy: 0.9514 - val_loss: 0.4089 - val_accuracy: 0.8919 Epoch 33/50 40199/40199 - 3s - loss: 0.1239 - accuracy: 0.9526 - val_loss: 0.4195 - val_accuracy: 0.8904 Epoch 34/50 40199/40199 - 3s - loss: 0.1228 - accuracy: 0.9529 - val_loss: 0.4848 - val_accuracy: 0.8873 Epoch 35/50 40199/40199 - 3s - loss: 0.1183 - accuracy: 0.9547 - val_loss: 0.4405 - val_accuracy: 0.8924 Epoch 36/50 40199/40199 - 3s - loss: 0.1176 - accuracy: 0.9555 - val_loss: 0.4518 - val_accuracy: 0.8895 Epoch 37/50 40199/40199 - 3s - loss: 0.1174 - accuracy: 0.9552 - val_loss: 0.4882 - val_accuracy: 0.8883 Epoch 38/50 40199/40199 - 3s - loss: 0.1101 - accuracy: 0.9581 - val_loss: 0.4646 - val_accuracy: 0.8947 Epoch 39/50 40199/40199 - 3s - loss: 0.1050 - accuracy: 0.9599 - val_loss: 0.4875 - val_accuracy: 0.8919 Epoch 40/50 40199/40199 - 3s - loss: 0.1089 - accuracy: 0.9599 - val_loss: 0.4824 - val_accuracy: 0.8900 Epoch 41/50 40199/40199 - 3s - loss: 0.1047 - accuracy: 0.9583 - val_loss: 0.5133 - val_accuracy: 0.8872 Epoch 42/50 40199/40199 - 3s - loss: 0.1056 - accuracy: 0.9598 - val_loss: 0.4922 - val_accuracy: 0.8908 Epoch 43/50 40199/40199 - 3s - loss: 0.0965 - accuracy: 0.9635 - val_loss: 0.5100 - val_accuracy: 0.8891 Epoch 44/50 40199/40199 - 3s - loss: 0.1012 - accuracy: 0.9617 - val_loss: 0.5292 - val_accuracy: 0.8891 Epoch 45/50 40199/40199 - 3s - loss: 0.0968 - accuracy: 0.9631 - val_loss: 0.5449 - val_accuracy: 0.8865 Epoch 46/50 40199/40199 - 3s - loss: 0.0926 - accuracy: 0.9642 - val_loss: 0.5455 - val_accuracy: 0.8904 Epoch 47/50 40199/40199 - 3s - loss: 0.0945 - accuracy: 0.9634 - val_loss: 0.5249 - val_accuracy: 0.8909 Epoch 48/50 40199/40199 - 3s - loss: 0.0915 - accuracy: 0.9649 - val_loss: 0.5712 - val_accuracy: 0.8861 Epoch 49/50 40199/40199 - 3s - loss: 0.0887 - accuracy: 0.9670 - val_loss: 0.5586 - val_accuracy: 0.8849 Epoch 50/50 40199/40199 - 3s - loss: 0.0873 - accuracy: 0.9670 - val_loss: 0.5836 - val_accuracy: 0.8896 CPU times: user 3min 50s, sys: 26.9 s, total: 4min 17s Wall time: 2min 21s #### Save the model You can save the model so you do not have `.fit` everytime you reset the kernel in the notebook. Network training is expensive! For more details on this see [https://www.tensorflow.org/guide/keras/save_and_serialize](https://www.tensorflow.org/guide/keras/save_and_serialize) ```python # save the model so you do not have to run the code everytime model.save('fashion_model.h5') # Recreate the exact same model purely from the file #model = tf.keras.models.load_model('fashion_model.h5') ``` #### 5. Evaluate the model on the test set. ```python test_loss, test_accuracy = model.evaluate(x_test, y_test, verbose=0) print(f'Test accuracy={test_accuracy}') ``` Test accuracy=0.8804000020027161 #### 6. We learn a lot by studying History! Plot metrics such as accuracy. You can learn a lot about neural networks by observing how they perform while training. You can issue `kallbacks` in `keras`. The networks's performance is stored in a `keras` callback aptly named `history` which can be plotted. ```python print(history.history.keys()) ``` dict_keys(['loss', 'accuracy', 'val_loss', 'val_accuracy']) ```python # plot accuracy and loss for the test set fig, ax = plt.subplots(1,2, figsize=(20,6)) ax[0].plot(history.history['accuracy']) ax[0].plot(history.history['val_accuracy']) ax[0].set_title('Model accuracy') ax[0].set_ylabel('accuracy') ax[0].set_xlabel('epoch') ax[0].legend(['train', 'val'], loc='best') ax[1].plot(history.history['loss']) ax[1].plot(history.history['val_loss']) ax[1].set_title('Model loss') ax[1].set_ylabel('loss') ax[1].set_xlabel('epoch') ax[1].legend(['train', 'val'], loc='best') ``` #### 7. Now let's use the Network for what it was meant to do: Predict! ```python predictions = model.predict(x_test) ``` ```python predictions[0] ``` array([2.3347583e-26, 5.1650537e-17, 7.8453709e-22, 4.0654924e-26, 6.6475328e-18, 1.3495652e-14, 1.9801722e-15, 1.6511808e-06, 1.7612138e-24, 9.9999833e-01], dtype=float32) ```python np.argmax(predictions[0]), class_names[np.argmax(predictions[0])] ``` (9, 'Ankle boot') Let's see if our network predicted right! Is the first item what was predicted? ```python plt.figure() plt.imshow(x_test[0], cmap=plt.cm.binary) plt.xlabel(class_names[y_test[0]]) plt.colorbar() ``` **Correct!!** Now let's see how confident our model is by plotting the probability values: ```python # code source: https://www.tensorflow.org/tutorials/keras/classification def plot_image(i, predictions_array, true_label, img): predictions_array, true_label, img = predictions_array, true_label[i], img[i] plt.grid(False) plt.xticks([]) plt.yticks([]) plt.imshow(img, cmap=plt.cm.binary) predicted_label = np.argmax(predictions_array) if predicted_label == true_label: color = 'blue' else: color = 'red' plt.xlabel("{} {:2.0f}% ({})".format(class_names[predicted_label], 100*np.max(predictions_array), class_names[true_label]), color=color) def plot_value_array(i, predictions_array, true_label): predictions_array, true_label = predictions_array, true_label[i] plt.grid(False) plt.xticks(range(10)) plt.yticks([]) thisplot = plt.bar(range(10), predictions_array, color="#777777") plt.ylim([0, 1]) predicted_label = np.argmax(predictions_array) thisplot[predicted_label].set_color('red') thisplot[true_label].set_color('blue') ``` ```python i = 406 plt.figure(figsize=(6,3)) plt.subplot(1,2,1) plot_image(i, predictions[i], y_test, x_test) plt.subplot(1,2,2) plot_value_array(i, predictions[i], y_test) plt.show() ``` #### 8. Predicting in the real world Let's see if our network can generalize beyond the MNIST fashion dataset. Let's give it an random googled image of a boot. Does it have to be a clothing item resembling the MNIST fashion dataset? Can it be a puppy? Download an image from the internet and resize it to 28x28. `tf.keras` models are optimized to make predictions on a batch, or collection, of examples at once. Accordingly, even though you're using a single image, you need to add it to a list: ```python ```

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Jupyter Notebook

content/labs/lab11/notes/lab11_MLP_solutions_part1.ipynb

chksi/2019-CS109A

4b925115f8a0ad5a4f5b95d3d616fabf60bfc3c0

content/labs/lab11/notes/lab11_MLP_solutions_part1.ipynb

chksi/2019-CS109A

4b925115f8a0ad5a4f5b95d3d616fabf60bfc3c0

content/labs/lab11/notes/lab11_MLP_solutions_part1.ipynb

chksi/2019-CS109A

4b925115f8a0ad5a4f5b95d3d616fabf60bfc3c0

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# `sympy` `scipy` 계열은 [`sympy`](https://www.sympy.org)라는 *기호 처리기*도 포함하고 있다.<br> `scipy` stack also includes [`sympy`](https://www.sympy.org), a *symbolic processor*. 2006년 이후 2019 까지 800명이 넘는 개발자가 작성한 코드를 제공하였다.<br> Since 2006, more than 800 developers contributed so far in 2019. ## 기호 연산 예<br>Examples of symbolic processing `sympy` 모듈을 `sy` 라는 이름으로 불러온다.<br>Import `sympy` module in the name of `sy`. ```python import sympy as sy sy.init_printing() ``` 비교를 위해 `numpy` 모듈도 불러온다.<br> Import `numpy` module to compare. ```python import numpy as np ``` ### 제곱근<br>Square root 10의 제곱근을 구해보자.<br>Let't find the square root of ten. ```python np.sqrt(10) ``` ```python sy.sqrt(10) ``` 10의 제곱근을 제곱해보자.<br>Let't square the square root of ten. ```python np.sqrt(10) ** 2 ``` ```python sy.sqrt(10) ** 2 ``` 위 결과의 차이에 대해 어떻게 생각하는가?<br> What do you think about the differences of the results above? ### 분수<br>Fractions 10 / 3 을 생각해보자.<br>Let't think about 10/3. ```python ten_over_three = 10 / 3 ``` ```python ten_over_three ``` ```python ten_over_three * 3 ``` ```python import fractions ``` ```python fr_ten_over_three = fractions.Fraction(10 ,3) ``` ```python fr_ten_over_three ``` ```python fr_ten_over_three * 3 ``` ```python sy_ten_over_three = sy.Rational(10, 3) ``` ```python sy_ten_over_three ``` ```python sy_ten_over_three * 3 ``` 위 결과의 차이에 대해 어떻게 생각하는가?<br> What do you think about the differences of the results above? ### 변수를 포함하는 수식<br>Expressions with variables 사용할 변수를 정의한다.<br>Define variables to use. ```python a, b, c, x = sy.symbols('a b c x') theta, phi = sy.symbols('theta phi') ``` 변수들을 한번 살펴보자.<br>Let's take a look at the variables ```python a, b, c, x ``` ```python theta, phi ``` 변수를 조합하여 새로운 수식을 만들어 보자.<br> Let's make equations using variables. ```python y = a * x + b ``` ```python y ``` ```python z = a * x * x + b * x + c ``` ```python z ``` ```python w = a * sy.sin(theta) ** 2 + b ``` ```python w ``` ```python p = (x - a) * (x - b) * (x - c) ``` ```python p ``` ```python sy.expand(p, x) ``` ```python sy.collect(_, x) ``` ### 미적분<br>Calculus ```python z.diff(x) ``` ```python sy.integrate(z, x) ``` ```python w.diff(theta) ``` ```python sy.integrate(w, theta) ``` ### 근<br>Root ```python z_sol_list = sy.solve(z, x) ``` ```python z_sol_list ``` ```python sy.solve(2* sy.sin(theta) ** 2 - 1, theta) ``` ### 코드 생성<br>Code generation ```python print(sy.python(z_sol_list[0])) ``` ```python import sympy.utilities.codegen as sc ``` ```python [(c_name, c_code), (h_name, c_header)] = sc.codegen( ("z_sol", z_sol_list[0]), "C89", "test" ) ``` ```python c_name ``` ```python print(c_code) ``` ```python h_name ``` ```python print(c_header) ``` ### 연립방정식<br>System of equations ```python a1, a2, a3 = sy.symbols('a1:4') b1, b2, b3 = sy.symbols('b1:4') c1, c2 = sy.symbols('c1:3') x1, x2 = sy.symbols('x1:3') ``` ```python eq1 = sy.Eq( a1 * x1 + a2 * x2, c1, ) ``` ```python eq1 ``` ```python eq2 = sy.Eq( b1 * x1 + b2 * x2, c2, ) ``` ```python eq_list = [eq1, eq2] ``` ```python eq_list ``` ```python sy.solve(eq_list, (x1, x2)) ``` ## 참고문헌<br>References * SymPy Development Team, SymPy 1.4 documentation, sympy.org, 2019 04 10. [Online] Available : https://docs/sympy.org/latest/index.html. * SymPy Development Team, SymPy Tutorial, SymPy 1.4 documentation, sympy.org, 2019 04 10. [Online] Available : https://docs/sympy.org/latest/tutorial/index.html. * d84_n1nj4, "How to keep fractions in your equation output", Stackoverflow.com, 2017 08 12. [Online] Available : https://stackoverflow.com/a/45651175. * Python developers, "Fractions", Python documentation, 2019 10 12. [Online] Available : https://docs.python.org/3.7/library/fractions.html. * SymPy Development Team, codegen, SymPy 1.4 documentation, sympy.org, 2019 04 10. [Online] Available : https://docs/sympy.org/latest/modules/utilities/codegen.html. ## Final Bell<br>마지막 종 ```python # stackoverfow.com/a/24634221 import os os.system("printf '\a'"); ``` ```python ```

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Jupyter Notebook

70_sympy/10_sympy.ipynb

kangwonlee/2009eca-nmisp-template

46a09c988c5e0c4efd493afa965d4a17d32985e8

70_sympy/10_sympy.ipynb

kangwonlee/2009eca-nmisp-template

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70_sympy/10_sympy.ipynb

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Qwen/Qwen-72B

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__label__kor_Hang

# PaBiRoboy dynamic equations First, import the necessary functions from SymPy that will allow us to construct time varying vectors in the reference frames. ```python from __future__ import print_function, division from sympy import symbols, simplify, Matrix from sympy import trigsimp from sympy.physics.mechanics import dynamicsymbols, ReferenceFrame, Point, Particle, inertia, RigidBody, KanesMethod from numpy import deg2rad, rad2deg, array, zeros, linspace from sympy.physics.vector import init_vprinting, vlatex import numpy as np from scipy.integrate import odeint from sympy.utilities.codegen import codegen from pydy.codegen.ode_function_generators import generate_ode_function from matplotlib.pyplot import plot, legend, xlabel, ylabel, rcParams rcParams['figure.figsize'] = (14.0, 6.0) ``` SymPy has a rich printing system. Here we initialize printing so that all of the mathematical equations are rendered in standard mathematical notation. ```python from sympy.physics.vector import init_vprinting init_vprinting(use_latex='mathjax', pretty_print=False) ``` ## Interial frames ```python inertial_frame = ReferenceFrame('I') lower_leg_left_frame = ReferenceFrame('R_1') upper_leg_left_frame = ReferenceFrame('R_2') hip_frame = ReferenceFrame('R_3') upper_leg_right_frame = ReferenceFrame('R_4') lower_leg_right_frame = ReferenceFrame('R_5') ``` ## Angles ```python theta0, theta1, theta2, theta3, phi = dynamicsymbols('theta0, theta1, theta2, theta3, phi') ``` ```python lower_leg_left_frame.orient(inertial_frame, 'Axis', (phi, inertial_frame.z)) simplify(lower_leg_left_frame.dcm(inertial_frame)) ``` $$\left[\begin{matrix}\operatorname{cos}\left(\phi\right) & \operatorname{sin}\left(\phi\right) & 0\\- \operatorname{sin}\left(\phi\right) & \operatorname{cos}\left(\phi\right) & 0\\0 & 0 & 1\end{matrix}\right]$$ ```python upper_leg_left_frame.orient(lower_leg_left_frame, 'Axis', (theta0, -lower_leg_left_frame.z)) simplify(upper_leg_left_frame.dcm(inertial_frame)) ``` $$\left[\begin{matrix}\operatorname{cos}\left(\phi - \theta_{0}\right) & \operatorname{sin}\left(\phi - \theta_{0}\right) & 0\\- \operatorname{sin}\left(\phi - \theta_{0}\right) & \operatorname{cos}\left(\phi - \theta_{0}\right) & 0\\0 & 0 & 1\end{matrix}\right]$$ ```python hip_frame.orient(upper_leg_left_frame, 'Axis', (theta1, -upper_leg_left_frame.z)) hip_frame.dcm(inertial_frame) ``` $$\left[\begin{matrix}\left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{sin}\left(\theta_{1}\right) + \operatorname{cos}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right)\right) \operatorname{cos}\left(\phi\right) - \left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right) - \operatorname{sin}\left(\theta_{1}\right) \operatorname{cos}\left(\theta_{0}\right)\right) \operatorname{sin}\left(\phi\right) & \left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{sin}\left(\theta_{1}\right) + \operatorname{cos}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right)\right) \operatorname{sin}\left(\phi\right) + \left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right) - \operatorname{sin}\left(\theta_{1}\right) \operatorname{cos}\left(\theta_{0}\right)\right) \operatorname{cos}\left(\phi\right) & 0\\- \left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{sin}\left(\theta_{1}\right) + \operatorname{cos}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right)\right) \operatorname{sin}\left(\phi\right) + \left(\operatorname{sin}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right) + \operatorname{sin}\left(\theta_{1}\right) \operatorname{cos}\left(\theta_{0}\right)\right) \operatorname{cos}\left(\phi\right) & \left(- \operatorname{sin}\left(\theta_{0}\right) \operatorname{sin}\left(\theta_{1}\right) + \operatorname{cos}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right)\right) \operatorname{cos}\left(\phi\right) + \left(\operatorname{sin}\left(\theta_{0}\right) \operatorname{cos}\left(\theta_{1}\right) + \operatorname{sin}\left(\theta_{1}\right) \operatorname{cos}\left(\theta_{0}\right)\right) \operatorname{sin}\left(\phi\right) & 0\\0 & 0 & 1\end{matrix}\right]$$ ```python upper_leg_right_frame.orient(hip_frame, 'Axis', (theta2, -hip_frame.z)) simplify(upper_leg_right_frame.dcm(inertial_frame)) ``` $$\left[\begin{matrix}\operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & - \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & 0\\\operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & 0\\0 & 0 & 1\end{matrix}\right]$$ ```python lower_leg_right_frame.orient(upper_leg_right_frame, 'Axis', (theta3, -upper_leg_right_frame.z)) simplify(lower_leg_right_frame.dcm(inertial_frame)) ``` $$\left[\begin{matrix}\operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & - \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & 0\\\operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & 0\\0 & 0 & 1\end{matrix}\right]$$ ## Points and Locations ```python origin = Point('Origin') ankle_left = Point('AnkleLeft') knee_left = Point('KneeLeft') hip_left = Point('HipLeft') hip_center = Point('HipCenter') hip_right = Point('HipRight') knee_right = Point('KneeRight') ankle_right = Point('AnkleRight') ``` ```python lower_leg_length, upper_leg_length, hip_length = symbols('l1, l2, l3') ``` ```python origin = Point('Origin') ankle_left = Point('AnkleLeft') knee_left = Point('KneeLeft') hip_left = Point('HipLeft') hip_center = Point('HipCenter') hip_right = Point('HipRight') knee_right = Point('KneeRight') ankle_right = Point('AnkleRight') # here go the lengths of your robots links lower_leg_length, upper_leg_length, hip_length = symbols('l1, l2, l3') ankle_left.set_pos(origin, (0 * inertial_frame.y)+(0 * inertial_frame.x)) #ankle_left.pos_from(origin).express(inertial_frame).simplify() knee_left.set_pos(ankle_left, lower_leg_length * lower_leg_left_frame.y) #knee_left.pos_from(origin).express(inertial_frame).simplify() hip_left.set_pos(knee_left, upper_leg_length * upper_leg_left_frame.y) #hip_left.pos_from(origin).express(inertial_frame).simplify() hip_center.set_pos(hip_left, hip_length/2 * hip_frame.x) #hip_center.pos_from(origin).express(inertial_frame).simplify() hip_right.set_pos(hip_center, hip_length/2 * hip_frame.x) #hip_right.pos_from(origin).express(inertial_frame).simplify() knee_right.set_pos(hip_right, upper_leg_length * -upper_leg_right_frame.y) #knee_right.pos_from(origin).express(inertial_frame).simplify() ankle_right.set_pos(knee_right, lower_leg_length * -lower_leg_right_frame.y) #ankle_right.pos_from(origin).express(inertial_frame).simplify() ``` ```python lower_leg_left_com_length = lower_leg_length/2 upper_leg_left_com_length = upper_leg_length/2 hip_com_length = hip_length/2 upper_leg_right_com_length = upper_leg_length/2 lower_leg_right_com_length = lower_leg_length/2 lower_leg_left_mass_center = Point('L_COMleft') upper_leg_left_mass_center = Point('U_COMleft') hip_mass_center = Point('H_COMleft') upper_leg_right_mass_center = Point('U_COMright') lower_leg_right_mass_center = Point('L_COMright') ``` ```python lower_leg_left_mass_center.set_pos(ankle_left, lower_leg_left_com_length * lower_leg_left_frame.y) #lower_leg_left_mass_center.pos_from(origin).express(inertial_frame).simplify() upper_leg_left_mass_center.set_pos(knee_left, upper_leg_left_com_length * upper_leg_left_frame.y) #upper_leg_left_mass_center.pos_from(origin).express(inertial_frame).simplify() hip_mass_center.set_pos(hip_center, 0 * hip_frame.x) #hip_mass_center.pos_from(origin).express(inertial_frame).simplify() upper_leg_right_mass_center.set_pos(knee_right, upper_leg_right_com_length * upper_leg_right_frame.y) #upper_leg_right_mass_center.pos_from(origin).express(inertial_frame).simplify() lower_leg_right_mass_center.set_pos(ankle_right, lower_leg_right_com_length * lower_leg_right_frame.y) lower_leg_right_mass_center.pos_from(origin).express(inertial_frame).simplify() ``` $$(- \frac{l_{1}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right))\mathbf{\hat{i}_x} + (- \frac{l_{1}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} \operatorname{cos}\left(\phi\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right))\mathbf{\hat{i}_y}$$ ## Kinematical Differential Equations ```python omega0, omega1, omega2, omega3, psi = dynamicsymbols('omega0, omega1, omega2, omega3, psi') kinematical_differential_equations = [omega0 - theta0.diff(), omega1 - theta1.diff(), omega2 - theta2.diff(), omega3 - theta3.diff(), psi - phi.diff(), ] kinematical_differential_equations ``` $$\left [ \omega_{0} - \dot{\theta}_{0}, \quad \omega_{1} - \dot{\theta}_{1}, \quad \omega_{2} - \dot{\theta}_{2}, \quad \omega_{3} - \dot{\theta}_{3}, \quad \psi - \dot{\phi}\right ]$$ ## Angular Velocities ```python lower_leg_left_frame.set_ang_vel(inertial_frame, 0 * inertial_frame.z) #lower_leg_left_frame.ang_vel_in(inertial_frame) upper_leg_left_frame.set_ang_vel(upper_leg_left_frame, omega0 * inertial_frame.z) #upper_leg_left_frame.ang_vel_in(inertial_frame) hip_frame.set_ang_vel(hip_frame, omega1 * inertial_frame.z) #hip_frame.ang_vel_in(inertial_frame) upper_leg_right_frame.set_ang_vel(upper_leg_right_frame, omega2 * inertial_frame.z) #upper_leg_right_frame.ang_vel_in(inertial_frame) lower_leg_right_frame.set_ang_vel(lower_leg_right_frame, omega3 * inertial_frame.z) lower_leg_right_frame.ang_vel_in(inertial_frame) ``` $$- \dot{\theta}_{3}\mathbf{\hat{r_4}_z} - \dot{\theta}_{2}\mathbf{\hat{r_3}_z} - \dot{\theta}_{1}\mathbf{\hat{r_2}_z} - \dot{\theta}_{0}\mathbf{\hat{r_1}_z}$$ # Linear Velocities Finally, the linear velocities of the mass centers are needed. Starting at the ankle_left which has a velocity of 0. ```python ankle_left.set_vel(inertial_frame, 0) origin.set_vel(inertial_frame, 0) ``` Working our way up the chain we can make use of the fact that the joint points are located on two rigid bodies. Any fixed point in a reference frame can be computed if the linear velocity of another point on that frame is known and the frame's angular velocity is known. $$^I\mathbf{v}^{P_2} = ^I\mathbf{v}^{P_1} + ^I\omega^A \times \mathbf{r}^{\frac{P_2}{P_1}}$$ The `Point.v2pt_theory()` method makes it easy to do this calculation. ```python knee_left.v2pt_theory(ankle_left, inertial_frame, lower_leg_left_frame) #knee_left.vel(inertial_frame) hip_left.v2pt_theory(knee_left, inertial_frame, upper_leg_left_frame) #hip_left.vel(inertial_frame) hip_center.v2pt_theory(hip_left, inertial_frame, hip_frame) #hip_center.vel(inertial_frame) hip_mass_center.v2pt_theory(hip_center, inertial_frame, hip_frame) #hip_mass_center.vel(inertial_frame) hip_right.v2pt_theory(hip_center, inertial_frame, hip_frame) #hip_right.vel(inertial_frame) knee_right.v2pt_theory(hip_right, inertial_frame, upper_leg_right_frame) #knee_right.vel(inertial_frame) ankle_right.v2pt_theory(knee_right, inertial_frame, lower_leg_right_frame) #ankle_right.vel(inertial_frame) lower_leg_left_mass_center.v2pt_theory(ankle_left, inertial_frame, lower_leg_left_frame) #lower_leg_left_mass_center.vel(inertial_frame) lower_leg_right_mass_center.v2pt_theory(ankle_right, inertial_frame, lower_leg_right_frame) #lower_leg_right_mass_center.vel(inertial_frame) upper_leg_left_mass_center.v2pt_theory(knee_left, inertial_frame, upper_leg_left_frame) #upper_leg_left_mass_center.vel(inertial_frame) upper_leg_right_mass_center.v2pt_theory(knee_right, inertial_frame, upper_leg_right_frame) upper_leg_right_mass_center.vel(inertial_frame) ``` $$l_{2} \dot{\theta}_{0}\mathbf{\hat{r_2}_x} + l_{3} \left(- \dot{\theta}_{0} - \dot{\theta}_{1}\right)\mathbf{\hat{r_3}_y} + \frac{l_{2}}{2} \left(- \dot{\theta}_{0} - \dot{\theta}_{1} - \dot{\theta}_{2}\right)\mathbf{\hat{r_4}_x}$$ # Masses, Inertia, Rigid Bodies ```python lower_leg_mass, upper_leg_mass, hip_mass = symbols('m_L, m_U, m_H') lower_leg_inertia, upper_leg_inertia, hip_inertia = symbols('I_Lz, I_Uz, I_Hz') lower_leg_left_inertia_dyadic = inertia(lower_leg_left_frame, lower_leg_inertia, lower_leg_inertia, lower_leg_inertia) lower_leg_left_central_inertia = (lower_leg_left_inertia_dyadic, lower_leg_left_mass_center) lower_leg_left_inertia_dyadic.to_matrix(lower_leg_left_frame) upper_leg_left_inertia_dyadic = inertia(upper_leg_left_frame, upper_leg_inertia, upper_leg_inertia, upper_leg_inertia) upper_leg_left_central_inertia = (upper_leg_left_inertia_dyadic, upper_leg_left_mass_center) upper_leg_left_inertia_dyadic.to_matrix(upper_leg_left_frame) hip_inertia_dyadic = inertia(hip_frame, hip_inertia, hip_inertia, hip_inertia) hip_central_inertia = (hip_inertia_dyadic, hip_mass_center) hip_inertia_dyadic.to_matrix(hip_frame) upper_leg_right_inertia_dyadic = inertia(upper_leg_right_frame, upper_leg_inertia, upper_leg_inertia, upper_leg_inertia) upper_leg_right_central_inertia = (upper_leg_right_inertia_dyadic, upper_leg_right_mass_center) upper_leg_right_inertia_dyadic.to_matrix(upper_leg_right_frame) lower_leg_right_inertia_dyadic = inertia(lower_leg_right_frame, lower_leg_inertia, lower_leg_inertia, lower_leg_inertia) lower_leg_right_central_inertia = (lower_leg_right_inertia_dyadic, lower_leg_right_mass_center) lower_leg_right_inertia_dyadic.to_matrix(lower_leg_right_frame) lower_leg_left = RigidBody('Lower Leg Left', lower_leg_left_mass_center, lower_leg_left_frame, lower_leg_mass, lower_leg_left_central_inertia) upper_leg_left = RigidBody('Upper Leg Left', upper_leg_left_mass_center, upper_leg_left_frame, upper_leg_mass, upper_leg_left_central_inertia) hip = RigidBody('Hip', hip_mass_center, hip_frame, hip_mass, hip_central_inertia) upper_leg_right = RigidBody('Upper Leg Right', upper_leg_right_mass_center, upper_leg_right_frame, upper_leg_mass, upper_leg_right_central_inertia) lower_leg_right = RigidBody('Lower Leg Right', lower_leg_right_mass_center, lower_leg_right_frame, lower_leg_mass, lower_leg_right_central_inertia) particles = [] particles.append(Particle('ankle_left', ankle_left, 0)) particles.append(Particle('knee_left', knee_left, 0)) particles.append(Particle('hip_left', hip_left, 0)) particles.append(Particle('hip_center', hip_center, 0)) particles.append(Particle('hip_right', hip_right, 0)) particles.append(Particle('knee_right', knee_right, 0)) particles.append(Particle('ankle_right', ankle_right, 0)) particles mass_centers = [] mass_centers.append(Particle('lower_leg_left_mass_center', lower_leg_left_mass_center, lower_leg_mass)) mass_centers.append(Particle('upper_leg_left_mass_center', upper_leg_left_mass_center, upper_leg_mass)) mass_centers.append(Particle('hip_mass_center', hip_mass_center, hip_mass)) mass_centers.append(Particle('hip_mass_center', hip_mass_center, hip_mass)) mass_centers.append(Particle('hip_mass_center', hip_mass_center, hip_mass)) mass_centers.append(Particle('lower_leg_left_mass_center', lower_leg_right_mass_center, lower_leg_mass)) mass_centers.append(Particle('upper_leg_left_mass_center', upper_leg_right_mass_center, upper_leg_mass)) mass_centers ``` [lower_leg_left_mass_center, upper_leg_left_mass_center, hip_mass_center, hip_mass_center, hip_mass_center, lower_leg_left_mass_center, upper_leg_left_mass_center] # Forces and Torques ```python g = symbols('g') ``` ```python lower_leg_left_grav_force = (lower_leg_left_mass_center, -lower_leg_mass * g * inertial_frame.y) upper_leg_left_grav_force = (upper_leg_left_mass_center, -upper_leg_mass * g * inertial_frame.y) hip_grav_force = (hip_mass_center, -hip_mass * g * inertial_frame.y) upper_leg_right_grav_force = (upper_leg_right_mass_center, -upper_leg_mass * g * inertial_frame.y) lower_leg_right_grav_force = (lower_leg_right_mass_center, -lower_leg_mass * g * inertial_frame.y) ankle_torque0, knee_torque0, hip_torque0, hip_torque1, knee_torque1, ankle_torque1 = dynamicsymbols('T_a0, T_k0, T_h0, T_h1, T_k1, T_a1') lower_leg_left_torque_vector = ankle_torque0 * inertial_frame.z - knee_torque0 * inertial_frame.z upper_leg_left_torque_vector = knee_torque0 * inertial_frame.z - hip_torque0 * inertial_frame.z hip_left_torque_vector = hip_torque0 * inertial_frame.z - hip_torque1 * inertial_frame.z hip_right_torque_vector = hip_torque1 * inertial_frame.z - knee_torque1 * inertial_frame.z upper_leg_right_torque_vector = knee_torque1 * inertial_frame.z - ankle_torque1 * inertial_frame.z lower_leg_right_torque_vector = ankle_torque1 * inertial_frame.z lower_leg_left_torque = (lower_leg_left_frame, lower_leg_left_torque_vector) upper_leg_left_torque = (upper_leg_left_frame, upper_leg_left_torque_vector) hip_left_torque = (hip_frame, hip_left_torque_vector) hip_right_torque = (hip_frame, hip_right_torque_vector) upper_leg_right_torque = (upper_leg_right_frame, upper_leg_right_torque_vector) lower_leg_right_torque = (lower_leg_right_frame, lower_leg_right_torque_vector) ``` # Equations of Motion ```python coordinates = [theta0, theta1, theta2, theta3, phi] coordinates ``` $$\left [ \theta_{0}, \quad \theta_{1}, \quad \theta_{2}, \quad \theta_{3}, \quad \phi\right ]$$ ```python speeds = [omega0, omega1, omega2, omega3, psi] speeds ``` $$\left [ \omega_{0}, \quad \omega_{1}, \quad \omega_{2}, \quad \omega_{3}, \quad \psi\right ]$$ ```python kinematical_differential_equations ``` $$\left [ \omega_{0} - \dot{\theta}_{0}, \quad \omega_{1} - \dot{\theta}_{1}, \quad \omega_{2} - \dot{\theta}_{2}, \quad \omega_{3} - \dot{\theta}_{3}, \quad \psi - \dot{\phi}\right ]$$ ```python kane = KanesMethod(inertial_frame, coordinates, speeds, kinematical_differential_equations) ``` ```python loads = [lower_leg_left_grav_force, upper_leg_left_grav_force, hip_grav_force, upper_leg_right_grav_force, lower_leg_right_grav_force, lower_leg_left_torque, upper_leg_left_torque, hip_left_torque, hip_right_torque, upper_leg_right_torque, lower_leg_right_torque] loads ``` [(L_COMleft, - g*m_L*I.y), (U_COMleft, - g*m_U*I.y), (H_COMleft, - g*m_H*I.y), (U_COMright, - g*m_U*I.y), (L_COMright, - g*m_L*I.y), (R_1, (T_a0 - T_k0)*I.z), (R_2, (-T_h0 + T_k0)*I.z), (R_3, (T_h0 - T_h1)*I.z), (R_3, (T_h1 - T_k1)*I.z), (R_4, (-T_a1 + T_k1)*I.z), (R_5, T_a1*I.z)] ```python bodies = [lower_leg_left, upper_leg_left, hip, upper_leg_right, lower_leg_right] bodies ``` [Lower Leg Left, Upper Leg Left, Hip, Upper Leg Right, Lower Leg Right] ```python fr, frstar = kane.kanes_equations(loads, bodies) ``` ```python trigsimp(fr + frstar) ``` $$\left[\begin{matrix}- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{H} \operatorname{sin}\left(\phi - \theta_{0}\right) - g l_{2} m_{L} \operatorname{sin}\left(\phi - \theta_{0}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{3 g}{2} l_{2} m_{U} \operatorname{sin}\left(\phi - \theta_{0}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + \frac{g l_{3}}{2} m_{H} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{L} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{U} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - \frac{l_{2}^{2} m_{U}}{2} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{H} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{H} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{L} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) + l_{2} l_{3} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{U} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} - 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right)\right) \dot{\omega}_{3} - \left(I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(- 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right)\right) \dot{\omega}_{2} - \left(I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3} m_{H}}{4} \left(- 2 l_{2} \operatorname{sin}\left(\theta_{1}\right) + l_{3}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- \frac{l_{2}^{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right)\right) \dot{\omega}_{1} - \left(I_{Hz} + I_{Lz} + 2 I_{Uz} + \frac{l_{2}^{2} m_{U}}{4} + m_{H} \left(l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) + \frac{l_{3}^{2}}{4}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - l_{1} l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - 2 l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + 2 l_{2}^{2} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{5 l_{2}^{2}}{4} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right)\right) \dot{\omega}_{0} - T_{k0}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + \frac{g l_{3}}{2} m_{H} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{L} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{U} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - \frac{l_{2} l_{3}}{2} m_{H} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{L} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{U} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right)\right) \dot{\omega}_{3} - \left(I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right)\right) \dot{\omega}_{2} - \left(I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3}^{2} m_{H}}{4} + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(\frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right)\right) \dot{\omega}_{1} - \left(I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3} m_{H}}{4} \left(- 2 l_{2} \operatorname{sin}\left(\theta_{1}\right) + l_{3}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- \frac{l_{2}^{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right)\right) \dot{\omega}_{0} - T_{h0}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right)\right)\right) \dot{\omega}_{3} - \left(I_{Lz} + I_{Uz} + \frac{l_{2}^{2} m_{U}}{4} + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) + l_{2}^{2}\right)\right) \dot{\omega}_{2} - \left(I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right)\right) \dot{\omega}_{1} - \left(I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(- 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right)\right) \dot{\omega}_{0} - T_{k1}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) - \left(I_{Lz} + \frac{l_{1}^{2} m_{L}}{4}\right) \dot{\omega}_{3} - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right)\right)\right) \dot{\omega}_{2} - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right)\right) \dot{\omega}_{1} - \left(I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} - 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right)\right) \dot{\omega}_{0} - T_{a1}\\0\end{matrix}\right]$$ Keep in mind that out utlimate goal is to have the equations of motion in first order form: $$ \dot{\mathbf{x}} = \mathbf{g}(\mathbf{x}, t) $$ The equations of motion are linear in terms of the derivatives of the generalized speeds and the `KanesMethod` class automatically puts the equations in a more useful form for the next step of numerical simulation: $$ \mathbf{M}(\mathbf{x}, t)\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x}, t) $$ Note that $$ \mathbf{g} = \mathbf{M}^{-1}(\mathbf{x}, t) \mathbf{f}(\mathbf{x}, t) $$ and that $\mathbf{g}$ can be computed analytically but for non-toy problems, it is best to do this numerically. So we will simply generate the $\mathbf{M}$ and $\mathbf{f}$ matrices for later use. The mass matrix, $\mathbf{M}$, can be accessed with the `mass_matrix` method (use `mass_matrix_full` to include the kinematical differential equations too. We can use `trigsimp` again to make this relatively compact: ```python mass_matrix = trigsimp(kane.mass_matrix_full) mass_matrix ``` $$\left[\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & I_{Hz} + I_{Lz} + 2 I_{Uz} + \frac{l_{2}^{2} m_{U}}{4} + m_{H} \left(l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) + \frac{l_{3}^{2}}{4}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - l_{1} l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - 2 l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + 2 l_{2}^{2} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{5 l_{2}^{2}}{4} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) & I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3} m_{H}}{4} \left(- 2 l_{2} \operatorname{sin}\left(\theta_{1}\right) + l_{3}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- \frac{l_{2}^{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) & I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(- 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} - 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right) & 0\\0 & 0 & 0 & 0 & 0 & I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3} m_{H}}{4} \left(- 2 l_{2} \operatorname{sin}\left(\theta_{1}\right) + l_{3}\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(- \frac{l_{2}^{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{1}\right) - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) & I_{Hz} + I_{Lz} + I_{Uz} + \frac{l_{3}^{2} m_{H}}{4} + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - l_{1} l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - 2 l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) + m_{U} \left(\frac{l_{2}^{2}}{4} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right) + l_{3}^{2}\right) & I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right) & 0\\0 & 0 & 0 & 0 & 0 & I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(- 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} - \frac{l_{1} l_{2}}{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} \operatorname{cos}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) & I_{Lz} + I_{Uz} + \frac{l_{2} m_{U}}{4} \left(l_{2} - 2 l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) - \frac{l_{1} l_{3}}{2} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} - l_{2} l_{3} \operatorname{sin}\left(\theta_{2}\right)\right) & I_{Lz} + I_{Uz} + \frac{l_{2}^{2} m_{U}}{4} + m_{L} \left(\frac{l_{1}^{2}}{4} + l_{1} l_{2} \operatorname{cos}\left(\theta_{3}\right) + l_{2}^{2}\right) & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right)\right) & 0\\0 & 0 & 0 & 0 & 0 & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} - 2 l_{2} \operatorname{cos}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right) & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right) - 2 l_{3} \operatorname{sin}\left(\theta_{2} + \theta_{3}\right)\right) & I_{Lz} + \frac{l_{1} m_{L}}{4} \left(l_{1} + 2 l_{2} \operatorname{cos}\left(\theta_{3}\right)\right) & I_{Lz} + \frac{l_{1}^{2} m_{L}}{4} & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right]$$ The right hand side, $\mathbf{f}$, is a vector function of all the non-inertial forces (gyroscopic, external, coriolis, etc): ```python forcing_vector = trigsimp(kane.forcing_full) forcing_vector ``` $$\left[\begin{matrix}\omega_{0}\\\omega_{1}\\\omega_{2}\\\omega_{3}\\\psi\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{H} \operatorname{sin}\left(\phi - \theta_{0}\right) - g l_{2} m_{L} \operatorname{sin}\left(\phi - \theta_{0}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{3 g}{2} l_{2} m_{U} \operatorname{sin}\left(\phi - \theta_{0}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + \frac{g l_{3}}{2} m_{H} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{L} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{U} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) - l_{2}^{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - \frac{l_{2}^{2} m_{U}}{2} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{H} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{H} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{L} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) + l_{2} l_{3} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{U} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - T_{k0}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + \frac{g l_{3}}{2} m_{H} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{L} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) + g l_{3} m_{U} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - \frac{l_{2} l_{3}}{2} m_{H} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{L} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) + \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - l_{2} l_{3} m_{U} \omega^{2}_{0} \operatorname{cos}\left(\theta_{1}\right) - T_{h0}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - g l_{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{g l_{2}}{2} m_{U} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2} - \omega_{3}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) + l_{2}^{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) + \frac{l_{2}^{2} m_{U}}{2} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2}\right) - l_{2} l_{3} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - \frac{l_{2} l_{3}}{2} m_{U} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2}\right) - T_{k1}\\- \frac{g l_{1}}{2} m_{L} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{2}}{2} m_{L} \left(- \omega_{0} - \omega_{1} - \omega_{2}\right)^{2} \operatorname{sin}\left(\theta_{3}\right) + \frac{l_{1} l_{2}}{2} m_{L} \omega^{2}_{0} \operatorname{sin}\left(\theta_{1} + \theta_{2} + \theta_{3}\right) - \frac{l_{1} l_{3}}{2} m_{L} \left(- \omega_{0} - \omega_{1}\right)^{2} \operatorname{cos}\left(\theta_{2} + \theta_{3}\right) - T_{a1}\\0\end{matrix}\right]$$ # Simulation ```python from scipy.integrate import odeint from sympy.utilities.codegen import codegen from pydy.codegen.ode_function_generators import generate_ode_function from matplotlib.pyplot import plot, legend, xlabel, ylabel, rcParams rcParams['figure.figsize'] = (14.0, 6.0) specified = [ankle_torque0, knee_torque0, hip_torque0, hip_torque1, knee_torque1, ankle_torque1] numerical_specified = zeros(6) x0 = zeros(10) x0 x0[0] = deg2rad(80) x0[1] = -deg2rad(80) x0[2] = -deg2rad(80) x0[3] = deg2rad(80) x0[4] = 0 x0 ``` array([ 1.3962634, -1.3962634, -1.3962634, 1.3962634, 0. , 0. , 0. , 0. , 0. , 0. ]) ```python #%% Jacobian for the right ankle, which we will try to leave where it is when # moving the hip center print ('calculating jacobian for the right ankle') F0 = ankle_right.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) F0 = Matrix([F0[0], F0[1]]) F0 ``` calculating jacobian for the right ankle $$\left[\begin{matrix}- l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\\- l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} \operatorname{cos}\left(\phi\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\end{matrix}\right]$$ ```python J_ankleRight = F0.jacobian([theta0, theta1, theta2, theta3, phi]) J_ankleRight ``` $$\left[\begin{matrix}- l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right) & - l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right) & - l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & - l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{1} \operatorname{cos}\left(\phi\right) - l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) + l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\\l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) + l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) & l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) & l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) & l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) & - l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\end{matrix}\right]$$ ```python F1 = hip_center.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) F1 = Matrix([F1[0], F1[1]]) F1 ``` $$\left[\begin{matrix}- l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) + \frac{l_{3}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\\l_{1} \operatorname{cos}\left(\phi\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - \frac{l_{3}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\end{matrix}\right]$$ ```python J_hipCenter = F1.jacobian([theta0, theta1, theta2, theta3, phi]) J_hipCenter ``` $$\left[\begin{matrix}l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - \frac{l_{3}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right) & - \frac{l_{3}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right) & 0 & 0 & - l_{1} \operatorname{cos}\left(\phi\right) - l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) + \frac{l_{3}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\\l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) - \frac{l_{3}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) & - \frac{l_{3}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right) & 0 & 0 & - l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) + \frac{l_{3}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\end{matrix}\right]$$ ```python #%% we stack the two jacobians J = J_ankleRight.col_join(J_hipCenter) J.shape ``` $$\left ( 4, \quad 5\right )$$ ```python #%% lets try the pseudo inverse with a couple of real values values = {lower_leg_length: 0.4, upper_leg_length: 0.54, hip_length: 0.2, theta0: x0[0], theta1: x0[1], theta2: x0[2], theta3: x0[3], phi: x0[4]} Jpinv = J.subs(values).evalf().pinv() Jpinv ``` $$\left[\begin{matrix}-0.0459332479170267 & 0.219478074655015 & -2.3994104929455 & -2.31969900118024\\0.235092856483776 & -1.12331981399207 & -0.514831317577474 & 1.87254742697749\\-0.235092856483778 & -0.757099837648197 & 0.514831317577473 & 1.88829187630304\\-2.45406675208297 & 2.10175902875642 & 2.39941049294551 & -2.32277520564262\\-0.00872301122043817 & 0.0416802598264825 & -2.48089742314152 & -0.440525356532326\end{matrix}\right]$$ ```python #%% we stack the two endeffektor points, which we will evaluate in the integration F2 = F0.col_join(F1) F2 ``` $$\left[\begin{matrix}- l_{1} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) - l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) + l_{3} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\\- l_{1} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2} + \theta_{3}\right) + l_{1} \operatorname{cos}\left(\phi\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - l_{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1} + \theta_{2}\right) - l_{3} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\\- l_{1} \operatorname{sin}\left(\phi\right) - l_{2} \operatorname{sin}\left(\phi - \theta_{0}\right) + \frac{l_{3}}{2} \operatorname{cos}\left(- \phi + \theta_{0} + \theta_{1}\right)\\l_{1} \operatorname{cos}\left(\phi\right) + l_{2} \operatorname{cos}\left(\phi - \theta_{0}\right) - \frac{l_{3}}{2} \operatorname{sin}\left(- \phi + \theta_{0} + \theta_{1}\right)\end{matrix}\right]$$ ```python #%% this defines how long you want to simulate, we simulate for one second with 30 fps # simulation can take quite a while... frames_per_sec = 30 final_time = 1 t = linspace(0.0, final_time, final_time * frames_per_sec) ``` ```python numerical_constants = array([0.42, # lower_leg_length [m] 0.54, # upper_leg_length [m] 0.2, # hip_length 1.0, # lower_leg_mass [kg] 1.5, # upper_leg_mass [kg] 2.0, # hip_mass [kg] 0.1, # lower_leg_inertia [kg*m^2] 0.2, # upper_leg_inertia [kg*m^2] 0.1, # hip_inertia [kg*m^2] 9.81], # acceleration due to gravity [m/s^2] ) numerical_constants ``` array([ 0.42, 0.54, 0.2 , 1. , 1.5 , 2. , 0.1 , 0.2 , 0.1 , 9.81]) ```python constants = [lower_leg_length, upper_leg_length, hip_length, lower_leg_mass, upper_leg_mass, hip_mass, lower_leg_inertia, upper_leg_inertia, hip_inertia, g] constants ``` $$\left [ l_{1}, \quad l_{2}, \quad l_{3}, \quad m_{L}, \quad m_{U}, \quad m_{H}, \quad I_{Lz}, \quad I_{Uz}, \quad I_{Hz}, \quad g\right ]$$ ```python #%% x0 = zeros(10) x0 x0[0] = deg2rad(80) x0[1] = -deg2rad(80) x0[2] = -deg2rad(80) x0[3] = deg2rad(80) x0[4] = 0 x0 ``` array([ 1.3962634, -1.3962634, -1.3962634, 1.3962634, 0. , 0. , 0. , 0. , 0. , 0. ]) ```python #%% Use this for full kinematics simulation #right_hand_side = generate_ode_function(forcing_vector, coordinates, speeds, # constants, mass_matrix=mass_matrix, # specifieds=specified, generator='cython') # #args = {'constants': numerical_constants, # 'specified': numerical_specified} # #right_hand_side(x0, 0.0, numerical_specified, numerical_constants) ##%% #y = odeint(right_hand_side, x0, t, args=(numerical_specified, numerical_constants)) #y #%% only useful when you want to simulate full kinematics, use in right_hand_side equation #from sympy import lambdify, solve #M_func = lambdify(coordinates + speeds + constants + specified, mass_matrix) # Create a callable function to evaluate the mass matrix #f_func = lambdify(coordinates + speeds + constants + specified, forcing_vector) # Create a callable function to evaluate the forcing vector #%% inverse kinematics gains, you can balance here how much the respective # targets should be tried to reach kp = np.eye(4) kp[0,0] = 10 kp[1,1] = 10 kp[2,2] = 10 kp[3,3] = 10 kp #%% kpN = np.zeros((4,4),dtype=float) kpN[1,1] = 10 kpN #%% These are our target points L = 1 # this defines how far apart the left and right ankle should be, y=0 so the # foot does not lift of the ground # the last two values are x and y coordinates for the hip center x_des = Matrix([L,0,0.7,0.1]) x_des.shape #%% This is the intergration function (the inverse kinematics) # I highly recommend you have a look at this textbook http://smpp.northwestern.edu/savedLiterature/Spong_Textbook.pdf # especially chapter 5.10 Inverse Velocity and Acceleration i = 0 def right_hand_side(x, t, args): """Returns the derivatives of the states. Parameters ---------- x : ndarray, shape(2 * (n + 1)) The current state vector. t : float The current time. args : ndarray The constants. Returns ------- dx : ndarray, shape(2 * (n + 1)) The derivative of the state. """ global i r = 0.0 # The input force is always zero arguments = np.hstack((x, args)) # States, input, and parameters values = {lower_leg_length: 0.4, upper_leg_length: 0.54, hip_length: 0.2, theta0: x[0], theta1: x[1], theta2: x[2], theta3: x[3], phi: x[4], omega0: 0, omega1: 0, omega2: 0, omega3: 0, psi: 0} Jpinv = J.subs(values).evalf().pinv() # use this for nullspace movements # N=np.eye(4)-Jpinv*J.subs(values).evalf() x_current = F2.subs(values).evalf() # use this for nullspace movements dq = np.array(Jpinv*(kp*( x_des - x_current ))).astype(float).T[0]# + N*(kpN*(Matrix([0,-deg2rad(80),0,0,0])-Matrix([x[0],x[1],x[2],x[3])))).astype(float).T[0] dq = np.hstack((dq,np.zeros(5,dtype=float))) if i%10==0: print(x_current) i = i+1 return dq # use this for full kinematic simulation # dq = np.array(np.linalg.solve(M_func(*arguments), # Solving for the derivatives # f_func(*arguments))).T[0] # return dq ``` ```python #%% Ok lets simulate print('integrating') args = (np.hstack((numerical_constants,)),) y = odeint(right_hand_side, x0, t, args)#,full_output = 1,mxstep=1000000 #y = np.hstack((y,np.zeros(y.shape))) print(y) print('done integrating') ``` integrating Matrix([[1.26359237325318], [0], [0.631796186626592], [0.493770015940142]]) Matrix([[1.26175159570018], [6.37010906447567e-9], [0.632272477736748], [0.491020164371748]]) Matrix([[1.25528230436820], [-8.25783376997690e-9], [0.633946400654344], [0.481355924543045]]) Matrix([[1.24204513938050], [-4.32124976079695e-9], [0.637371479067839], [0.461581463804779]]) Matrix([[1.22016985269966], [-7.65121452371391e-9], [0.643031651745630], [0.428902850571606]]) Matrix([[1.19347469825923], [-4.40686484179387e-9], [0.649938948994721], [0.389024040678037]]) Matrix([[1.16441873966249], [-1.32369283719833e-9], [0.657457097886850], [0.345618520687556]]) Matrix([[1.13086159526863], [3.19119596654989e-9], [0.666139915607050], [0.295488862780702]]) Matrix([[1.09762332427629], [8.60721838030418e-9], [0.674740225069592], [0.245835558050327]]) Matrix([[1.06845152046108], [1.41858201431304e-8], [0.682288346687779], [0.202256984915244]]) Matrix([[1.04481151791494], [1.23655330506317e-8], [0.688405133486092], [0.166942137274830]]) Matrix([[1.02579277469106], [2.67137921592525e-9], [0.693326187989628], [0.138530789666744]]) Matrix([[1.01358973421518], [-1.82801794187948e-9], [0.696483696012692], [0.120301150837070]]) Matrix([[1.00627090789950], [-6.78787654556645e-10], [0.698377421556476], [0.109367853168821]]) Matrix([[1.00242339361369], [-1.00974011790766e-9], [0.699372955849698], [0.103620206981665]]) Matrix([[1.00085360468088], [-8.53671905734488e-10], [0.699779133163587], [0.101275165217524]]) Matrix([[1.00024778277731], [2.14709708346028e-10], [0.699935886932838], [0.100370152121145]]) Matrix([[1.00006920332609], [4.67516945296120e-11], [0.699982093856457], [0.100103379898465]]) Matrix([[1.00001656522374], [-1.12818834135942e-11], [0.699995713807771], [0.100024746112718]]) [[ 1.3962634 -1.3962634 -1.3962634 1.3962634 0. 0. 0. 0. 0. 0. ] [ 1.58082794 -1.62016678 -1.57321165 1.84765989 -0.03033541 0. 0. 0. 0. 0. ] [ 1.67079324 -1.76233809 -1.68716838 2.12830504 -0.088468 0. 0. 0. 0. 0. ] [ 1.71531961 -1.85578284 -1.76218806 2.31695656 -0.14682271 0. 0. 0. 0. 0. ] [ 1.73730152 -1.91741631 -1.81150631 2.44799534 -0.19618586 0. 0. 0. 0. 0. ] [ 1.74806591 -1.9579694 -1.84371585 2.54052487 -0.23486065 0. 0. 0. 0. 0. ] [ 1.75326896 -1.98459453 -1.86464227 2.60641834 -0.26395708 0. 0. 0. 0. 0. ] [ 1.75573295 -2.00208225 -1.87822052 2.65352561 -0.28534559 0. 0. 0. 0. 0. ] [ 1.75686168 -2.01361142 -1.8870601 2.68723902 -0.30085218 0. 0. 0. 0. 0. ] [ 1.7573498 -2.02126402 -1.89285726 2.71135407 -0.31199988 0. 0. 0. 0. 0. ] [ 1.75753831 -2.0263886 -1.89669765 2.7285809 -0.31997206 0. 0. 0. 0. 0. ] [ 1.75759247 -2.02985352 -1.8992705 2.74086793 -0.32565456 0. 0. 0. 0. 0. ] [ 1.75759082 -2.0322183 -1.90101329 2.74961855 -0.32969648 0. 0. 0. 0. 0. ] [ 1.75756949 -2.03384574 -1.90220558 2.75584256 -0.33256757 0. 0. 0. 0. 0. ] [ 1.75754423 -2.03497359 -1.90302811 2.76026484 -0.33460516 0. 0. 0. 0. 0. ] [ 1.75752123 -2.0357596 -1.90359937 2.7634044 -0.33605036 0. 0. 0. 0. 0. ] [ 1.75750238 -2.03630975 -1.90399819 2.76563194 -0.33707498 0. 0. 0. 0. 0. ] [ 1.75748773 -2.03669608 -1.90427773 2.76721166 -0.33780123 0. 0. 0. 0. 0. ] [ 1.75747671 -2.03696802 -1.90447424 2.7683316 -0.33831589 0. 0. 0. 0. 0. ] [ 1.75746858 -2.03715978 -1.90461268 2.76912538 -0.33868056 0. 0. 0. 0. 0. ] [ 1.75746266 -2.03729517 -1.90471035 2.76968789 -0.33893892 0. 0. 0. 0. 0. ] [ 1.75745838 -2.03739086 -1.90477935 2.77008646 -0.33912196 0. 0. 0. 0. 0. ] [ 1.75745531 -2.03745853 -1.90482812 2.77036885 -0.33925163 0. 0. 0. 0. 0. ] [ 1.75745311 -2.03750641 -1.90486262 2.77056891 -0.33934349 0. 0. 0. 0. 0. ] [ 1.75745155 -2.03754029 -1.90488704 2.77071064 -0.33940856 0. 0. 0. 0. 0. ] [ 1.75745043 -2.03756428 -1.90490432 2.77081103 -0.33945465 0. 0. 0. 0. 0. ] [ 1.75744964 -2.03758126 -1.90491655 2.77088215 -0.3394873 0. 0. 0. 0. 0. ] [ 1.75744908 -2.03759329 -1.90492521 2.77093253 -0.33951043 0. 0. 0. 0. 0. ] [ 1.75744868 -2.03760181 -1.90493135 2.77096822 -0.33952681 0. 0. 0. 0. 0. ] [ 1.7574484 -2.03760784 -1.90493569 2.77099349 -0.33953841 0. 0. 0. 0. 0. ]] done integrating ```python #%% Plot # here we plot a little print("Plotting") plot(t, rad2deg(y[:, :5])) xlabel('Time [s]') ylabel('Angle [deg]') legend(["${}$".format(vlatex(c)) for c in coordinates]) ``` Plotting <matplotlib.legend.Legend at 0x7f46cc386790> # Visualization ```python from pydy.viz.shapes import Cylinder, Sphere from pydy.viz.scene import Scene from pydy.viz.visualization_frame import VisualizationFrame # stack the lengths and use the bodies from kinematics lengths = [lower_leg_length, upper_leg_length, hip_length, hip_length, hip_length, upper_leg_length, lower_leg_length] bodies = [lower_leg_left, upper_leg_left, hip, hip, hip, upper_leg_right, lower_leg_right] viz_frames = [] colors = ['yellow','green','red','red','red','green','blue'] for i, (body, particle, mass_center) in enumerate(zip(bodies, particles, mass_centers)): # body_shape = Cylinder(name='cylinder{}'.format(i), # radius=0.05, # length=lengths[i], # color='red') # # viz_frames.append(VisualizationFrame('link_frame{}'.format(i), body, # body_shape)) particle_shape = Sphere(name='sphere{}'.format(i), radius=0.06, color=colors[i]) viz_frames.append(VisualizationFrame('particle_frame{}'.format(i), body.frame, particle, particle_shape)) mass_center_shape = Sphere(name='sphere{}'.format(i), radius=0.02, color='black') viz_frames.append(VisualizationFrame('mass_center_frame{}'.format(i), body.frame, mass_center, mass_center_shape)) target_shape = Sphere(name='sphere{}'.format(i), radius=0.02, color='green') target_right_leg = Point('target_right_leg') target_right_leg.set_pos(origin, (x_des[2] * inertial_frame.x)+(x_des[3] * inertial_frame.y)) viz_frames.append(VisualizationFrame('target_frame_right', inertial_frame, target_right_leg, target_shape)) ## Now the visualization frames can be passed in to create a scene. scene = Scene(inertial_frame, origin, *viz_frames) # Provide the data to compute the trajectories of the visualization frames. scene.constants = dict(zip(constants, numerical_constants)) scene.states_symbols = coordinates+speeds scene.states_trajectories = y scene.display() ``` /home/letrend/workspace/roboy-ros-control/python/pydy-resources ('Serving HTTP on', '127.0.0.1', 'port', 8001, '...') To view visualization, open: http://localhost:8001/index.html?load=2017-06-28_02-51-33_scene_desc.json Press Ctrl+C to stop server... # export to c header ```python t = symbols('t') a0 = ankle_left.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) k0 = knee_left.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) hl = hip_left.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) hc = hip_center.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) hr = hip_right.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) k1 = knee_right.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) a1 = ankle_right.pos_from(origin).express(inertial_frame).simplify().to_matrix(inertial_frame) [(c_name, c_code), (h_name, c_header)] = codegen( [("Jacobian",J), ("ankle_right_hip_center",F2), ("ankle_left", a0), ("knee_left", k0), ("hip_left", hl), ("hip_center", hc), ("hip_right", hr), ("knee_right", k1), ("ankle_right",a1) ] , "C", "PaBiRoboy_DanceControl", project='PaBiRoboy_DanceControl',global_vars=(t,lower_leg_length, upper_leg_length, hip_length, theta0,theta1,theta2,theta3), header=True, empty=False) print(c_code) ``` 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /index.html?load=2017-06-28_02-51-33_scene_desc.json HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /css/bootstrap.min.css HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /css/slider.css HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /css/main.css HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /css/codemirror/codemirror.css HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /css/codemirror/blackboard.css HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/jquery/jquery.min.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/jquery/jquery-ui.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/bootstrap/bootstrap.min.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/codemirror/codemirror.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/codemirror/javascript-mode.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/three/three.min.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/three/TrackballControls.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/utils/bootstrap-slider.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/utils/modernizr-2.0.6.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/external/utils/prototype.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/dv.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/scene.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/parser.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/param_editor.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/materials.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:40] "GET /js/dyviz/main.js HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:41] code 404, message File not found 127.0.0.1 - - [28/Jun/2017 02:51:41] "GET /fonts/glyphicons-halflings-regular.woff2 HTTP/1.1" 404 - 127.0.0.1 - - [28/Jun/2017 02:51:41] "GET /2017-06-28_02-51-33_scene_desc.json HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:41] "GET /fonts/glyphicons-halflings-regular.woff HTTP/1.1" 200 - 127.0.0.1 - - [28/Jun/2017 02:51:41] "GET /2017-06-28_02-51-33_simulation_data.json HTTP/1.1" 200 - /****************************************************************************** * Code generated with sympy 1.0 * * * * See http://www.sympy.org/ for more information. * * * * This file is part of 'PaBiRoboy_DanceControl' * ******************************************************************************/ #include "PaBiRoboy_DanceControl.h" #include <math.h> void Jacobian(double *out_7040956801295068946) { out_7040956801295068946[0] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l2*cos(phi(t) - theta0(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[1] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[2] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)); out_7040956801295068946[3] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)); out_7040956801295068946[4] = l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l1*cos(phi(t)) - l2*cos(phi(t) - theta0(t)) + l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) + l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[5] = l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l2*sin(phi(t) - theta0(t)) + l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*cos(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[6] = l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*cos(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[7] = l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)); out_7040956801295068946[8] = l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)); out_7040956801295068946[9] = -l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) - l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) + l3*cos(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[10] = l2*cos(phi(t) - theta0(t)) - 1.0L/2.0L*l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[11] = -1.0L/2.0L*l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[12] = 0; out_7040956801295068946[13] = 0; out_7040956801295068946[14] = -l1*cos(phi(t)) - l2*cos(phi(t) - theta0(t)) + (1.0L/2.0L)*l3*sin(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[15] = l2*sin(phi(t) - theta0(t)) - 1.0L/2.0L*l3*cos(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[16] = -1.0L/2.0L*l3*cos(-phi(t) + theta0(t) + theta1(t)); out_7040956801295068946[17] = 0; out_7040956801295068946[18] = 0; out_7040956801295068946[19] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) + (1.0L/2.0L)*l3*cos(-phi(t) + theta0(t) + theta1(t)); } void ankle_right_hip_center(double *out_916151544320618253) { out_916151544320618253[0] = -l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) - l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) + l3*cos(-phi(t) + theta0(t) + theta1(t)); out_916151544320618253[1] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_916151544320618253[2] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) + (1.0L/2.0L)*l3*cos(-phi(t) + theta0(t) + theta1(t)); out_916151544320618253[3] = l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - 1.0L/2.0L*l3*sin(-phi(t) + theta0(t) + theta1(t)); } void ankle_left(double *out_8826841557315653786) { out_8826841557315653786[0] = 0; out_8826841557315653786[1] = 0; out_8826841557315653786[2] = 0; } void knee_left(double *out_2247889337447503075) { out_2247889337447503075[0] = -l1*sin(phi(t)); out_2247889337447503075[1] = l1*cos(phi(t)); out_2247889337447503075[2] = 0; } void hip_left(double *out_9064110474024648940) { out_9064110474024648940[0] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)); out_9064110474024648940[1] = l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)); out_9064110474024648940[2] = 0; } void hip_center(double *out_659177576507600807) { out_659177576507600807[0] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) + (1.0L/2.0L)*l3*cos(-phi(t) + theta0(t) + theta1(t)); out_659177576507600807[1] = l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - 1.0L/2.0L*l3*sin(-phi(t) + theta0(t) + theta1(t)); out_659177576507600807[2] = 0; } void hip_right(double *out_8420808845996663653) { out_8420808845996663653[0] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) + l3*cos(-phi(t) + theta0(t) + theta1(t)); out_8420808845996663653[1] = l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_8420808845996663653[2] = 0; } void knee_right(double *out_2877649314090428462) { out_2877649314090428462[0] = -l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) - l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) + l3*cos(-phi(t) + theta0(t) + theta1(t)); out_2877649314090428462[1] = l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_2877649314090428462[2] = 0; } void ankle_right(double *out_8643290169570541467) { out_8643290169570541467[0] = -l1*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) - l1*sin(phi(t)) - l2*sin(phi(t) - theta0(t)) - l2*sin(-phi(t) + theta0(t) + theta1(t) + theta2(t)) + l3*cos(-phi(t) + theta0(t) + theta1(t)); out_8643290169570541467[1] = -l1*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t) + theta3(t)) + l1*cos(phi(t)) + l2*cos(phi(t) - theta0(t)) - l2*cos(-phi(t) + theta0(t) + theta1(t) + theta2(t)) - l3*sin(-phi(t) + theta0(t) + theta1(t)); out_8643290169570541467[2] = 0; } ```python ```

9cca2b44019c33fcf24e519985f511a0801b3ec4

ipynb

Jupyter Notebook

python/PaBiRoboy_dynamics.ipynb

Roboy/roboy_dynamics

a0a0012bad28029d01b6aead507faeee4509dd62

python/PaBiRoboy_dynamics.ipynb

Roboy/roboy_dynamics

a0a0012bad28029d01b6aead507faeee4509dd62

python/PaBiRoboy_dynamics.ipynb

Roboy/roboy_dynamics

a0a0012bad28029d01b6aead507faeee4509dd62

Qwen/Qwen-72B

1. YES 2. YES

__label__kor_Hang

<a href="https://colab.research.google.com/github/robfalck/dymos_tutorial/blob/main/01_dymos_simple_driver_boundary_value_problem.ipynb" target="_parent"></a> # Dymos: Using an Optimizer to Solve a Simple Boundary Value Problem In the previous notebook, we demonstrated - how to install Dymos - how to define a simple ODE system - how to use that system to propagate a simple trajectory Using the `scipy.itegrate.solve_ivp` functionality from within Dymos will provide a trajectory assuming we know the initial state, control profile, and time duration of that trajectory. In our next use case we want to solve a simple boundary value problem: _How far will the cannonball fly?_ In order to determine this, we need to allow the duration of the flight to vary such that the final height of the cannonball is zero. There are two general approaches to this. 1. Use an optimizer and pose the problem as an optimal control problem. 2. Use a nonlinear solver and pose the problem with residuals to be driven to zero. # Posing the cannonball boundary value problem as an optimal control problem. The optimal control problem can be stated as: \begin{align} \mathrm{Minimize}\;J &= t_f \\ \mathrm{subject\;to\!:} \\ x(t_0) &= 0 \;\mathrm{m} \\ y(t_0) &= 0 \;\mathrm{m}\\ vx(t_0) &= 100 \;\mathrm{m/s} \\ vy(t_0) &= 100 \;\mathrm{m/s} \\ y(t_f) &= 0 \;\mathrm{m} \end{align} Traditionally, the collocation techniques in Dymos have used an optimizer to satisfy the _defect constraints_ ($\Delta$). These defects are the difference between the slope of each state polynomial representation in each segment and the rate of change as given by the ODE. In Dymos, these defects are measured at a certain subset of our nodes called the _collocation nodes_. For the Radau pseudospectral transcription, each 3rd-order segment has 4 state input nodes, and 3 collocation nodes. Dymos handles this underlying transcription from an optimal control problem into a nonlinear programming (NLP) problem. The equivalent NLP problem for our optimal control problem above is: \begin{align} \mathrm{Minimize}\;J &= t[-1] \\ \mathrm{subject\;to\!:} \\ x_{lb} &\le x_i \le x_{ub} \\ y_{lb} &\le y_i \le y_{ub} \\ vx_{lb} &\le vx_i \le vx_{ub} \\ vy_{lb} &\le vy_i \le vy_{ub} \\ \Delta x_j &= 0 \\ \Delta y_j &= 0 \\ \Delta vx_j &= 0 \\ \Delta vy_j &= 0 \\ vy[-1] &= 0 \\ i &= 1\;...n-1 \\ j &= 0\;...n-2 \\ \end{align} In the case of our single 3rd-order Radau segment, we have 4 nodes (n=4). Since each initial state is fixed, we have 12 design variables and 12 constraints. There is, at most, a single solution to this problem. This makes sense. Since there are no control variables or changes in the initial state allowed, there should be only one possible physical path for the cannonball. Most optimizers will be perfectly capable of solving this problem, despite the fact that it only has a single feasible solution. # Solving the optimal control problem First, lets bring in our definition of the ODE and the other general setup that we used in the previous notebook. ``` !pip install dymos ``` Requirement already satisfied: dymos in /usr/local/lib/python3.6/dist-packages (0.17.0) Requirement already satisfied: openmdao>=3.3.0 in /usr/local/lib/python3.6/dist-packages (from dymos) (3.5.0) Requirement already satisfied: numpy>=1.14.1 in /usr/local/lib/python3.6/dist-packages (from dymos) (1.18.5) Requirement already satisfied: scipy>=1.0.0 in /usr/local/lib/python3.6/dist-packages (from dymos) (1.4.1) Requirement already satisfied: pyparsing in /usr/local/lib/python3.6/dist-packages (from openmdao>=3.3.0->dymos) (2.4.7) Requirement already satisfied: pyDOE2 in /usr/local/lib/python3.6/dist-packages (from openmdao>=3.3.0->dymos) (1.3.0) Requirement already satisfied: requests in /usr/local/lib/python3.6/dist-packages (from openmdao>=3.3.0->dymos) (2.23.0) Requirement already satisfied: networkx>=2.0 in /usr/local/lib/python3.6/dist-packages (from openmdao>=3.3.0->dymos) (2.5) Requirement already satisfied: chardet<4,>=3.0.2 in /usr/local/lib/python3.6/dist-packages (from requests->openmdao>=3.3.0->dymos) (3.0.4) Requirement already satisfied: urllib3!=1.25.0,!=1.25.1,<1.26,>=1.21.1 in /usr/local/lib/python3.6/dist-packages (from requests->openmdao>=3.3.0->dymos) (1.24.3) Requirement already satisfied: idna<3,>=2.5 in /usr/local/lib/python3.6/dist-packages (from requests->openmdao>=3.3.0->dymos) (2.10) Requirement already satisfied: certifi>=2017.4.17 in /usr/local/lib/python3.6/dist-packages (from requests->openmdao>=3.3.0->dymos) (2020.12.5) Requirement already satisfied: decorator>=4.3.0 in /usr/local/lib/python3.6/dist-packages (from networkx>=2.0->openmdao>=3.3.0->dymos) (4.4.2) ``` import numpy as np import openmdao.api as om import dymos as dm ``` ``` class ProjectileODE(om.ExplicitComponent): def initialize(self): self.options.declare('num_nodes', types=int, desc='the number of points at which this ODE is simultaneously evaluated') def setup(self): nn = self.options['num_nodes'] self.add_input('vx', shape=(nn,), units='m/s') self.add_input('vy', shape=(nn,), units='m/s') self.add_output('x_dot', shape=(nn,), units='m/s', tags=['state_rate_source:x', 'state_units:m']) self.add_output('y_dot', shape=(nn,), units='m/s', tags=['state_rate_source:y', 'state_units:m']) self.add_output('vx_dot', shape=(nn,), units='m/s**2', tags=['state_rate_source:vx', 'state_units:m/s']) self.add_output('vy_dot', shape=(nn,), units='m/s**2', tags=['state_rate_source:vy', 'state_units:m/s']) self.declare_partials(of='*', wrt='*', method='fd') def compute(self, inputs, outputs): outputs['x_dot'][:] = inputs['vx'] outputs['y_dot'][:] = inputs['vy'] outputs['vx_dot'][:] = 0.0 outputs['vy_dot'][:] = -9.81 ``` ``` prob = om.Problem() traj = prob.model.add_subsystem('traj', dm.Trajectory()) phase = traj.add_phase('phase', dm.Phase(ode_class=ProjectileODE, transcription=dm.Radau(num_segments=1, order=3))) phase.set_time_options(fix_initial=True, duration_bounds=(1, 1000), units='s') phase.set_state_options('x', fix_initial=True) phase.set_state_options('y', fix_initial=True) phase.set_state_options('vx', fix_initial=True) phase.set_state_options('vy', fix_initial=True) ``` # Imposing state boundary constraints There are two ways in which we can impose boundary constraints on the state. The first is to use the arguments `fix_initial` and `fix_final`. Using `fix_final` removes the final value of the state as a design variable. That is, our final value of `y` will always be satisfied and the optimizer will work to find the rest of the state history. The other option is to use a nonlinear boundary constraint. This is a constraint applied after-the-fact. The optimizer is free to suggest a final value of `y` that violates the constraint, but then will work to satisfy it as it iterates. This approach gives the optimizer a bit more freedom to explore the design space and might therefore be more robust. Using `fix_initial` and `fix_final` also impose limitations when using shooting methods in Dymos, which we'll discuss later. ``` phase.add_boundary_constraint('y', loc='final', equals=0) ``` # Adding an objective Using an optimization driver **requires** that an objective be specified. Again, in this case it's somewhat meaningless, there should only be a single valid solution that satisfies all of our constraints. To satisfy the driver, we'll set the objective to be the final time. Note that the Dymos `Phase` object has an `add_objective` method that's similar to the standard OpenMDAO `add_objective` method, but also allows us to specify the _locaiton_ in time where the objective should be evaluated (`'initial'` or `'final'`). The variable `'time'` is recognized by Dymos as the one of the special variables in the phase (along with states, controls, and parameters). If we wanted to, we could also make any output of the ODE an objective by specifying the ODE-relative path to that variable. ``` phase.add_objective('time', loc='final', ref=1.0) ``` # Specifying a driver Since we're performing optimization now, we'll need a driver. The optimizers in `scipy.optimize` will work in this case. To use them, we use OpenMDAO's `ScipyOptimizeDriver`. ``` prob.driver = om.ScipyOptimizeDriver() ``` Next we call OpenMDAO's `Problem.setup` method. This works out the various connections between the different systems that Dymos uses to solve this problem and allocates memory. You can think of this step a bit like compiling our model before we run it. ``` prob.setup() ``` <openmdao.core.problem.Problem at 0x7f3e99f6e320> As before, we'll specify an initial guess for the initial time and duration of the phase, along with linear guesses to the state-time history. ``` prob.set_val('traj.phase.t_initial', 0.0, units='s') prob.set_val('traj.phase.t_duration', 15.0, units='s') prob.set_val('traj.phase.states:x', phase.interpolate(ys=[0, 100], nodes='state_input'), units='m') prob.set_val('traj.phase.states:y', phase.interpolate(ys=[0, 0], nodes='state_input'), units='m') prob.set_val('traj.phase.states:vx', phase.interpolate(ys=[100, 100], nodes='state_input'), units='m/s') prob.set_val('traj.phase.states:vy', phase.interpolate(ys=[100, -100], nodes='state_input'), units='m/s') ``` # Solving the Problem To solve this problem, we need to allow the driver to iterate. The native OpenMDAO option to do so is `Problem.run_driver`. Dymos provides its own function, `dymos.run_problem`, which is basically a wrapper around the OpenMDAO method with some additional conveniences built in. Though not necessary yet, we'll use that method just to get in the habit of doing so. ``` dm.run_problem(prob, run_driver=True, simulate=True, make_plots=True) ``` Model viewer data has already has already been recorded for Driver. Optimization terminated successfully. (Exit mode 0) Current function value: 20.387359836901105 Iterations: 2 Function evaluations: 2 Gradient evaluations: 2 Optimization Complete ----------------------------------- Simulating trajectory traj /usr/local/lib/python3.6/dist-packages/openmdao/recorders/recording_manager.py:266: UserWarning:The model is being run again, if the options or scaling of any components has changed then only their new values will be recorded. Done simulating trajectory traj ``` # TODO: Make these more automatic (and use an interactive plot library like bokeh) from IPython.display import display from ipywidgets import widgets, GridBox items = [widgets.Image(value=open(f'plots/states_{state}.png', 'rb').read()) for state in ['x', 'y', 'vx', 'vy']] + \ [widgets.Image(value=open(f'plots/state_rates_{state}.png', 'rb').read()) for state in ['x', 'y', 'vx', 'vy']] widgets.GridBox(items, layout=widgets.Layout(grid_template_columns="repeat(2, 500px)")) ``` GridBox(children=(Image(value=b'\x89PNG\r\n\x1a\n\x00\x00\x00\rIHDR\x00\x00\x01\xb0\x00\x00\x01 \x08\x06\x00\x… # The Solution The plots above show that the implicit solution and the simulated trajectory are now in agreement (the simulated trajectory is a reasonably accurate interpolation of the solution). To extract the solution, we can pull values for the special variables in the phase from the _timeseries_ output. ``` t = prob.get_val('traj.phase.timeseries.time') x = prob.get_val('traj.phase.timeseries.states:x') y = prob.get_val('traj.phase.timeseries.states:y') print(f'The time of flight is {t[-1, 0]} sec') print(f'The range flown is {x[-1, 0]} m') print(f'The final altitude of the cannonball is {y[-1, 0]} m') ``` The time of flight is 20.387359836901105 sec The range flown is 2038.7359836901096 m The final altitude of the cannonball is -5.684341886080802e-14 m

c3b255e1d5fb313dc718cbd98022aa960a12249d

ipynb

Jupyter Notebook

01_dymos_simple_driver_boundary_value_problem.ipynb

robfalck/dymos_tutorial

04ec3b4804c601818503b3aa10679a42ab13fece

01_dymos_simple_driver_boundary_value_problem.ipynb

robfalck/dymos_tutorial

04ec3b4804c601818503b3aa10679a42ab13fece

01_dymos_simple_driver_boundary_value_problem.ipynb

robfalck/dymos_tutorial

04ec3b4804c601818503b3aa10679a42ab13fece

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Modeling Stock Movement Brian Bahmanyar *** ```python import numpy as np import pandas as pd import scipy.optimize as opt import seaborn as sns import sys sys.path.append('./src/') from plots import * ``` ```python %matplotlib inline ``` ```python tech = pd.read_csv('data/tech_bundle.csv', index_col=0) tech.index = pd.to_datetime(tech.index) ``` Recall neither Facebook's and Apple's stock had an obvious autocorrelation structure in their adjusted daily close prices. Thus we conclude that these series follow a random walk. It can be shown that the following holds: $\frac{y_t}{y_{t-1}} \sim LogN(\mu, \sigma^2)$, where $y_t$ represents the adjusted close price of a stock. Notice below when we plot $\frac{y_t}{y_{t-1}}$ (daily return ratio) for Facebook's historical stock price the distibution we observe can be approximated by a log-normal distribution. > We prefer a log-normal to a normal distribution because the daily return ratio can never be negative ```python def get_daily_return_ratio(series): """ Args: series (ndarray)----the time series to get the lagged return ratios window_size (int)---number of trading weeks to use Return: (ndarray) y(t)/y(t-1) over the weeks specified """ return (series[1:]/series[:-1]) ``` ```python assert(tech.FB.count()-1 == len(get_daily_return_ratio(tech.FB.values))) assert(tech.FB[1]/tech.FB[0] == get_daily_return_ratio(tech.FB.values)[0]) ``` ```python plt.figure(figsize=(10,3)); plt.title("Facebook's Daily Return Ratio"); plt.xlabel('Return Ratio'); plt.ylabel('Count'); sns.distplot(get_daily_return_ratio(tech['FB'].values), kde=False); ``` We will use Maximum Likelyhood Estimation to estimate the parameters of the distribution. ```python def neg_log_llh(theta, data): """ Args theta (list)-----the params [mu, sigma**2] data (ndarray)---data points to be fit by log normal distribution Return: (double) negative log-likelihood for the log normal. """ mu, sigma = theta[0], np.sqrt(theta[1]) neg_log_llhs = np.log(data*sigma*np.sqrt(2*np.pi)) + ((np.log(data)-mu)**2/(2*(sigma**2))) return neg_log_llhs.sum() ``` ```python def mle_log_norm(data, init_theta=[1,1]): """ Args: theta (list)-----the params [mu, sigma**2] data (ndarray)---data points to be fit by log normal distribution Return: (list) mu, sigma**2 for fitted log normal params """ fit = opt.minimize(neg_log_llh, init_theta, data, method='Nelder-Mead') return fit.x ``` ```python fit = mle_log_norm(get_daily_return_ratio(tech['FB'].values)) fit ``` array([ 0.00173078, 0.00054006]) This tells us for Facebook: $\frac{y_t}{y_{t-1}} \sim LogN(0.0017, 0.00054)$ Lets vizualize what this theoretical distribution looks like: ```python plt.figure(figsize=(10,3)); plt.title(r'$LogN(0.0017, 0.00054)$'); plt.xlabel('Return Ratio'); plt.ylabel('Count'); sns.distplot(np.random.lognormal(fit[0], np.sqrt(fit[1]), size=1000000), hist=False); ``` Now we should be convinced that daily return ratios follow a log-normal distribution we can exploit this to forcast future days prices with a log normal random walk. Quite simply if $\quad \frac{y_t}{y_{t-1}} \sim LogN(0.0017, 0.00054) \quad\Rightarrow\quad y_{t+1} \sim y_{t}*LogN(0.0017, 0.00054) $ ### Modeling Facebook's Daily Adjusted Close Prices ```python plot_stocks(tech.index, [tech['FB']], ['FB'], label_annually=False) ``` Lets simulate this random walk and see what the movement looks like. ```python def simulate_random_walk(series, window_size=10, ahead=50): """ Args: series (ndarray)----time series to use window_size (int)---number of trading weeks to use for log normal estimation ahead (int)---------how many days do you want to forcast head for Return (ndarray): simulated future price """ forcasts = np.zeros(ahead) window = series[-((window_size*5)+1):].values for step in range(ahead): mu, sigma2 = mle_log_norm( get_daily_return_ratio(window) ) forcast = window[-1]*np.random.lognormal(mu, np.sqrt(sigma2), 1) forcasts[step] = forcast window = np.roll(window, -1) window[-1] = forcast return forcasts ``` ```python def plot_simulated_forcast(series, window, ahead, train_on=None, n=10): """ Plots Simulated Forcast and Original Series Args: series (ndarray)---time series for which to forcast window (int)-------window size for MLE of log normal ahead (int)--------number of days to forcast for train_on (int)-----optional, number of days from series to use to forcast n (int)------------number of simulations to plot Returns: None, plots inline """ if not train_on: train_on = series.count() train_x_space = np.arange(0,series.count()) test_x_space = np.arange(train_on, train_on+ahead) plt.figure(figsize=(15,6)); plt.plot(train_x_space, series) for _ in range(n): walk = simulate_random_walk(series[:train_on], window, ahead) plt.plot(test_x_space, walk, linestyle='-', color='salmon', alpha=0.7) plt.xlabel('Time Index (Days)'); plt.ylabel('Adjusted Close Price'); plt.title('Simulated Log Normal Random Walk Forcast'); ``` Here we are using data from the first 650 prices and a daily return window of 100 days to predict Facebook's price 135 days into the future. We can notice a general positive slope in all our predictions (in pink). However when we approximate our forcast we will want to employ some math to calculate an expected value and variance at each step. ```python plot_simulated_forcast(tech['FB'], window=100, ahead=135, train_on=650, n=10); ``` Simulation is great but what we really want is the expected value and variance of the process and any time $t+k$. ##### Deriving the k-step ahead expected value $$ \begin{align} \hat{y_{t+k}} &= y_{t+(k-1)} \cdot \mathrm{E}[LogN(\mu, \sigma^2)] \\ &= y_{t} \cdot \mathrm{E}[LogN(\mu, \sigma^2)]^k \\ &= y_{t} \cdot {(e^{\mu+\sigma^2/2})}^k \end{align} $$ The expected value of a log-normal: $\mathrm{E}[LogN(\mu, \sigma^2)] = e^{\mu+\sigma^2/2}$ as a python function below ```python def get_expected_value(mu, sigma2): """ Returns expected value for log normal with params mu, sigma2 Args: mu (float)-------parameter mu sigma2 (float)---parameter sigma squared Returns: expected value (float) of the log normal """ return (np.exp((mu+sigma2)/2)) ``` The variance of a log normal: $\mathrm{Var}(LogN(\mu, \sigma^2)) = (e^{\sigma^2}\!\!-1) e^{2\mu+\sigma^2}$ as a python function below ```python def get_variance(mu, sigma2): """ Returns variance for log normal with params mu, sigma2 Args: mu (float)-------parameter mu sigma2 (float)---parameter sigma squared Returns: variance (float) of the log normal """ return (np.exp(sigma2)-1)*np.exp((2*mu)+sigma2) ``` **Deriving the k-step ahead variance** $$ \begin{aligned} \mathrm{V}[y_{t+k}|y_t] &= \mathrm{E}[y^2_{t+k}|y_t] - \mathrm{E}[y_{t+k}|y_t]^2 && \textit{by definition of Variance} \\ &= y^2_t \cdot \mathrm{E}\big[Z^2_{t+1}\big]^k - y^2_t \cdot \mathrm{E}\big[Z_{t+1}\big]^{2k} && \textit{let } Z \sim \mathrm{E}[LogN(\mu, \sigma^2)]\\ &= y^2_t \cdot \Big(\mathrm{E}\big[Z^2_{t+1}\big]^k - \mathrm{E}\big[Z_{t+1}\big]^{2k}\Big) \\ &= y^2_t \cdot \Big[\Big(\mathrm{V}\big[Z_{t+1}\big] + \mathrm{E}\big[Z_{t+1}\big]^{2}\Big)^k - \cdot \mathrm{E}\big[Z_{t+1}\big]^{2k}\Big] \end{aligned} $$ ```python def get_k_step_variance(mu, sigma2, k, yt): """ Returns a k step ahead variance for log normal with params mu, sigma2 Args: mu (float)-------parameter mu sigma2 (float)---parameter sigma squared k (int)----------number of steps ahead yt (float)-------price at time t (end of sample) Returns: variance (float) of the log normal """ return (yt**2) * ((get_variance(mu, sigma2) + get_expected_value(mu, sigma2)**2)**k - (get_expected_value(mu, sigma2)**2)**k) ``` We can now use these functions to create an expected forcast. ```python def get_expected_walk(series, window_size=10, ahead=50): """ Args: series (ndarray)----time series to use window_size (int)---number of trading weeks to use for log normal estimation ahead (int)---------how many days do you want to forcast head for Return (ndarray): simulated future price """ E_Xs = np.zeros(ahead) V_Xs = np.zeros(ahead) yt = series[-1] last_price = series[-1] window = series[-((window_size*5)+1):].values for step in range(ahead): mu, sigma2 = mle_log_norm( get_daily_return_ratio(window) ) E_Xs[step] = last_price * get_expected_value(mu, sigma2) V_Xs[step] = get_k_step_variance(mu, sigma2, step+1, yt) last_price = last_price * get_expected_value(mu, sigma2) return E_Xs, V_Xs ``` ```python def plot_expected_forcast(series, window, ahead, train_on=None, error=2): """ Plots Expected Forcast and Original Series Args: series (ndarray)---time series for which to forcast window (int)-------window size for MLE of log normal ahead (int)--------number of days to forcast for train_on (int)-----optional, number of days from series to use to forcast error (int)--------number of standard devations to use for confidence bands Returns: None, plots inline """ if not train_on: train_on = series.count() mu, sigma2 = get_expected_walk(series[:train_on], window, ahead) low_bound = mu-(error*np.sqrt(sigma2)) high_bound = mu+(error*np.sqrt(sigma2)) train_x_space = np.arange(0,series.count()) test_x_space = np.arange(train_on, train_on+ahead) plt.figure(figsize=(15,6)); plt.plot(train_x_space, series); plt.plot(test_x_space, mu, 'maroon'); plt.plot(test_x_space, high_bound, 'olive'); plt.plot(test_x_space, low_bound, 'olive'); plt.fill_between(test_x_space, low_bound, high_bound, alpha=0.2, color='y') plt.xlabel('Time Index (Days)'); plt.ylabel('Adjusted Close Price'); plt.title('Expected Log Normal Random Walk Forcast'); ``` Below we use the equations above to create an expected forcast using the first 635 data points and a return ratio window of 30 days. We can see that the forcast is doing a great job. The yellow are the 95% confidence interval bands which incease as we get farther from our sample. ```python plot_expected_forcast(tech.FB, 30, 150, 635, error=2) ``` #### Please ignore, this was an ARMA(p,d,q) I chose to remove ```python # def acf_objective(order, series): # try: # model = arima_model.ARIMA(series, order).fit() # return model.aic # except: # return np.inf # if the model is not stationary return inf ``` ```python # def get_optimal_arima(series, cost): # grid = (slice(0, 4, 1), slice(1,2,1), slice(0, 4, 1)) # orders = brute(cost, grid, args=(series,), finish=None) # orders = [int(order) for order in orders] # arima = arima_model.ARIMA(series, order=orders).fit() # print({'p':orders[0], 'd':orders[1], 'q':orders[2]}) # return arima ``` ```python # def fit_arima(series, p, d, q): # model = arima_model.ARIMA(series, [p,d,q], method='css').fit() # return model.aic ``` ```python # fit = opt.minimize(hat, -2, (a, b)) ``` ```python # fig, ax = plt.subplots(figsize=(10,5)) # ax.plot(np.arange(1,787), tech.FB.values) # fig = arima.plot_predict(649, 785, dynamic=True, ax=ax, # plot_insample=False) ``` ___ ```python from IPython.display import HTML HTML("""<style>@import "http://fonts.googleapis.com/css?family=Lato|Source+Code+Pro|Montserrat:400,700";@font-face{font-family:"Computer Modern";src:url('http://mirrors.ctan.org/fonts/cm-unicode/fonts/otf/cmunss.otf')}.rendered_html h1,h2,h3,h4,h5,h6,p{font-family:'Computer Modern'}p,ul{font-family:'Computer Modern'}div#notebook-container{-webkit-box-shadow:none;box-shadow:none}h1{font-size:70pt}h2{font-size:50pt}h3{font-size:40pt}h4{font-size:35pt}h5{font-size:30pt}h6{font-size:25pt}.rendered_html p{font-size:11pt;line-height:14pt}.CodeMirror pre{font-family:'Source Code Pro', monospace;font-size:09pt}div.input_area{border:none;background:#f5f5f5}ul{font-size:10.5pt;font-family:'Computer Modern'}ol{font-size:11pt;font-family:'Computer Modern'}.output_png img{display:block;margin-left:auto;margin-right:auto}</style>""") ``` <style>@import "http://fonts.googleapis.com/css?family=Lato|Source+Code+Pro|Montserrat:400,700";@font-face{font-family:"Computer Modern";src:url('http://mirrors.ctan.org/fonts/cm-unicode/fonts/otf/cmunss.otf')}.rendered_html h1,h2,h3,h4,h5,h6,p{font-family:'Computer Modern'}p,ul{font-family:'Computer Modern'}div#notebook-container{-webkit-box-shadow:none;box-shadow:none}h1{font-size:70pt}h2{font-size:50pt}h3{font-size:40pt}h4{font-size:35pt}h5{font-size:30pt}h6{font-size:25pt}.rendered_html p{font-size:11pt;line-height:14pt}.CodeMirror pre{font-family:'Source Code Pro', monospace;font-size:09pt}div.input_area{border:none;background:#f5f5f5}ul{font-size:10.5pt;font-family:'Computer Modern'}ol{font-size:11pt;font-family:'Computer Modern'}.output_png img{display:block;margin-left:auto;margin-right:auto}</style>

78a03238bda446e724c8a9d5b313cfbba2341529

ipynb

Jupyter Notebook

Example-Project/03-LogNormalRandomWalk.ipynb

wileong/data_science_projects_directory

1a2e018bd6e8e0b97a8b6df1fa074f1a369d4318

Example-Project/03-LogNormalRandomWalk.ipynb

wileong/data_science_projects_directory

1a2e018bd6e8e0b97a8b6df1fa074f1a369d4318

Example-Project/03-LogNormalRandomWalk.ipynb

wileong/data_science_projects_directory

1a2e018bd6e8e0b97a8b6df1fa074f1a369d4318

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Tutorial ## Regime-Switching Model `regime_switch_model` is a set of algorithms for learning and inference on regime-switching model. Let $y_t$ be a $p\times 1$ observed time series and $h_t$ be a homogenous and stationary hidden Markov chain taking values in $\{1, 2, \dots, m\}$ with transition probabilities \begin{equation} w_{kj} = P(h_{t+1}=j\mid h_t=k), \quad k,j=1, \dots, m \end{equation} where the number of hidden states $m$ is known. It is assumed that the financial market in each period can be realized as one of $m$ regime. Furthermore, the regimes are characterized by a set of $J$ risk factors, which represent broad macro and micro economic indicators. Let $F_{tj}$ be the value of the $j$th risk factor $(j=1, \dots, J)$ in period $t$. Correspondingly, $F_t$ is the vector of risk factors in period $t$. We assumes that, for $t=1, \dots, n$, when the market is in regime $h_t$ in period $t$, \begin{equation} y_t = u_{h_t} + B_{h_t}F_t + \Gamma_{h_t}\epsilon_t, \end{equation} where $\epsilon_t \sim N(0,I)$. The model parameters $\{u_{h_t}, B_{h_t}, \Gamma_{h_t}\}$ depend on the regime $h_t$. $u_{h_t}$ is the state-dependent intercepts of the linear factor model. The matrix $B_{h_t}$ defines the sensitivities of asset returns to the common risk factors in state $h_t$ and is often called the loading matrix. `regime_switch_model` solves the following fundamental problems: * Given the observed data, estimate the model parameters * Given the model parameters and observed data, estimate the optimal sequence of hidden states The implementation of code is based on the well-known Baum-Welch algorithm and Viterbi algorithm that are widely used in hidden Markov model. ```python import numpy as np import pandas as pd from regime_switch_model.rshmm import * ``` ## Generate samples based on the regime-switching model ```python model = HMMRS(n_components=2) # startprob model.startprob_ = np.array([0.9, 0.1]) # transition matrix model.transmat_ = np.array([[0.9, 0.1], [0.6, 0.4]]) # risk factor matrix # read file from Fama-French three-factor data Fama_French = pd.read_csv('Global_ex_US_3_Factors_Daily.csv', skiprows=3) Fama_French.rename(columns={'Unnamed: 0': 'TimeStamp'}, inplace=True) Fama_French.replace(-99.99, np.nan); Fama_French.replace(-999, np.nan); # select data #Fama_French_subset = Fama_French[(Fama_French['TimeStamp'] >= 20150101) & (Fama_French['TimeStamp'] <= 20171231)] Fama_French_subset = Fama_French Fama_French_subset.drop(['TimeStamp', 'RF'], axis=1, inplace=True) F = np.hstack((np.atleast_2d(np.ones(Fama_French_subset.shape[0])).T, Fama_French_subset)) # loading matrix with intercept loadingmat1 = np.array([[0.9, 0.052, -0.02], [0.3, 0.27, 0.01], [0.12, 0.1, -0.05], [0.04, 0.01, -0.15], [0.15, 0.04, -0.11]]) intercept1 = np.atleast_2d(np.array([-0.015, -0.01, 0.005, 00.1, 0.02])).T model.loadingmat_ = np.stack((np.hstack((intercept1, loadingmat1)), np.hstack((0.25*intercept1, -0.5* loadingmat1))), axis=0) # covariance matrix n_stocks = 5 rho = 0.2 Sigma1 = np.full((n_stocks, n_stocks), rho) + np.diag(np.repeat(1-rho, n_stocks)) model.covmat_ = np.stack((Sigma1, 10*Sigma1), axis=0) save = True # sample Y, Z = model.sample(F) ``` ## Split data into training and test ```python # Use the last 300 day as the test data Y_train = Y[:-300,:] Y_test = Y[-300:,:] F_train = F[:-300,:] F_test = F[-300:,:] ``` ## Fitting Regime-Switch Model ```python remodel = HMMRS(n_components=2, verbose=True) remodel.fit(Y_train, F_train) Z2, logl, viterbi_lattice = remodel.predict(Y_train, F_train) ``` 1 -63535.1590 nan 2 -60972.1968 2562.9622 3 -59533.7367 1438.4601 4 -56005.9127 3527.8240 5 -54584.0500 1421.8628 6 -54259.0186 325.0314 7 -54199.8384 59.1802 8 -54192.7580 7.0804 9 -54192.0477 0.7103 10 -54191.9793 0.0684 11 -54191.9727 0.0065 12 -54191.9721 0.0006 13 -54191.9720 0.0001 ### Examine model parameters ```python np.set_printoptions(precision=2) print("Number of data points = ", Y_train.shape[0]) print(" ") print("Starting probability") print(remodel.startprob_) print(" ") print("Transition matrix") print(remodel.transmat_) print(" ") print("Means and vars of each hidden state") for i in range(remodel.n_components): print("{0}th hidden state".format(i)) print("loading matrix = ", remodel.loadingmat_[i]) print("covariance = ", remodel.covmat_[i]) print(" ") ``` ('Number of data points = ', 6723) Starting probability [ 1.78e-13 1.00e+00] Transition matrix [[ 0.37 0.63] [ 0.11 0.89]] Means and vars of each hidden state 0th hidden state ('loading matrix = ', array([[-0.14, -0.53, -0.15, -0.03], [-0.04, -0.04, -0.17, 0.06], [ 0.04, -0.15, -0.23, -0.02], [ 0.02, 0.02, 0.19, -0. ], [ 0.01, 0.07, -0.04, 0.24]])) ('covariance = ', array([[ 10.55, 1.92, 2.14, 1.56, 1.37], [ 1.92, 9.53, 1.68, 2.19, 1.74], [ 2.14, 1.68, 9.44, 1.85, 1.71], [ 1.56, 2.19, 1.85, 9.89, 2.62], [ 1.37, 1.74, 1.71, 2.62, 10.24]])) 1th hidden state ('loading matrix = ', array([[ -6.79e-03, 8.93e-01, 3.71e-02, 4.66e-02], [ -2.98e-02, 2.93e-01, 2.42e-01, 1.15e-01], [ 7.01e-04, 1.09e-01, 6.24e-02, -2.15e-02], [ 9.63e-02, 4.09e-02, -1.29e-02, -1.78e-01], [ 1.62e-02, 1.48e-01, -2.12e-04, -1.60e-01]])) ('covariance = ', array([[ 0.98, 0.2 , 0.24, 0.19, 0.2 ], [ 0.2 , 0.98, 0.21, 0.21, 0.17], [ 0.24, 0.21, 1.01, 0.19, 0.21], [ 0.19, 0.21, 0.19, 0.94, 0.19], [ 0.2 , 0.17, 0.21, 0.19, 0.98]])) ### Examine the predicted hidden state ```python print("Prediction accuracy of the hidden states = ", np.mean(np.equal(Z[:-300], 1-Z2))) ``` ('Prediction accuracy of the hidden states = ', 0.9840844860925182)

df3ebad12407026b9c838ae10451ec067ca44092

ipynb

Jupyter Notebook

examples/tutorial.ipynb

Liuyi-Hu/regime_switch_model

1da6ab9cf989f3b6363f628c88138eebf3215277

2018-04-16T20:44:01.000Z

2022-03-27T13:03:37.000Z

examples/tutorial.ipynb

arita37/regime_switch_model

1da6ab9cf989f3b6363f628c88138eebf3215277

2019-06-29T18:56:13.000Z

2020-04-06T04:04:57.000Z

examples/tutorial.ipynb

arita37/regime_switch_model

1da6ab9cf989f3b6363f628c88138eebf3215277

2018-02-01T07:44:10.000Z

2021-07-03T12:25:05.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Bayesian Linear Regression ## What is the problem? Given inputs $X$ and outputs $\mathbf{y}$, we want to find the best parameters $\boldsymbol{\theta}$, such that predictions $\hat{\mathbf{y}} = X\boldsymbol{\theta}$ can estimate $\mathbf{y}$ very well. In other words, we want L2 norm of errors $||\hat{\mathbf{y}} - \mathbf{y}||_2$, as low as possible. ## Applying Bayes Rule In this problem, we will {ref}`model the distribution of parameters <parameters-framework>`. \begin{equation} \underbrace{p(\boldsymbol{\theta}|X, \mathbf{y})}_{\text{Posterior}} = \frac{\overbrace{p(\mathbf{y}|X, \boldsymbol{\theta})}^{\text{Likelihood}}}{\underbrace{p(\mathbf{y}|X)}_{\text{Evidence}}}\underbrace{p(\boldsymbol{\theta})}_{\text{Prior}} \end{equation} \begin{equation} p(\mathbf{y}|X) = \int_{\boldsymbol{\theta}}p(\mathbf{y}|X, \boldsymbol{\theta})p(\boldsymbol{\theta})d\boldsymbol{\theta} \end{equation} We are interested in posterior $p(\boldsymbol{\theta}|X, \mathbf{y})$ and to derive that, we need prior, likelihood and evidence terms. Let us look at them one by one. ### Prior Let's assume a multivariate Gaussian prior over the $\boldsymbol{\theta}$ vector. $$ p(\theta) \sim \mathcal{N}(\boldsymbol{\mu}_0, \Sigma_0) $$ ### Likelihood Given a $\boldsymbol{\theta}$, our prediction is $X\boldsymbol{\theta}$. Our data $\mathbf{y}|X$ will have some irreducible noise which needs to be incorporated in the likelihood. Thus, we can assume the likelihood distribution over $\mathbf{y}$ to be centered at $X\boldsymbol{\theta}$ with random i.i.d. homoskedastic noise with variance $\sigma^2$: $$ p(\mathbf{y}|X, \theta) \sim \mathcal{N}(X\boldsymbol{\theta}, \sigma^2I) $$ ### Maximum Likelihood Estimation (MLE) Let us find the optimal parameters by differentiating likelihood $p(\mathbf{y}|X, \boldsymbol{\theta})$ w.r.t $\boldsymbol{\theta}$. \begin{equation} p(\mathbf{y}|X, \boldsymbol{\theta}) = \frac{1}{\sqrt{(2\pi)^n |\sigma^2I|}}\exp \left( (\mathbf{y} - X\boldsymbol{\theta})^T(\sigma^2I)^{-1}(\mathbf{y} - X\boldsymbol{\theta}) \right) \end{equation} Simplifying the above equation: \begin{equation} p(\mathbf{y}|X, \boldsymbol{\theta}) = \frac{1}{(2\pi\sigma^2)^{\frac{n}{2}}}\exp \left( \sigma^{-2}(\mathbf{y} - X\boldsymbol{\theta})^T(\mathbf{y} - X\boldsymbol{\theta}) \right) \end{equation} Taking log to simplify further: \begin{align} \log p(\mathbf{y}|X, \boldsymbol{\theta}) &= (\mathbf{y} - X\boldsymbol{\theta})^T(\mathbf{y} - X\boldsymbol{\theta}) + \log \sigma^{-2} + \log \frac{1}{(2\pi\sigma^2)^{\frac{n}{2}}}\\ \frac{d}{d\boldsymbol{\theta}} \log p(\mathbf{y}|X, \boldsymbol{\theta}) &= \frac{d}{d\boldsymbol{\theta}}(\mathbf{y} - X\boldsymbol{\theta})^T(\mathbf{y} - X\boldsymbol{\theta})\\ &= \frac{d}{d\boldsymbol{\theta}}(\mathbf{y}^T - \boldsymbol{\theta}^TX^T)(\mathbf{y} - X\boldsymbol{\theta})\\ &= \frac{d}{d\boldsymbol{\theta}} \left[ \mathbf{y}^T\mathbf{y} - \mathbf{y}^TX\boldsymbol{\theta} - \boldsymbol{\theta}^TX^T\mathbf{y} + \boldsymbol{\theta}^TX^TX\boldsymbol{\theta}\right]\\ &= -(\mathbf{y}^TX)^T - X^T\mathbf{y} + 2X^TX\boldsymbol{\theta} = 0\\ \therefore X^TX\boldsymbol{\theta} &= X^T\mathbf{y}\\ \therefore \boldsymbol{\theta}_{MLE} &= (X^TX)^{-1}X^T\mathbf{y} \end{align} We used some of the formulas from [this cheatsheet](http://www.gatsby.ucl.ac.uk/teaching/courses/sntn/sntn-2017/resources/Matrix_derivatives_cribsheet.pdf) but they can also be derived from scratch. ### Maximum a posteriori estimation (MAP) We know from {ref}`the previous discussion <MAP-1>` that: $$ \arg \max_{\boldsymbol{\theta}} p(\boldsymbol{\theta}|X, \mathbf{y}) = \arg \max_{\boldsymbol{\theta}} p(\mathbf{y}|X, \boldsymbol{\theta})p(\boldsymbol{\theta}) $$ Now, differentiating $p(\mathbf{y}|X, \boldsymbol{\theta})p(\boldsymbol{\theta})$ w.r.t $\theta$ by reusing some of the steps from MLE:

0d05335fe186302c056309541eb50d3aca9f16c0

ipynb

Jupyter Notebook

linear-regression.ipynb

patel-zeel/bayesian-ml

2b7657f22fbf70953a91b2ab2bc321bb451fa5a5

linear-regression.ipynb

patel-zeel/bayesian-ml

2b7657f22fbf70953a91b2ab2bc321bb451fa5a5

linear-regression.ipynb

patel-zeel/bayesian-ml

2b7657f22fbf70953a91b2ab2bc321bb451fa5a5

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Node Embeddings and Skip Gram Examples **Purpose:** - to explore the node embedding methods used for methods such as Word2Vec. **Introduction-** one of the key methods used in node classification actually draws inspiration from natural language processing. This based in the fact that one approach for natural language processing views the ordering of words in a manner similar to a graph since each n-gram has a set of words that follow it. Strategies that treat text this way are naturally amenable to domains where we are explicitly working on a network structure. Methods which employ node embeddings have several fundamental steps: 1. Create a "corpus" of node connections using a random walk. 2. Define a transformation on the list of node connections from **1** which groups node values that are close together with a high number, and nodes that have less of a relationship with a small number. 3. Run a standard machine learning method on the new set of factors from step **2**. ## Random Walks: Here we explore the first step in this process: The random choosing of node values in the graph structure. This step is taken to approximate the connections each node has as a list. This carries two advantages: 1. Each node similarity measure has both local (direct) connections, and also expresses higher order connections (indirect). This is known as **Expressivity**. 2. All node pairs don't need to be encoded; we don't have to worry about coding the zero probabilities. This is **Efficiency**. We will discuss some of the methods used for random walks in the sections below in reference to the paper where they were originally discussed. ### DeepWalk Method *DeepWalk: Online Learning of Social Representations* uses short random walks. In this case, we define a random walk starting at vertex $V_i$ as $W_i$. This random walk is a stochastic process composed of random variables $W_i^k$ where k denotes the step in the sequence of each random walk. For this method, a stream of random walks is created. This method has the added advantage of being easy to parallelize and is also less sensitive to changes in the underlying graph than using a larger length random walk. The implementation of the DeepWalk method is used in the function below: ```python import pandas as pd, numpy as np, os, random from IPython.core.debugger import set_trace np.random.seed(13) dat = pd.read_csv("../Data/soc-sign-bitcoinalpha.csv", names = ["SOURCE", "TARGET", "RATING", "TIME"]) ``` ```python len(pd.unique(dat.SOURCE)) ``` 3286 ```python len(pd.unique(dat.TARGET) ) ``` 3754 ```python #from_vals = pd.unique(dat.SOURCE) #a = dat.TARGET[dat.SOURCE == from_vals[1]] # Generate list comprehension using from values as a key; to values are saved as a list. #node_lists = {x:dat.TARGET[dat.SOURCE == x].values for x in from_vals } # Generate a step by selecting one value randomly from the list of "to" nodes: def gen_step(key_val,dict_vals): # print(dict_vals[key_val]) return( dict_vals[key_val][random.randint(0,len(dict_vals[key_val])-1)] ) def gen_walk(key_val,dict_vals,steps): walk_vals = [key_val] for i in range(0,steps-1): walk_vals.append(gen_step(walk_vals[-1],dict_vals) ) return(walk_vals) def RW_DeepWalk( orig_nodes, to_vals, walk_length=3): from_vals = pd.unique(orig_nodes) node_lists = {x:to_vals[orig_nodes == x].values for x in from_vals} start_nodes = [* node_lists] start_nodes=[x for x in start_nodes if x in node_lists.keys()] walks = {x:gen_walk(key_val= x,dict_vals = node_lists,steps=walk_length) for x in start_nodes} return(walks) ``` ```python # In order to sort these values, we need to make a full list of "from" and "to" for the random walk. This is performed in the script below: # Identify values in "to" column that might not be in the from column: f = dat.SOURCE t = dat.TARGET unique_t = [x for x in pd.unique(t) if not(x in pd.unique(f))] x_over = dat[dat['TARGET'].isin( unique_t)] # Add entries from the "to" column to the from column; add corresponding entries from the "from" column. This way, we include mappings of nodes in the "to" column as part of the random walk. full_from = f.append(x_over.TARGET) full_to = t.append(x_over.SOURCE) ``` ```python random_walk = RW_DeepWalk( full_from, full_to, walk_length=10) ``` An example of one of the arrays obtained using a random walk: ```python random_walk[1] ``` [1, 2273, 2202, 1134, 35, 1385, 114, 1202, 605, 230] The choice of the random walk method provides a way of representing the network that can be performed quickly. This method is also simple to parallelize. Finally, this method and the speed it can be used allows for a quick way to update calculations due to changes in the graph structure. ### Node2vec Method The paper "Scalable Feature Learning for Networks" uses a separate method called a "biased random walk". One of the points made in the paper is the type of sampling strategies that can be used to try to approximate the neighborhood around some node (this is denoted as $N_s$ in the paper). There are two extremes for sampling strategies that can be employed: * Breadh-first sampling (BFS) - The neighborhood is restricted to nodes which are immediate neighbors of the source node. For this, we define the neighborhood **only** with directly adjacent nodes. * Depth-first sampling (DFS) - The neighborhood consists of nodes sequentially sampled at increasing distances from the source node. This is represented in the random walk algorithm that was shown in the last section. A biased random walk as expressed by the authors is an interpolation between the two strategies mentioned above. Let $u$ be the source node, and $l$ be the length of the random walk. Let $c_i$ be the $i$th node in the walk where $c_0 = u$. Then, $c_i$ is generated as follows: $$ P(c_i = x | c_{i-1} =v) = \frac{\pi_{v,x} }{Z} $$ and 0 otherwise. Where $\pi_{v,x}$ is the unnormalized transition probability between nodes $v$ and $x$, and $Z$ is some constant that normalizes the probability between the two nodes. This is very similar to the formulation that was desecribed earlier for DeepWalk. The simplest way to introduce bias to the random walks is to sample based onthe static edge weights: $w_{v,x} = \pi_{v,x} $. In the case of an unweighted graph like the one used in the example above, $w_{v,x} =1$. We will define a $2$nd order random walk with parameters $p,q$. We will set the unnoramlized transition probability to $\pi_{v,x} = \alpha_{p,q}(t,x)*w_{v,x}$ where \alpha_{p,q}(t,x) is defined as: \begin{equation} \alpha_{p,q}(t,x) = \begin{cases} \frac{1}{p} & \text{if $d_{t,x}=0$ }\\ 1 & \text{if $d_{t,x}=1$ }\\ \frac{1}{q} & \text{if $d_{t,x}=2$ } \end{cases} \end{equation} Where $d_{t,x}$ defines the shortest path distance between nodes $t$ and $x$ Also note that $d_{t,x} \in \{0,1,2\}$ Changing parameters $p$ and $q$ will impact the speed that the walk leaves the current neighborhood. In the example provided in the paper, the authors consider a process which as just transitioned to node *v* from node *t*. It has three potential choices for its next step: * Transition back to *t* with the bias of $\alpha_{t,v} = \frac{1}{p}$ being applied. * Transition to a shared node with a bias of 1 being applied. * Transition to an unshared node with a bias of $\alpha_{t,v} = \frac{1}{q}$ being applied. Then - a lower q-value and higher p-value will increase the likelihood of leaving the initial neighborhood of *t*. At the extreme, you would get the original random walk implementation described above by letting $p =1$ and $q=1$. A higher q value will decrease the likelihood of the current step moving to a node that neig ```python from_vals = pd.unique(full_from) node_lists = {x:full_to[full_from == x].values for x in from_vals} node_lists ``` {7188: array([1]), 430: array([ 1, 13, 59, 247, 831, 817, 1055, 7595, 7509]), 3134: array([ 1, 22, 27, 617]), 3026: array([1]), 3010: array([1]), 804: array([ 1, 25, 26, 85, 204, 7583, 1020]), 160: array([ 1, 18, 57, 89, 294, 7579, 952, 1845, 817, 945]), 95: array([ 1, 3, 4, 6, 7, 8, 11, 19, 24, 25, 26, 29, 31, 32, 33, 36, 38, 40, 41, 42, 43, 47, 56, 62, 67, 73, 75, 82, 92, 93, 188, 394, 1829, 493, 526, 391, 315, 242, 331, 5679, 179, 221, 966, 345, 411, 278, 2410, 3403, 245, 464, 1065, 2336, 191, 205, 105, 1889, 154, 2953, 373, 3302, 1370, 666, 5342, 1874, 136, 3246, 413, 246, 2358, 553, 3179, 1045, 332, 244, 1278, 104, 174, 2330, 1307, 241, 7432, 7550, 172, 643, 2304, 111, 752, 941, 1171, 318, 1348, 123, 185, 882, 813, 228, 396, 362, 428, 7497, 103, 3134, 2257, 177, 7339, 145, 424, 124, 7580, 288, 125, 2369, 7602, 7601, 7599, 7598, 7604, 7574, 7581, 7584, 7600, 7548]), 377: array([ 1, 363, 399, 3413]), 888: array([ 1, 58, 213, 261, 1886]), 89: array([ 1, 2, 10, 28, 29, 37, 41, 48, 77, 85, 373, 395, 160, 369, 175, 124, 2269, 699, 519, 122, 945, 587, 946, 154, 629, 2097, 516, 1708, 2096, 253, 90, 1690, 406, 1267, 2628, 2625, 2623, 302, 931, 1453, 254, 132, 3774, 2043, 7439, 7564]), 1901: array([1]), 161: array([ 1, 2, 7, 10, 31, 41, 43, 69, 77, 78, 103, 156, 550, 635, 260, 1322, 1022, 7600, 695, 1324, 1039, 623, 454, 2213, 592, 1765, 2209, 1770, 1505, 520, 690, 1494, 204, 1140, 1313, 798]), 256: array([ 1, 9, 18, 38, 40, 42, 57, 60, 65, 67, 69, 75, 121, 156, 175, 236, 1739, 329, 672, 485, 632, 947, 479, 4934, 1970, 943, 2792, 2790, 346, 7592]), 351: array([ 1, 3, 12, 34, 36, 58, 82, 96, 105, 120, 222, 432, 1893, 1594, 1924, 759]), 3329: array([1]), 3341: array([1]), 649: array([ 1, 12, 33, 34, 105, 118, 123, 125, 310, 1192, 1885, 650, 1197]), 1583: array([1]), 87: array([ 1, 2, 3, 7, 15, 19, 27, 31, 33, 42, 64, 67, 85, 86, 397, 126, 1070, 206, 707, 424, 602, 454, 2080, 241, 158, 96, 291, 192, 171, 1045, 157, 526, 3318, 1884, 702, 259, 280, 2155, 163, 7412, 309, 154, 3306, 1863, 1880, 1156, 288, 1381, 572, 592, 395, 133, 214, 491, 7415, 7552, 7591, 7473, 7542]), 37: array([ 1, 2, 4, 9, 10, 14, 15, 16, 20, 21, 25, 28, 31, 1019, 7552, 312, 297, 54, 159, 314, 657, 113, 89, 479, 380, 601, 41, 168, 7403, 185, 594, 166, 66, 570, 610, 561, 1926, 990, 7400, 1076, 1764, 75, 1025, 65, 140, 97, 1399, 91, 40, 1632, 474, 7371, 7564, 7448]), 309: array([ 1, 3, 5, 11, 42, 87, 92, 125, 248, 292, 416, 3334, 1885, 1353, 637, 702, 1180, 894, 759, 3274, 3172, 3254, 1372]), 821: array([ 1, 92, 334, 648]), 1496: array([ 1, 330]), 637: array([ 1, 7, 52, 58, 309, 432, 759]), 964: array([ 1, 30, 46, 162]), 594: array([ 1, 18, 37, 1582, 1522]), 2249: array([1]), 554: array([ 1, 8, 600]), 20: array([ 1, 2, 6, 9, 11, 14, 15, 17, 22, 31, 37, 47, 188, 65, 1042, 159, 231, 730, 489, 372, 66, 366, 52, 152, 176, 145, 1301, 211, 38, 892, 491, 122, 90, 241, 212, 586, 272, 729, 41, 83, 807, 1771, 45, 108, 26, 131, 155, 870, 1269, 198, 1145, 60, 1127, 177, 7565, 7577, 7547]), 2227: array([1]), 1315: array([ 1, 793]), 519: array([ 1, 6, 27, 49, 89, 110, 142, 146, 453, 745, 1164, 1288, 2899, 946]), 1316: array([1]), 2149: array([ 1, 35, 943]), 1724: array([ 1, 14]), 18: array([ 1, 11, 243, 74, 294, 57, 160, 91, 860, 1170, 594, 220, 1306, 455, 690, 872, 7565, 868, 346, 194, 1564, 1800, 72, 1130, 729, 1703, 945, 1164, 2228, 260, 2200, 117, 256, 7603, 543, 736, 384, 7372, 1442, 733, 386, 2091, 1271, 2693, 2086, 866, 79, 151, 454, 1526, 1525, 3053, 2247, 954, 2961, 2199, 1305, 38, 2803, 121, 2690, 2118, 40, 50, 1284, 613, 2758, 542, 2097, 630, 1477, 26, 154, 274, 33, 2106, 2710, 437]), 57: array([ 1, 10, 18, 35, 38, 50, 294, 160, 630, 1135, 361, 1480, 389, 1719, 664, 2160, 1733, 1689, 2137, 1728, 304, 793, 2120, 2107, 323, 252, 1963, 150, 2656, 583, 1742, 7579, 117, 874, 256, 346, 872, 1442, 2766, 2098, 2106, 2093, 75, 614, 2576, 79, 2508, 357, 2637, 2552, 2595, 7564, 994, 109, 716, 65, 407, 1657]), 118: array([ 1, 5, 10, 12, 19, 27, 32, 34, 36, 43, 58, 72, 82, 85, 103, 105, 115, 765, 1592, 292, 526, 141, 125, 375, 238, 528, 222, 603, 158, 173, 649, 163, 822, 214, 1580, 706, 398, 7597]), 3254: array([ 1, 3, 5, 19, 36, 191, 309, 432, 433, 641, 1147]), 1177: array([ 1, 220, 243]), 112: array([ 1, 6, 8, 13, 26, 30, 33, 56, 73, 427, 524, 177, 701, 301, 748, 396, 124, 116, 2276, 7595, 7600, 3031, 1043, 697]), 11: array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 93, 34, 13, 31, 21, 47, 122, 24, 12, 26, 7562, 17, 20, 90, 29, 82, 1159, 133, 255, 14, 211, 15, 22, 734, 58, 105, 87, 128, 32, 19, 228, 42, 196, 41, 43, 411, 304, 171, 190, 103, 25, 482, 70, 318, 51, 156, 153, 77, 7328, 253, 45, 398, 207, 762, 3345, 198, 1887, 7446, 326, 125, 310, 1883, 647, 309, 155, 36, 290, 174, 1871, 248, 185, 1379, 3269, 1068, 242, 7514, 280, 216, 30, 123, 7434, 3224, 2356, 1562, 96, 966, 116, 124, 95, 1342, 74, 2257, 162, 320, 801, 205, 487, 295, 593, 488, 491, 78, 694, 83, 7377, 166, 372, 260, 1156, 623, 1512, 1775, 552, 1508, 2217, 88, 2918, 2915, 360, 550, 2167, 145, 639, 197, 1034, 276, 347, 188, 345, 272, 1493, 264, 949, 1755, 1145, 1754, 739, 1311, 277, 274, 2865, 2180, 1748, 1030, 799, 69, 100, 2168, 65, 316, 1292, 193, 2791, 7563, 942, 735, 135, 72, 39, 870, 1716, 7603, 66, 1697, 732, 1116, 408, 187, 79, 154, 777, 131, 327, 6878, 7445, 7597, 7525, 85, 7595, 7588, 7538, 680, 183, 7583, 177, 7593, 7564, 7499, 7577, 7573, 798, 7592, 4910]), 586: array([ 1, 14, 20, 135, 592, 1283, 2753]), 35: array([ 1, 2, 14, 15, 86, 64, 233, 1385, 126, 192, 104, 114, 3441, 558, 375, 491, 94, 2289, 7384, 7562, 279, 70, 7415, 80, 801, 7580, 85]), 15: array([ 1, 2, 6, 9, 10, 11, 13, 14, 352, 94, 233, 86, 64, 98, 35, 953, 114, 126, 192, 104, 605, 337, 2336, 498, 1894, 1386, 2436, 897, 1202, 766, 1074, 1590, 1596, 281, 558, 230, 96, 376, 87, 164, 958, 272, 91, 48, 137, 3451, 2437, 3443, 3448, 1396, 3447, 2287, 7325, 2434, 2433, 3442, 3382, 3439, 2432, 1075, 1390, 1599, 1906, 3438, 2422, 604, 906, 2431, 902, 1905, 3437, 653, 179, 3434, 7310, 3433, 269, 2289, 530, 1134, 88, 234, 95, 698, 159, 438, 20, 37, 70, 67, 681, 40, 155, 3410, 3430, 3431, 1208, 1904, 3427, 2428, 1903, 6921, 3091, 3286, 7369, 764, 905, 2312, 1393, 2426, 3407, 2293, 1205, 651, 358, 7161, 984, 186, 832, 2417, 2633, 1895, 900, 1070, 1387, 375, 1385, 2828, 427, 279, 824, 1891, 1873, 165, 1162, 3120, 1351, 967, 518, 767, 122, 24, 190, 354, 17, 196, 51, 29, 75, 274, 211, 1471, 853, 33, 682, 7335, 7330, 7321, 3372, 7312, 3428, 400, 19, 47, 85, 7368, 801, 7366, 7560, 7393, 55, 185, 3369, 80, 7552, 1249, 374, 7415, 491, 7400, 7580, 7603, 369, 7379, 7577, 200, 31, 469, 7562, 113, 267, 7518, 5342, 1921, 176, 188, 145]), 1445: array([1, 4]), 152: array([ 1, 9, 20, 22, 37, 38, 44, 45, 48, 54, 75, 91, 101, 113, 148, 234, 251, 1691, 216, 672, 253, 168, 626, 188, 908, 236, 988, 7552, 153, 769, 512, 1628]), 2: array([ 1, 37, 38, 168, 4, 20, 767, 23, 40, 91, 113, 54, 108, 469, 250, 271, 471, 185, 188, 31, 312, 74, 119, 8, 381, 223, 257, 47, 17, 9, 150, 7400, 263, 42, 109, 489, 145, 107, 22, 473, 99, 165, 24, 7, 189, 787, 264, 66, 562, 270, 871, 15, 137, 75, 353, 21, 658, 45, 121, 10, 164, 1380, 2404, 1691, 14, 13, 139, 206, 19, 318, 561, 51, 90, 241, 1494, 77, 133, 70, 211, 681, 272, 862, 140, 217, 182, 251, 1098, 106, 503, 1409, 446, 769, 1209, 917, 710, 1632, 2006, 2698, 1127, 647, 126, 43, 242, 87, 5, 111, 1156, 16, 412, 5342, 95, 85, 3, 219, 93, 18, 350, 268, 89, 1785, 306, 25, 161, 52, 97, 83, 117, 146, 2839, 1487, 67, 175, 587, 1475, 1709, 672, 39, 424, 1693, 35, 449, 1998, 1929, 1977, 236, 777, 63, 1957, 1635, 915, 65, 508, 2499, 2497, 299, 1421, 1225, 1232, 2488, 1936, 382, 355, 238, 1090, 1222, 564, 2459, 774, 420, 237, 209, 1922, 532, 504, 7401, 987, 2446, 200, 44, 986, 148, 1603, 379, 436, 512, 402, 11, 7508, 7552, 7500, 7482, 7583, 798, 7577, 7603]), 113: 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125, 457]), 599: array([ 5, 13, 61, 68, 127, 190, 191, 229, 244, 1188, 2320, 1537]), 245: array([ 5, 6, 11, 12, 19, 26, 32, 88, 95, 111, 120, 173, 177, 185, 244, 308, 526, 1890, 708, 2380, 1368, 1889, 7398, 397, 1186, 7560, 7600, 7604, 7497, 7562, 7559]), 1819: array([ 5, 13]), 3309: array([5]), 310: array([ 5, 11, 36, 49, 58, 116, 120, 707, 649, 7435, 7391, 759]), 695: array([ 5, 10, 103, 158, 161, 166, 315, 330, 451, 803]), 280: array([ 5, 6, 11, 12, 27, 29, 34, 47, 49, 58, 72, 87, 97, 106, 115, 142, 174, 244, 1032, 288, 704, 1173]), 3009: array([5]), 2386: array([ 5, 13]), 484: array([ 5, 7, 8, 12, 14, 21, 39, 52, 65, 93, 105, 125, 188, 208, 526, 1568, 1059, 1322]), 2354: array([ 5, 206, 597]), 1194: array([ 5, 972, 3311, 7531, 1869]), 2333: array([5, 8]), 825: array([ 5, 12, 27, 59, 82, 229, 3188, 3268]), 292: array([ 5, 11, 12, 19, 34, 36, 58, 72, 105, 116, 118, 125, 128, 141, 526, 321, 894, 462, 432, 1526, 309]), 1378: array([ 5, 26, 34, 36, 49, 145]), 1868: array([ 5, 1050, 3195, 3272, 1869]), 1854: array([ 5, 457]), 1376: array([ 5, 12, 59, 123, 372]), 692: array([ 5, 156, 2225]), 7444: array([ 5, 6, 12, 92, 3129, 3227]), 1859: array([ 5, 81, 518, 585]), 3236: array([5]), 2363: array([ 5, 206]), 3231: array([5]), 2320: array([ 5, 13, 599]), 706: array([ 5, 6, 27, 93, 116, 118, 128, 286, 398, 765]), 2335: array([ 5, 663]), 141: array([ 5, 19, 34, 49, 53, 82, 85, 92, 93, 96, 103, 110, 111, 118, 120, 458, 614, 292, 2325, 1180, 602, 751, 1370, 3389, 288, 321, 2406, 173, 1353, 7481, 7591]), 7531: array([ 5, 58, 127, 1194, 7521]), 3305: array([6]), 524: array([ 6, 112]), 809: array([ 6, 7, 317]), 1057: array([6, 7]), 1541: array([ 6, 177, 490, 7595]), 1371: array([6]), 1571: array([ 6, 8, 138]), 1185: array([ 6, 7, 12, 115]), 1566: array([ 6, 8, 27]), 810: array([6]), 1350: array([6]), 549: array([ 6, 7, 8, 17, 60, 107, 7603, 1579]), 526: array([ 6, 29, 49, 59, 73, 87, 92, 95, 110, 118, 125, 245, 292, 7591, 3216, 904, 1826]), 431: array([ 6, 30, 33, 45, 56, 156, 262, 278, 861, 1552, 610, 1549, 2340, 3201, 493]), 1546: array([ 6, 138]), 1461: array([ 6, 101, 2073]), 458: array([ 6, 8, 12, 125, 141, 525, 974]), 1374: array([ 6, 12, 19, 59]), 7420: array([ 6, 59]), 2376: array([ 6, 116, 149]), 3222: array([6]), 3158: array([6]), 2326: array([6]), 728: array([ 6, 17, 239, 2615]), 278: array([ 6, 29, 30, 31, 41, 62, 73, 95, 166, 177, 227, 320, 431, 380, 523, 1792, 876, 360, 1038, 1638, 7565, 7603, 7588]), 1830: array([ 6, 25, 138, 244, 362, 7554]), 7415: array([ 6, 863]), 104: array([ 6, 12, 15, 27, 35, 48, 49, 64, 86, 87, 94, 498, 281, 279, 2336, 192, 233, 114, 766, 900, 1879, 2437, 2293, 3445, 3443, 2432, 7321, 1397, 1396, 246, 966, 1597, 1388, 1596, 497, 378, 375, 376, 709, 984, 1351, 269, 832, 3353, 7603, 145, 7560, 491, 185, 801, 288, 7552]), 1157: array([ 6, 509, 548, 7595, 7527]), 2384: array([6]), 2371: array([ 6, 138, 372]), 1578: array([ 6, 8, 19]), 3176: array([6]), 3155: array([6]), 1254: array([ 6, 8, 21, 7564]), 1179: array([ 6, 27, 40, 56, 177]), 1055: array([ 6, 27, 430]), 7554: array([ 6, 27, 177, 1035, 1830]), 522: array([ 6, 8, 63, 171, 177, 883, 1337, 5342, 1156]), 1152: array([ 6, 7, 21, 295]), 3112: array([ 6, 138]), 3096: array([ 6, 15, 101]), 3066: array([6]), 188: array([ 6, 11, 15, 17, 54, 73, 99, 108, 110, 113, 124, 139, 145, 157, 161, 165, 177, 185, 200, 573, 196, 483, 469, 234, 489, 334, 241, 941, 484, 943, 251, 815, 1762, 7579, 7551, 7594, 7602, 7601, 7598, 7599, 7334, 7604, 7550, 7540, 7460, 7600]), 3092: array([6]), 750: array([ 6, 7, 30, 124, 177]), 1346: array([ 6, 7, 27, 30, 56]), 149: array([ 6, 16, 45, 50, 70, 99, 1072, 2294, 565, 467, 531, 974, 640, 485, 1969, 226, 169, 217, 155, 182, 1951, 660, 536, 911, 293, 1094, 2451, 237, 711, 419]), 2292: array([ 6, 962, 1349]), 1776: array([ 6, 25, 317, 876]), 7528: array([6, 7]), 1450: array([ 6, 28, 328, 407, 786, 2050]), 408: array([ 6, 11, 28, 33, 39, 48, 76, 77, 138, 195, 240, 313, 328, 365, 7564, 1453, 542, 986, 467, 622, 619]), 7517: array([ 7, 13, 20, 62, 145, 159, 166, 177, 5533, 7565, 7536, 7553]), 7536: array([ 7, 11, 62, 145, 159, 166, 5533, 7517, 7565, 7553]), 1151: array([7]), 429: array([ 7, 31, 34, 53, 125]), 956: array([ 7, 21, 2218]), 638: array([ 7, 9, 52, 424]), 5342: array([ 7, 8, 13, 15, 21, 29, 30, 41, 53, 67, 75, 76, 84, 103, 109, 135, 142, 154, 166, 172, 226, 294, 303, 320, 323, 328, 353, 357, 383, 394, 406, 412, 447, 451, 478, 507, 516, 522, 536, 643, 680, 686, 689, 754, 777, 813, 815, 994, 1235, 1328, 1401, 1611, 1789, 1790, 2280, 4934, 7482, 7500, 7584, 7392, 7405, 7407, 7556, 7509, 7418, 7595, 7530, 7581, 7551, 7594, 7430, 5389, 7600, 7427, 7538, 7373, 7586, 7590, 7565, 7588]), 1178: array([ 7, 8, 497]), 640: array([ 7, 42, 92, 149, 177, 2831]), 2211: array([7]), 98: array([ 7, 9, 15, 17, 35, 40, 51, 64, 66, 69, 70, 86, 87, 94, 114, 233, 192, 1272, 126, 337, 279, 1597, 352, 906, 558, 1074, 375, 164, 1387, 2336, 953, 281, 900, 634, 687, 211, 940]), 696: array([ 7, 8, 25, 32, 143, 179, 214, 350, 2299]), 878: array([ 7, 21, 177, 509, 5392]), 1782: array([7]), 1768: array([7]), 1310: array([7]), 417: array([ 7, 53, 58, 82, 110, 111, 412, 901, 520, 1199, 1582, 7591, 754, 7512]), 666: array([ 7, 19, 49, 88, 95, 125, 173, 207, 208, 212, 253, 397, 463, 494, 704]), 980: array([ 7, 19, 34, 82, 248, 336, 895, 7512]), 645: array([ 7, 8, 27, 81, 100, 108, 142, 155, 315, 596, 2343, 1060, 754, 1838, 3165, 3160]), 371: array([ 7, 24, 27, 32, 41, 47, 51, 70, 80, 133, 177, 199, 361, 417, 523, 1173, 622, 798, 1272, 672]), 7562: array([ 7, 11, 41, 185, 190, 245, 941, 1143]), 1171: array([ 7, 95, 177, 7595]), 2981: array([7]), 946: array([ 7, 10, 17, 21, 41, 52, 89, 103, 142, 146, 296, 519, 738, 798, 1741, 1746]), 506: array([ 7, 82, 90, 117, 145, 177, 201, 205, 7375]), 318: array([ 7, 8, 10, 11, 25, 26, 133, 145, 190, 228, 288, 656, 937, 7604]), ...} ```python gen_step(430,node_lists) ``` 1 ```python cur_node = gen_step(430,node_lists) prev_node_list = node_lists[cur_node] cur_node_list = node_lists[430] shared_nodes = list(set(prev_node_list) & set(cur_node_list)) unshared_nodes = list(set(prev_node_list) ^ set(cur_node_list)) prev_node = 430 ``` ```python def gen_biased_step(cur_val, prev_val,dict_vals,p = 1, q = 1): # set_trace() prev_node_list = node_lists[prev_val] cur_node_list = node_lists[cur_val] shared_nodes = list(set(prev_node_list) & set(cur_node_list)) unshared_nodes = list( set(prev_node_list) ^ set(cur_node_list)^set([prev_val]) ) all_nodes = shared_nodes + unshared_nodes + [prev_val] shared_weights = [1/p]*len(shared_nodes) unshared_weights = [1/q]*len(unshared_nodes) all_weights = shared_weights +unshared_weights + [1] # set_trace() node_step = random.choices(all_nodes,all_weights) return( node_step ) ``` ```python test = gen_biased_step(cur_val = 59, prev_val = 430,dict_vals = node_lists,p = 1, q = 1) test ``` [247] ```python def gen_walk_biased(key_val,dict_vals,steps,p=1,q=1): walk_vals = [key_val] for i in range(0,steps-1): if i==0: prev_val = key_val else: prev_val =walk_vals[-1] # set_trace() walk_vals.append( gen_biased_step( cur_val = key_val, prev_val = prev_val,dict_vals = dict_vals,p = p, q = q)[0] ) # gen_biased_step(cur_val = 59, prev_val = 430,dict_vals = node_lists,p = 1, q = 1) # walk_vals.append(gen_step(walk_vals[-1],dict_vals) ) # set_trace() return(walk_vals) # Split the node values into three different groups # Apply weightings to each edge to change the likelihood of leaving the neighborhood. # A biased random walk as described in the node2vec paper. The p and q values are defaulted to 1 which will make this the same as the RW_DeepWalk paper described earlier. def RW_Biased( orig_nodes, to_vals, walk_length=3,p = 1,q =1): from_vals = pd.unique(orig_nodes) node_lists = {x:to_vals[orig_nodes == x].values for x in from_vals} start_nodes = [* node_lists] start_nodes=[x for x in start_nodes if x in node_lists.keys()] # set_trace() # walks = {x:gen_walk_biased(key_val= start_nodes[x], prev_key = start_nodes[x-1],dict_vals = node_lists,steps=walk_length,p = p, q = q) for x in range(1,len(start_nodes))} # walks = {x:gen_walk(key_val= x,dict_vals = node_lists,steps=walk_length) for x in start_nodes} walks = [gen_walk(key_val= x,dict_vals = node_lists,steps=walk_length) for x in start_nodes] return(walks) ``` ```python full_from = full_from.astype(str ) full_to = full_to.astype(str ) ``` ```python test = RW_Biased(full_from, full_to,walk_length =10,p = .5, q = .7) test ``` [['7188', '1', '7557', '542', '48', '2114', '48', '270', '4', '131'], 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'2387', '27', '67'], ['371', '80', '269', '35', '1385', '35', '7580', '689', '7580', '689'], ['7562', '11', '360', '350', '1740', '255', '17', '176', '20', '241'], ['1171', '7', '768', '148', '768', '446', '4', '156', '9', '733'], ['2981', '7', '7501', '7', '2361', '2412', '96', '2412', '96', '269'], ['946', '10', '1031', '19', '88', '7404', '88', '10', '492', '115'], ['506', '90', '203', '89', '77', '4', '1088', '537', '45', '26'], ['318', '26', '7604', '7334', '145', '51', '7', '1731', '14', '1158'], ...] ## Creating a Node Embedding Now that we've created a representation of the likelihood of getting to different nodes in each graph, we can the methods which we will use to represent the network as an embedding vector. Note that this is an alternative to other methods such as one-hot encoding of the results which are extremely memory/computation intensive. In principle, what we want to do is represent the "context" or relationship of each of these nodes to all other nodes by mapping each node into an $N$ dimensional vector space. The length of the vector is arbitrary; As it is increased the precision will rise while the speed of the computation will fall.Nodes which are in the immediate neighborhood of the current node will be heavily favored, second order connections, less so, and those that are completely unconnected, not at all. This method was first explored in [Efficient Estimation of Word Representations in Vector Space](https://arxiv.org/pdf/1301.3781.pdf). The paper that was just mentioned provides two methods for natural language processing: 1. Continuous bag-of-words 2. Skip-gram models. Both methods are valid and have their strengths and weaknesses, but we will rely on skip-gram models in this discussion. For skip-gram models, the node embedding is generated using a simple neural network. We will step through an independent implementation of this below which leans on Tensorflow, but [Stellargraph](https://www.stellargraph.io/) provides a good straightforward interface to it as well. ### Step 1: Identify neighborhood for each node This is the step that we discussed above by implementing the biased random walk and the random walk methods. This has a key impact: The longer and more biased our random walk, the greater of range of connections we will identify, but we will possibly draw in more tenuous connections. ### Step 2: Map neighborhood values to one-hot autoencoders: The neighborhoods are used to generate vectors which encode the relationship of nodes. This includes a one-hot autoencoder for the target node, and a set of autoencoders for neighboring nodes. ### Step 3: Perform Optimization: The following procedure is used for each one-hot autoencoders: 1. The $1 \times N$ encoder multiplies an $N \times w $ matrix. 2. ```python ``` ```python ``` ```python ``` ```python ``` ```python ``` ```python def gen_auto_encoders(node_lists): ``` [['7188', '1', '7557', '542', '48', '2114', '48', '270', '4', '131'], ['430', '817', '430', '1055', '430', '817', '125', '249', '25', '1031'], ['3134', '27', '754', '755', '3', '227', '8', '107', '7565', '18'], ['3026', '1', '35', '192', '171', '73', '163', '59', '494', '11'], ['3010', '1', '472', '1235', '2588', '1235', '328', '661', '1007', '448'], ['804', '7583', '804', '7583', '1020', '51', '40', '51', '1020', '166'], ['160', '945', '18', '613', '226', '75', '411', '67', '69', '256'], ['95', '179', '88', '461', '88', '15', '80', '698', '64', '133'], ['377', '363', '399', '363', '377', '3413', '377', '3413', '377', '399'], ['888', '1', '725', '1661', '725', '10', '1040', '10', '2204', '874'], ['89', '3774', '1684', '3774', '1247', '102', '1005', '102', '62', '42'], ['1901', '1', '296', '203', '70', 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biased_rw_for_training[0].append(biased_rw_for_training[1]) final_results = [] for i in biased_rw_for_training: final_results = final_results + i ``` ```python ``` ```python biased_rw_for_training[0][0] ``` ```python from stellargraph.data import BiasedRandomWalk from stellargraph import StellarGraph from stellargraph import datasets from IPython.display import display, HTML from gensim.models import Word2Vec model = Word2Vec(biased_rw_for_training[0] , size=128, window=5, min_count=0, sg=1, workers=2, iter=1) ``` ## References: 1. [NRL Totorial Part 1](http://snap.stanford.edu/proj/embeddings-www/files/nrltutorial-part1-embeddings.pdf) ```python ```

b32e0d1c240dd77a99edca86372a5cf9ee8e0833

ipynb

Jupyter Notebook

Notes/Node Embeddings and Skip Gram Examples.ipynb

poc1673/ML-for-Networks

201ca30ab51954a7b1471740eb404b98f1d26213

Notes/Node Embeddings and Skip Gram Examples.ipynb

poc1673/ML-for-Networks

201ca30ab51954a7b1471740eb404b98f1d26213

Notes/Node Embeddings and Skip Gram Examples.ipynb

poc1673/ML-for-Networks

201ca30ab51954a7b1471740eb404b98f1d26213

Qwen/Qwen-72B

1. YES 2. YES

__label__krc_Cyrl

# Worksheet 5 ``` %matplotlib inline ``` ## Question 1 Explain when multistep methods such as Adams-Bashforth are useful and when multistage methods such as RK methods are better. ### Answer Question 1 Multistep methods are more computationally efficient (fewer function evaluations) and more accurate than multistage methods. However, they are not self-starting, difficult to adapt to use variable step sizes, and the theory to show that they are stable and convergent is more complex. They are most useful when efficiency is the primary concern and the system is sufficiently well controlled that equally spaced steps can be taken. In other situations, as discussed on worksheet 4, the self-starting simplicity combined with adaptive stepping means that multistage methods are preferrable. ## Question 2 Compute the coefficients of the AB3 algorithm. ### Answer Question 2 For Adams-Bashforth methods we have \begin{equation} y_{n+1} = y_n + b_{k−1} f_n + b_{k−2} f_{n−1} + \dots + b_0 f_{n+1−k}. \end{equation} Here we have $k = 3$ and so we have \begin{equation} y_{n+1} = y_n + h \left[ b_2 f_n + b_1 f_{n−1} + b_0 f_{n−2} \right] . \end{equation} We want to ensure that this gives an exact approximation of the integral form for polynomials of order s = $0, \dots, 2$. That is, we want \begin{equation} \int^{x_{n+1}}_{x_n} p_s (x) = h \left[ b_2 p_s (x_n) + b_1 p_s (x_{n−1}) + b_0 p_s (x_{n−2}) \right]. \end{equation} For simplicity, and without loss of generality, we set $x_n = 0$, and use the polynomials \begin{align} p_0(x) & = 1, \\ p_1(x) & = x, \\ p_2(x) & = x ( x + h ) \end{align} We then see that we get \begin{align} s & = 0: & \int_0^h 1 & = h \left[ b_2 \times 1 + b_1 \times 1 + b_0 \times 1 \right] \\ \implies && 1 & = b_2 + b_1 + b_0. \\ s & = 1: & \int_0^h x & = h \left[ b_2 \times 0 + b_1 \times (-h) + b_0 \times (-2 h) \right] \\ \implies && \frac{1}{2} & = -b_1 - 2 b_0. \\ s & = 2: & \int_0^h x ( x + h ) & = h \left[ b_2 \times 0 + b_1 \times 0 + b_0 \times (-2 h) (-h) \right] \\ \implies && \frac{5}{6} & = 2 b_0. \end{align} By back-substitution we find \begin{equation} b_0 = \frac{5}{12}, \quad b_1 = -\frac{4}{3}, \quad b_2 = \frac{23}{12}. \end{equation} This means the algorithm is \begin{equation} y_{n+1} = y_n + \frac{h}{12} \left[ 23 f_n − 16 f_{n−1} + 5 f_{n−2} \right]. \end{equation} ## Question 3 Explain the meaning of stability, consistency and convergence when applied to numerical methods for IVPs. State the theorem connecting these. ### Answer Question 3 *Stability*: The numerical solution is bounded at all iterations over a finite interval. I.e., if the true solution is $y(x)$ and $x \in [0, X]$ with $X$ finite, and we use $N + 1$ steps with $x_0 = 0$ and $x_N = X$, then $|y_i|$ is finite for all $i = 0, 1, \dots , N$, irrespective of the value of $N$. *Consistency*: The numerical method is a faithful representation of the differential equation to lowest order in $h$. That is, if you Taylor expand the numerical difference scheme and let $h \to 0$ you recover the original differential equation independent of the limiting process. *Convergence*: If $y(x)$ is the exact solution and $y(x; h)$ the numerical solution using step size $h$, in the limit as $h \to 0$ the numerical solution is the exact solution: \begin{equation} \lim_{h \to 0} y(x; h) = y(x). \end{equation} The theorem states that consistency and stability are equivalent to convergence. ## Question 4 Using the stability polynomial and your results above, check the order of accuracy and the stability of the 3 step Adams-Bashforth method. ### Answer Question 4 The coefficients of AB3 in the standard $k$-step formula notation are \begin{align} a_3 & = 1 & a_2 & = -1 & a_1 & = 0 & a_0 & = 0 \\ b_3 & = 0 & b_2 & = \frac{23}{12} & b_1 & = -\frac{4}{3} & b_0 & = \frac{5}{12}. \end{align} Therefore the stability polynomial is \begin{equation} p(z) = z^3 - z^2 \end{equation} with derivative \begin{equation} p'(z) = 3 z^2 - 2 z \end{equation} and the other required polynomial is \begin{equation} q(z) = \frac{1}{12} \left( 23 z^2 - 16 z + 5 \right). \end{equation} To check consistency we need that $p(1) = 0$ and $p'(1) = q(1)$, which we check: \begin{align} p(1) & = 1 - 1 \\ & = 0. \\ p'(1) - q(1) & = (3 - 2) - \frac{1}{12} (23 -16 + 5) \\ & = 1 - \frac{12}{12} \\ & = 0. \end{align} So the method is consistent. To check stability we have to find the roots of the stability polynomial $p(z)$. We write \begin{equation} p(z) = z^2 (z − 1) \end{equation} to see that the roots are 0 (twice) and 1, which means that the method satisfies the *strong* root condition implying both stability and relative stability, meaning it is a useful method. ## Coding Question 1 Apply the 2-step Adams-Bashforth method to the ODE from Worksheet 4, \begin{equation} y' + 2 y = 2 − e^{−4x}, \quad y(0) = 1. \end{equation} Use the Euler or Euler predictor-corrector method to start. Again, find the value of y(1) (analytic answer is $1 − (e^{−2} − e^{−4} )/2)$ and see how your method converges with resolution. ### Answer Coding Question 1 ``` def AB2(f, y0, interval, N = 100, start = 'Euler'): """Solve the IVP y' = f(x, y) on the given interval using N+1 points (counting the initial point) with initial data y0.""" import numpy as np h = (interval[1] - interval[0]) / N x = np.linspace(interval[0], interval[1], N+1) y = np.zeros((len(y0), N+1)) ff = np.zeros((len(y0), N+1)) y[:, 0] = y0 ff[:, 0] = f(x[0], y[:, 0]) if (start == 'Euler'): y[:, 1] = y0 + h * ff[:, 0] elif (start == 'Euler PC'): yp = y0 + h * ff[:, 0] y[:, 1] = y0 + h * ( ff[:, 0] + f(x[1], yp) ) / 2.0 else: raise Exception("Only allowed values for start are ['Euler', 'Euler PC']") ff[:, 1] = f(x[1], y[:, 1]) for i in range(1, N): y[:, i+1] = y[:, i] + h * ( 3.0 * ff[:, i] - ff[:, i-1] ) / 2.0 ff[:, i+1] = f(x[i+1], y[:, i+1]) return x, y def fn_q1(x, y): """Function defining the IVP in question 1.""" import numpy as np return 2.0 - np.exp(-4.0*x) - 2.0*y # Now do the test import numpy as np exact_y_end = 1.0 - (np.exp(-2.0) - np.exp(-4.0)) / 2.0 # Test at default resolution x, y = AB2(fn_q1, np.array([1.0]), [0.0, 1.0]) print "Error at the end point is ", y[:, -1] - exact_y_end import matplotlib.pyplot as plt fig = plt.figure(figsize = (12, 8), dpi = 50) plt.plot(x, y[0, :], 'b-+') plt.xlabel('$x$', size = 16) plt.ylabel('$y$', size = 16) # Now do the convergence test levels = np.array(range(4, 10)) Npoints = 2**levels abs_err_Euler = np.zeros(len(Npoints)) abs_err_EulerPC = np.zeros(len(Npoints)) for i in range(len(Npoints)): x, y = AB2(fn_q1, np.array([1.0]), [0.0, 1.0], Npoints[i]) abs_err_Euler[i] = abs(y[0, -1] - exact_y_end) x, y = AB2(fn_q1, np.array([1.0]), [0.0, 1.0], Npoints[i], 'Euler PC') abs_err_EulerPC[i] = abs(y[0, -1] - exact_y_end) # Best fit to the errors h = 1.0 / Npoints p_Euler = np.polyfit(np.log(h), np.log(abs_err_Euler), 1) p_EulerPC = np.polyfit(np.log(h), np.log(abs_err_EulerPC), 1) fig = plt.figure(figsize = (12, 8), dpi = 50) plt.loglog(h, abs_err_Euler, 'kx') plt.loglog(h, np.exp(p_Euler[1]) * h**(p_Euler[0]), 'k-') plt.loglog(h, abs_err_EulerPC, 'bo') plt.loglog(h, np.exp(p_EulerPC[1]) * h**(p_EulerPC[0]), 'b--') plt.xlabel('$h$', size = 16) plt.ylabel('$|$Error$|$', size = 16) plt.legend(('AB2 Errors (Euler start)', "Best fit line slope {0:.3}".format(p_Euler[0]), 'AB2 Errors (Euler PC start)', "Best fit line slope {0:.3}".format(p_EulerPC[0])), loc = "upper left") plt.show() ``` Both converge at order two: oddly, the results starting with the Euler predictor-corrector are noticeably worse. ## Coding Question 2 Apply the 2-step implicit Adams-Moulton method to the above ODE, using the 2-step Adams-Bashforth method as a predictor. Use the Euler or Euler predictor-corrector method to start. See how your method converges with resolution. ### Answer Coding Question 3 ``` def AM2(f, y0, interval, N = 100, start = 'Euler'): """Solve the IVP y' = f(x, y) on the given interval using N+1 points (counting the initial point) with initial data y0.""" import numpy as np h = (interval[1] - interval[0]) / N x = np.linspace(interval[0], interval[1], N+1) y = np.zeros((len(y0), N+1)) ff = np.zeros((len(y0), N+1)) y[:, 0] = y0 ff[:, 0] = f(x[0], y[:, 0]) if (start == 'Euler'): y[:, 1] = y0 + h * ff[:, 0] elif (start == 'Euler PC'): yp = y0 + h * ff[:, 0] y[:, 1] = y0 + h * ( ff[:, 0] + f(x[1], yp) ) / 2.0 else: raise Exception("Only allowed values for start are ['Euler', 'Euler PC']") ff[:, 1] = f(x[1], y[:, 1]) for i in range(1, N): # Adams-Bashforth 2 for the predictor step yp = y[:, i] + h * ( 3.0 * ff[:, i] - ff[:, i-1] ) / 2.0 # Adams-Moulton 2 for the corrector step y[:, i+1] = h * (ff[:, i] + f(x[i+1], yp)) / 2.0 ff[:, i+1] = f(x[i+1], y[:, i+1]) return x, y def fn_q2(x, y): """Function defining the IVP in question 2.""" import numpy as np return 2.0 - np.exp(-4.0*x) - 2.0*y # Now do the test import numpy as np exact_y_end = 1.0 - (np.exp(-2.0) - np.exp(-4.0)) / 2.0 # Test at default resolution x, y = AB2(fn_q2, np.array([1.0]), [0.0, 1.0]) print "Error at the end point is ", y[:, -1] - exact_y_end import matplotlib.pyplot as plt fig = plt.figure(figsize = (12, 8), dpi = 50) plt.plot(x, y[0, :], 'b-+') plt.xlabel('$x$', size = 16) plt.ylabel('$y$', size = 16) # Now do the convergence test levels = np.array(range(4, 10)) Npoints = 2**levels abs_err_AM2_Euler = np.zeros(len(Npoints)) abs_err_AM2_EulerPC = np.zeros(len(Npoints)) for i in range(len(Npoints)): x, y = AB2(fn_q2, np.array([1.0]), [0.0, 1.0], Npoints[i]) abs_err_AM2_Euler[i] = abs(y[0, -1] - exact_y_end) x, y = AB2(fn_q2, np.array([1.0]), [0.0, 1.0], Npoints[i], 'Euler PC') abs_err_AM2_EulerPC[i] = abs(y[0, -1] - exact_y_end) # Best fit to the errors h = 1.0 / Npoints p_Euler = np.polyfit(np.log(h), np.log(abs_err_AM2_Euler), 1) p_EulerPC = np.polyfit(np.log(h), np.log(abs_err_AM2_EulerPC), 1) fig = plt.figure(figsize = (12, 8), dpi = 50) plt.loglog(h, abs_err_AM2_Euler, 'kx') plt.loglog(h, np.exp(p_Euler[1]) * h**(p_Euler[0]), 'k-') plt.loglog(h, abs_err_AM2_EulerPC, 'bo') plt.loglog(h, np.exp(p_EulerPC[1]) * h**(p_EulerPC[0]), 'b--') plt.xlabel('$h$', size = 16) plt.ylabel('$|$Error$|$', size = 16) plt.legend(('AM2 Errors (Euler start)', "Best fit line slope {0:.3}".format(p_Euler[0]), 'AM2 Errors (Euler PC start)', "Best fit line slope {0:.3}".format(p_EulerPC[0])), loc = "upper left") plt.show() ``` The results are essentially identical to Adams-Bashforth 2. ``` from IPython.core.display import HTML def css_styling(): styles = open("../../IPythonNotebookStyles/custom.css", "r").read() return HTML(styles) css_styling() ``` <style> @font-face { font-family: "Computer Modern"; src: url('http://mirrors.ctan.org/fonts/cm-unicode/fonts/otf/cmunss.otf'); } div.cell{ width:800px; margin-left:16% !important; margin-right:auto; } h1 { font-family: Verdana, Arial, Helvetica, sans-serif; } h2 { font-family: Verdana, Arial, Helvetica, sans-serif; } h3 { font-family: Verdana, Arial, Helvetica, sans-serif; } div.text_cell_render{ font-family: Gill, Verdana, Arial, Helvetica, sans-serif; line-height: 110%; font-size: 120%; width:700px; margin-left:auto; margin-right:auto; } .CodeMirror{ font-family: "Source Code Pro", source-code-pro,Consolas, monospace; } /* .prompt{ display: None; }*/ .text_cell_render h5 { font-weight: 300; font-size: 12pt; color: #4057A1; font-style: italic; margin-bottom: .5em; margin-top: 0.5em; display: block; } .warning{ color: rgb( 240, 20, 20 ) } </style> > (The cell above executes the style for this notebook. It closely follows the style used in the [12 Steps to Navier Stokes](http://lorenabarba.com/blog/cfd-python-12-steps-to-navier-stokes/) course.)

2962f79baf57212f929725c9439fd61d611649ff

ipynb

Jupyter Notebook

Worksheets/Worksheet5_Notebook.ipynb

alistairwalsh/NumericalMethods

fa10f9dfc4512ea3a8b54287be82f9511858bd22

2021-12-01T09:15:04.000Z

2021-12-01T09:15:04.000Z

Worksheets/Worksheet5_Notebook.ipynb

indranilsinharoy/NumericalMethods

989e0205565131057c9807ed9d55b6c1a5a38d42

Worksheets/Worksheet5_Notebook.ipynb

indranilsinharoy/NumericalMethods

989e0205565131057c9807ed9d55b6c1a5a38d42

2021-04-13T02:58:54.000Z

2021-04-13T02:58:54.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import odeint ``` ### O Metódo de Euler Explicito ou Forward Euler #### Expansão em Taylor de uma função y Expansão em Série de Taylor de $y(t)$ centrada em $t_0$ é dada por $$ y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)}(t_0)}{n!}(t-t_0)^n $$ ##### Expansão de y até a primeira derivada Seja $h = t_n - t_{n-1}$ $$ y(t_{k+1}) = y(t_k) + y'(t_k)h + \mathcal{O}(h^2)\\ $$ O metódo de Euler explicito é um método recursivo de solução de equações diferenciais ordinárias, e consiste em utilizar a aproximação por Taylor e ignorar o erro $\mathcal{O}(h^2)$, nos dando: <br> $$ y_{n+1} \approx u_{n+1} = u_n + f(u_n,t_n) \cdot (t_{n+1} - t_n) $$ Com $y_n = y(t_n)$ a solução analitica no ponto $t_n$ e $u_n$ a aproximação númerica, $f(a,b)$ a derivada de $a$ em $b$ ```python ## F: Função derivada da função que queremos encontrar ## t0: tempo inicial ## y0: ponto inicial ## ts: range de tempo ## p: parametros particulares de cada modelo def f_euler(F, y0, ts, p = 0): ys = [y0] h = ts[1]-ts[0] for tnext in ts[1:]: ynext = ys[-1] + F(ys[-1],tnext,p)*h ys.append(ynext) t = tnext return np.array(ys) ``` ### O metódo de Runge-Kutta de segunda ordem Enquanto o metódo de Euler aproxima a solução em um ponto andando na tangente daquele ponto, o metódo de Runge-Kutta de segunda ordem aproxima o mesmo ponto andando na média entre a tangente no ponto e a tangente no ponto futuro.<br> Seja $$ k_1 = f(u_n,t_n)\\ k_2 = f(u_n + hk_1,t_{n+1}) $$ Então $k_1$ é a derivada no ponto e $k_2$ é a derivada no ponto futuro, aproximando a mesma pelo metódo de Euler<br> O passo para Runge-Kutta será então a média entre estas duas, e ficará: $$ y_{n+1} \approx u_{n+1} = u_n + h \frac{k_1 + k_2}{2} $$ ```python def rk_2(F, y0, ts, p = 0): ys = [y0] t = ts[0] h = ts[1] - ts[0] for tnext in ts: k1 = F(ys[-1], t, p) k2 = F(ys[-1] + h*k1,tnext, p) ynext = ys[-1] + h * (k1+k2) / 2.0 ys.append(ynext) t = tnext return np.array(ys[:-1]) ``` Testando para a EDO $$ \begin{cases} y'(t) = - y(t) + 2\sin(t^2) \\ y(0) = 1.2\end{cases} $$ ```python def F(y,t,p = 0): return -y + 2*np.sin(t**2) ``` ```python ## Definindo o dominio ts = np.linspace(-5,5,500) y0 = 1.2 ## Criando a lista para Runge-Kutta 2nd order ys = rk_2(F,y0,ts) ## Criando a lista para Euler Explicito ys2 = f_euler(F,y0,ts) #ans = f(y,ts) plt.plot(ts,ys,label='RK') plt.plot(ts,ys2,label='Explicito') plt.legend() plt.show() ``` ### O metódo de Euler converge - Um pouco de Análise ### Definições: Seja $\frac{\mathrm{d}y}{\mathrm{d}t} = f(y,t)$<br> Seja $t \in \mathbb{N}$ o número de 'tempos' no dominio e $t^*$ o tempo final e $\lfloor \cdot \rfloor$ a função `floor`, que retorna a parte inteira.<br> Seja $h \in \mathbb{R}, h > 0$, o 'tamanho' de cada partição, ou seja $h = t_{n+1} - t_n$ <br> Podemos então definir $n$, tal que $n$ assume valores no conjunto $\{0, \dots , \lfloor \frac{t^*}{h} \rfloor\}$<br> Seja $\lVert \cdot \rVert$ uma norma definida no espaço Seja $y_n$ o valor real (analitico) da função **$y$** no ponto $t_n$, ou seja $y_n = y(t_n)$<br> Seja $u_n$ o valor númerico aproximado da função $y$ no ponto $t_n$ pelo método de Euler, ou seja $u_{n+1} = u_{n} + f(u_{n},t_{n})\cdot h$<br> Uma função $f$ é dita Lipschitz se satisfaz a condição de Lipschitz: $\exists M \in \mathbb{R}; \lVert f(x_1) - f(x_2) \rVert \leq M\cdot\lVert x_1 - x_2\rVert$ Um metódo é dito convergente se: $$ \lim_{h\to 0^+} \max_{n=0, \dots , \lfloor \frac{t^*}{h} \rfloor} \lVert u_n - y_n \rVert = 0 $$ Ou seja, sempre que a malha for refinida, a solução númerica em um ponto se aproxima da solução analitica neste ponto ### Teorema: O metódo de Euler converge #### Prova Tomemos $f(y,t)$ analitica, ou seja, pode ser representada pela série de Taylor centrada em um ponto $t_0$ e é Lipschitz.<br> $f(y,t)$ analitica implica $y$ analitica.<br> Vamos definir $err_n = u_n - y_n$, nosso erro númerico, então queremos provar $$ \lim_{h\to 0^+} \max_{n=0, \dots , \lfloor \frac{t^*}{h} \rfloor} \lVert err_n \rVert = 0 $$ Expandindo nossa solução $y$ da equação diferencial por Taylor: $$ y_{n+1} = y_n + hf(y_n,t_n)+\mathcal{O}(h^2) \tag{1} $$ Como $y$ é analitica, então sua derivada é contínua, logo pelo `Teorema do Valor Extremo`, dado uma vizinhança em torno de $t_n$ o termo $\mathcal{O}(h^2)$ é limitado $\forall h>0$ e $n \leq \lfloor t^*/h \rfloor$ por $M>0, M \in \mathbb{R}$, e pela propriedade arquimediana do corpo do reais $\exists c \in \mathbb{R}, c>0; c\cdot h^2 \geq M$, portanto podemos limitar $\mathcal{O}(h^2)$ por $ch^2, c>0$.<br> Agora vamos fazer $err_{n+1} = u_{n+1} - y_{n+1}$ usando a expansão em Taylor $y_{n+1}$ e Euler em $u_{n+1}$ $$ \begin{align} err_{n+1} &= u_{n+1} - y_{n+1}\\ &= u_n + h(f(u_n,t_n)) - y_n - h(f(yn,tn) + \mathcal{O}(h^2)\\ &= \underbrace{u_n - y_n}_{err_n} + h\left(f(u_n,t_n) - f(y_n,t_n)\right) + \mathcal{O}(h^2)\\ &= err_n + h\left(f(u_n,t_n) - f(y_n,t_n)\right) + \mathcal{O}(h^2)\\ \end{align} $$ Daqui podemos perceber que o erro no passo seguinte depende também do erro anterior já cometido<br> E segue do fato de que $\mathcal{O}(h^2)$ é limitada, com uma cota superior $ch^2$ e da desigualdade triangular $$ \lVert err_{n+1} \rVert \leq \lVert err_n\rVert + \lVert h\left(f(u_n,t_n) - f(y_n,t_n)\right)\rVert + \lVert ch^2 \rVert $$ E pela condição de Lipschitz $$ \lVert f(u_n,t_n) - f(y_n,t_n) \rVert \leq \lambda\lVert u_n - y_n \rVert = \lambda\lVert err_n \rVert, \lambda > 0 $$ Então temos $$ \lVert err_{n+1} \rVert \leq \lVert err_n\rVert + \lVert h\left(f(u_n,t_n) - f(y_n,t_n)\right)\rVert + \lVert ch^2 \rVert \leq \lVert err_n\rVert + \lambda h\lVert err_n \rVert + ch^2\\$$ $\therefore$ $$ \lVert err_{n+1} \rVert \leq (1+h\lambda)\lVert err_n \rVert + ch^2 \tag{2} $$ --- Agora vamos propor: $$ \lVert err_n \rVert \leq \frac{c}{\lambda}h[(1+h\lambda)^n - 1] $$ #### Demonstração: Indução em n Para $n = 0$ $$ \lVert err_0 \rVert \leq \frac{c}{\lambda}h[(1+h\lambda)^0 - 1] = \frac{c}{\lambda}h[1 - 1] = 0\\ err_0 = u_0 - y_0 = 0, \text{pois é a condição inicial} $$ Temos portanto nossa hipotese de indução, vale para $n=k$, vamos para o passo indutivo: $n = k+1$. Da equação 2, temos: $$ \lVert err_{k+1}\rVert \leq (1+h\lambda)\lVert err_k \rVert + ch^2 $$ E pela hipotese de indução $$ \lVert err_k \rVert\leq \frac{c}{\lambda}h[(1+h\lambda)^k - 1] $$ Logo $$ \lVert err_{k+1} \rVert \leq (1+h\lambda)\frac{c}{\lambda}h[(1+h\lambda)^k - 1] + ch^2 $$ Desenvolvendo o termo da direita: $$ \begin{align} (1+h\lambda)\frac{c}{\lambda}h[(1+h\lambda)^k - 1] + ch^2 &= \frac{c}{\lambda}h[(1+h\lambda)^{k+1} - (1+h\lambda)] +ch^2\\ &= \frac{c}{\lambda}h(1+h\lambda)^{k+1} - \frac{c}{\lambda}h(1+h\lambda) +ch^2\\ &= \frac{c}{\lambda}h(1+h\lambda)^{k+1} - \frac{c}{\lambda}h - \frac{c}{\lambda}h^2\lambda + ch^2\\ &= \frac{c}{\lambda}h(1+h\lambda)^{k+1} - \frac{c}{\lambda}h\\ &= \frac{c}{\lambda}h[(1+h\lambda)^{k+1} - 1] \end{align} $$ Portanto $$ \lVert err_{k+1} \rVert \leq \frac{c}{\lambda}h[(1+h\lambda)^{k+1} - 1] $$ E o passo indutivo vale. Logo pelo principio de indução finita temos: $$ \lVert err_n \rVert \leq \frac{c}{\lambda}h[(1+h\lambda)^n - 1] \tag{3} $$ --- Como $h\lambda >0$, então temos $(1+h\lambda) < e^{h\lambda}$ e portanto $(1+h\lambda)^n < e^{nh\lambda}$, e n assume valor máximo em $n = \lfloor t^*/h \rfloor $, portanto: $$(1+h\lambda)^n < e^{\lfloor t^*/h \rfloor h\lambda} \leq e^{t^*\lambda}$$ Substituindo na inequação 3 para $err_n$, teremos: $$ \lVert err_n \rVert \leq \frac{c}{\lambda}h[e^{t^*\lambda} - 1] $$ Passando o limite $h\to 0$, teremos: $$ \lim_{h\to 0}\lVert err_n \rVert \leq \frac{c}{\lambda}h[e^{t^*\lambda} - 1] = 0\\ \therefore \lim_{h\to 0}\lVert err_n \rVert = 0 $$ Portanto o Metódo de Euler converge para toda função Lipschitz. Q.E.D. ### Visualizando o teorema Vamos plotar a solução da equação diferencial $y' = sin(t^2) - y$ com um refinamento da malha cada vez melhor e visualizar a convergência do metódo<br> Plotaremos também um gráfico com a evolução do erro relativo entre a solução de malha mais fina e todas as soluções anteriores ```python ## Equação Diferencial def F(y,t,p=0): return -y + 2*np.sin(t**2) # Criação dos dominios com vários h diferentes ts = np.array([np.linspace(-10,10,i) for i in np.arange(50,300,63)]) # Condição inicial y0 = 1.2 # Preparação da listas para plotagem ys_e = np.array([f_euler(F,y0,i) for i in ts ]) # Estilo das curvas lstyle = ['--','-.',':','-'] # Plot do gráfico de solução plt.figure(figsize=(15,7)) for i in range(len(ts)): plt.plot(ts[i],ys_e[i], ls = lstyle[i], label='$h = '+ str("{0:.2f}".format(20.0/len(ts[i])) +'$')) plt.title('Visualização da convergência do Metódo de Euler') plt.xlabel('t') plt.ylabel('y(t)') plt.legend() plt.show() ## Criando os arrays de erro hs = [0.4,0.18,0.11] ans = [[],[],[]] for i in range(len(ys_e[:-1])): n = np.floor(hs[i]/0.08) for j in range(len(ys_e[i])): try: ans[i].append(ys_e[-1][n*j]) except: ans[i].append(ys_e[-1][-1]) for i in range(len(ans)): ans[i] = np.array(ans[i]) err = np.array([abs(j - i) for i,j in zip(ys_e,ans)]) plt.figure(figsize=(15,7)) for i in range(len(ts)-1): plt.plot(ts[i],err[i], ls = lstyle[i], label='$h = '+ str("{0:.2f}".format(20.0/len(ts[i])) +'$')) plt.title('Visualização do Erro da solução mais convergida em relação às outras soluções') plt.xlabel('t') plt.ylabel('err(y)') plt.legend() plt.show() ``` ### Gráficos de Convergência de Runge-Kutta ```python # Preparação da listas para plotagem ys_rk = np.array([rk_2(F,y0,i) for i in ts ]) # Estilo das curvas lstyle = ['--','-.',':','-'] # Plot do gráfico de solução plt.figure(figsize=(15,7)) for i in range(len(ts)): plt.plot(ts[i],ys_rk[i], ls = lstyle[i], label='$h = '+ str("{0:.2f}".format(20.0/len(ts[i])) +'$')) plt.title('Visualização da convergência de Runge-Kutta') plt.xlabel('t') plt.ylabel('y(t)') plt.legend() plt.show() ``` ```python plt.plot(ts[-1],abs(ys_e[-1]-ys_rk[-1])) ``` ##### Teorema: O metódo de Euler converge ###### Prova Tomemos $f(y,t)$ analitica, ou seja, pode ser representada pela série de Taylor centrada em um ponto $t_0$ e é Lipschitz continua.<br> $f(y,t)$ analitica implica $y$ analitica.<br> Vamos definir $err_n = u_n - y_n$, nosso erro númerico, então queremos provar $$ \lim_{h\to 0^+} \max_{n=0, \dots , \lfloor \frac{t^*}{h} \rfloor} \lVert err_n \rVert = 0 $$ Expandindo nossa solução $y$ da equação diferencial por Taylor: $$ y_{n+1} = y_n + hf(y_n,t_n)+\mathcal{O}(h^2) $$ Como $y$ é analitica, então sua derivada é contínua, logo pelo `Teorema do Valor Extremo`, dado uma vizinhança em torno de $t_n$ o termo $\mathcal{O}(h^2)$ é limitado $\forall h>0$ e $n \leq \lfloor t^*/h \rfloor$ por $M>0, M \in \mathbb{R}$, e pela propriedade arquimediana do corpo do reais $\exists c \in \mathbb{R}, c>0; c\cdot h^2 \geq M$, portanto podemos limitar $\mathcal{O}(h^2)$ por $ch^2, c>0$.<br> Agora vamos fazer $err_{n+1} = u_{n+1} - y_{n+1}$ usando a expansão em Taylor e Euler em $u_n$ $$ \begin{align} err_{n+1} &= u_{n+1} - y_{n+1}\\ &= u_n + h(f(u_n,t_n)) - y_n - h(f(yn,tn) + \mathcal{O}(h^2)\\ &= \underbrace{u_n - y_n}_{err_n} + h\left(f(u_n,t_n) - f(y_n,t_n)\right) + \mathcal{O}(h^2)\\ &= err_n + h\left(f(u_n,t_n) - f(y_n,t_n)\right) + \mathcal{O}(h^2)\\ \end{align} $$ Daqui podemos perceber que o erro no passo seguinte depende também do erro anterior já cometido<br> E segue do limite superior para $\mathcal{O}(h^2)$ e da desigualdade triangular $$ \lVert err_{n+1} \rVert \leq \lVert err_n\rVert + \lVert h\left(f(u_n,t_n) - f(y_n,t_n)\right)\rVert + \lVert ch^2 \rVert $$ E pela condição de Lipschitz $$ \lVert f(u_n,t_n) - f(y_n,t_n) \rVert \leq \lambda\lVert u_n - y_n \rVert = \lambda\lVert err_n \rVert, \lambda > 0 $$ Então temos $$ \lVert err_{n+1} \rVert \leq \lVert err_n\rVert + \lVert h\left(f(u_n,t_n) - f(y_n,t_n)\right)\rVert + \lVert ch^2 \rVert \leq \lVert err_n\rVert + \lambda h\lVert err_n \rVert + ch^2\\ \therefore \lVert err_{n+1} \rVert \leq \lVert err_n\rVert + \lVert h\left(f(u_n,t_n) - f(y_n,t_n)\right)\rVert + \lVert ch^2 \rVert \leq (1+h\lambda)\lVert err_n \rVert + ch^2 $$ Assumiremos (provar mais tarde) $$ \lVert err_n \rVert \leq \frac{c}{\lambda}h[(1+h\lambda)^n - 1] $$ Como $h\lambda >0$, então temos $(1+h\lambda) < e^{h\lambda}$ e portanto $(1+h\lambda)^n < e^{nh\lambda}$, e n assume valor máximo em $n = \lfloor t^*/h \rfloor $, portanto: $(1+h\lambda)^n < e^{\lfloor t^*/h \rfloor h\lambda} = e^{t^*\lambda}$<br> Substituindo na inequação anterior para $err_n$, teremos: $$ \lVert err_n \rVert \leq \frac{c}{\lambda}h[e^{t^*\lambda} - 1] $$ Passando o limite $h\to 0$, teremos: $$ \lim_{h\to 0}\lVert err_n \rVert \leq \frac{c}{\lambda}h[e^{t^*\lambda} - 1] = 0\\ \therefore \lim_{h\to 0}\lVert err_n \rVert = 0 $$ O Metódo de Euler converger. Q.E.D. ```python ``` ```python ``` ### O modelo presa-predador de Lotka-Volterra (em construção) O modelo é dado pelas EDOs $$ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = (\lambda - by)x\\ \frac{\mathrm{d}y}{\mathrm{d}t} = (-\mu + cx)y\\ \end{cases} $$ Com $\lambda, \mu, b, c$ todos reais positivos e $x$ representando a população de presas e $y$ a população de predadores<br> Como já visto anteriormente, vamos tratar de forma vetorial este problema, sendo: $$ v = \begin{bmatrix} \frac{\mathrm{d}x}{\mathrm{d}t} \\ \frac{\mathrm{d}y}{\mathrm{d}t} \end{bmatrix} $$ E sendo $D$ o operador linear de derivada, teremos: $$ Dv = \begin{bmatrix} (\lambda - by)x\\ (-\mu + cx)y \end{bmatrix} $$ E então podemos encontrar a solução aplicando algum metódo númerico ```python ## Parametros: ## v: vetor dos pontos iniciais ## p[l,b,m,c]: uma lista com os parametros do modelo ### l: lambda ### b: b ### m: mu ### c: c def model(v,t,p = 0): if p == 0: p = [1,1,1,1] return np.array([(p[0]-p[1]*v[1])*v[0],(p[3]*v[0]-p[2])*v[1]]) ``` ```python # Parametros de ajusate ts = np.linspace(0,30,500) y0 = [2,1] ``` ```python ys_e = f_euler(model,y0,ts) plt.figure(figsize=(10,5)) plt.plot(ts,ys_e) plt.title('Modelo Presa-Predador - Solução com Euler') plt.legend(['Presa', 'Predador']) plt.xlabel('$t$') plt.ylabel('População $y(t)$') plt.grid(alpha = 0.5) plt.show() ``` ```python ys_rk = rk_2(model,y0,ts) plt.figure(figsize=(10,5)) plt.plot(ts,ys_rk) plt.title('Modelo Presa-Predador - Solução com Runge-Kutta') plt.legend(['Presa', 'Predador']) plt.xlabel('$t$') plt.ylabel('População $y(t)$') plt.grid(alpha = 0.5) plt.show() ``` ```python ## F: Função derivada da função que queremos encontrar ## t0: tempo inicial ## y0: ponto inicial ## ts: range de tempo ## p: parametros particulares de cada modelo def f2_euler(F, y0, ts, p = 0): ys = [y0] h = ts[1]-ts[0] for tnext in ts[1:]: ynext = ys[-1] + F(ys[-1],tnext,p)*h ys.append(ynext) t = tnext return np.array(ys) ``` ```python ## Parametros: ## v: vetor dos pontos iniciais ## p[l,b,m,c]: uma lista com os parametros do modelo ### l: lambda ### b: b ### m: mu ### c: c def model(v,t,p = 0): if p == 0: p = [1,1,1,1] return np.array([(p[0]-p[1]*v[1])*v[0],(p[3]*v[0]-p[2])*v[1]]) ``` ```python # Parametros de ajusate ts = np.linspace(0,30,500) y0 = [2,1] ys_e = f2_euler(model,y0,ts) plt.figure(figsize=(10,5)) plt.plot(ts,ys_e) plt.title('Modelo Presa-Predador - Solução com Euler') plt.legend(['Presa', 'Predador']) plt.xlabel('$t$') plt.ylabel('População $y(t)$') plt.grid(alpha = 0.5) plt.show() ``` ```python def model2(y, t, p = 0): return -0.5*y ``` ```python ts = np.linspace(0,30,500) plt.plot(ts,f2_euler(model2,2,ts)) ``` ```python def EulerFW(F,y0,ts,p=0): ys=[y0] h=ts[1]-ts[0] tc=ts[0] for t in ts[1:]: yn=ys[-1]+h*F(ys[-1],ts,p) ys.append(yn) tc=t return ys ``` ```python ts = np.linspace(0,30,500) plt.plot(ts,EulerFW(model2,2,ts)) ``` ```python ts = np.linspace(0,30,500) plt.plot(ts,EulerFW(model,[2,1],ts)) ``` ```python ts = np.linspace(0,5,500) plt.plot(ts,f2_euler(model,[2,1],ts)) ``` ```python def model3(v,t,p=0): if p == 0: p = [1.5,1.2,1,1] return np.array([p[0]*np.log(v[1])-p[1]*v[1]+p[2]*np.log(v[0])+p[3]*v[0]]) ``` ```python plt.plot(ts,rk_2(model3,[2,1],ts)) ``` ```python model3([2,1],ts) ``` array([1.49314718]) ### Análise qualitativa da EDO Não temos uma solução analítica do sistema de EDO, mas podemos encontrar uma relação entre as variaveis do problema, olhando para a taxa de variação de cada uma das populações da seguinte forma \begin{equation} \dfrac{\mathrm{d}y}{\mathrm{d}x} = \frac{\dfrac{\mathrm{d}y}{\mathrm{d}t}}{\dfrac{\mathrm{d}x}{\mathrm{d}t}} = \dfrac{(-\mu + cx)y}{(\lambda - by)x} \end{equation} E esta é uma equação separavel, podemos seguir com a solução: \begin{equation} \dfrac{(-\mu + cx)y}{y}\mathrm{d}y = \dfrac{(\lambda - by)x}{x}\mathrm{d}x \end{equation} Obtendo então \begin{equation} \int \bigg( \dfrac{\lambda - by}{y}\bigg) dy = \int \bigg( \dfrac{-\mu + cx}{x}\bigg) dx \end{equation} Resolvendo, temos a solução geral para o modelo: \begin{equation} \lambda\ln(|y|) - by = -\mu\ln(|x|) + cx + K \end{equation} Como as populações $x,y$ são sempre positivas, podemos reescrever \begin{equation} \lambda\ln(y) - by + \mu\ln(x) - cx = K \end{equation} Sendo $K \in \mathbb{R}$, constante em relação a cada solução.\\ Temos assim uma relação entre cada parametro e variavel do nosso problema ```python ys_rk = f_euler(model,[40,20],ts,p=[3,1.3,2.7,0.5]) plt.figure(figsize=(10,5)) plt.plot(ts,ys_rk) plt.title('Modelo Presa-Predador - Solução com Runge-Kutta') plt.legend(['Presa', 'Predador']) plt.xlabel('$t$') plt.ylabel('População $y(t)$') plt.grid(alpha = 0.5) plt.show() ``` ```python ## Parametros: ## v: vetor dos pontos iniciais ## p[l,b,m,c]: uma lista com os parametros do modelo ### l: lambda ### b: b ### m: mu ### c: c def model4(v,t,p = 0): if p == 0: p = [3,1.3,2.7,0.5] return np.array([(p[0]-p[1]*v[1])*v[0],(p[3]*v[0]-p[2])*v[1]]) ``` ```python ans = odeint(model4,[2,1],ts) plt.plot(ts,ans) ``` ```python ans2 = rk_2(model4,[2,1],ts) plt.plot(ts,ans2) ``` ```python ```

6d91e40d70885540198327df5fd837201f0950b4

ipynb

Jupyter Notebook

analise-numerica-edo-2019-1/RK e Eulers.ipynb

mirandagil/university-courses

e70ce5262555e84cffb13e53e139e7eec21e8907

2019-12-23T16:39:01.000Z

2019-12-23T16:39:01.000Z

analise-numerica-edo-2019-1/RK e Eulers.ipynb

mirandagil/university-courses

e70ce5262555e84cffb13e53e139e7eec21e8907

analise-numerica-edo-2019-1/RK e Eulers.ipynb

mirandagil/university-courses

e70ce5262555e84cffb13e53e139e7eec21e8907

Qwen/Qwen-72B

1. YES 2. YES

__label__por_Latn

# Principal Component Analysis: lecture ## 1. Introduction Up until now, We have focused on supervised learning. This group of methods aims at predicting labels based on training data that is labeled as well. Principal Componant Analysis is our first so-called "unsupervised" estimator. Generally, the aim of unsupervised estimators is to reveal interesting data patterns without having any reference labels. The first unsupervised learning algorithm, is Principal Component Analysis, also referred to as PCA. PCA is a dimensionality reduction technique which is often used in practice for visualization, feature extraction, noise filtering, etc. Generally, PCA would be applied on data sets with many variables. PCA creates new variables that are linear combinations of the original variables. The idea is to reduce the dimension of the data considerably while maintaining as much information as possible. While the purpose is to significantly reduce the dimensionality, the maximum amount of new variables that can possibly be created is equal to the number of original variables. A nice feature of PCA is that the newly created variables are uncorrelated. **A simple example start** : Imagine that a data set consists of the height and weight of a group of people. One could imagine that these 2 metrics are heavily correlated, so we could basically summarize these 2 metrics in one variable, a linear combination of the two. This one variable will contain most of the information from 2 variables in one. It is important to note that the effectiveness of PCA strongly depends on the structure of the correlation matrix of the existing variables! ## 2. Intermezzo: Eigenvalues and eigenvectors An eigenvector is a vector that after transformation hasn't changed, except by a scalar value known as the *eigenvalue*. ### 2.1 Definition If there exists a square matrix $A$ (an n x n matrix) , then a scalar $\lambda$ is called the **eigenvalue** of $A$ if there is a non-zero vector $v$ such that $$Av = \lambda v$$. This vector $v$ is thene called the **eigenvector** of A corresponding to $\lambda$. Eigenvalues and eigenvectors are very useful and have tons of applications! Imagine you have a matrix \begin{equation} A = \begin{bmatrix} 3 & 2 \\ 3 & -2 \end{bmatrix} \end{equation} We have an eigenvector \begin{equation} v = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \end{equation} Let's perform the multiplication $A v$ \begin{equation} Av = \begin{bmatrix} 3 & 2 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 3*2+2*1 \\ 3*2+(-2*1) \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix} \end{equation} Now we want to see if we can find a $\lambda$ such that \begin{equation} Av = \begin{bmatrix} 8 \\ 4 \end{bmatrix}= \lambda \begin{bmatrix} 2 \\ 1 \end{bmatrix} \end{equation} Turns out $\lambda =4$ is the eigenvalue for our proposed eigenvector! ### 2.2 But how can you find values of eigenmatrices? An $n xn$ matrix has n eigenvectors and n eigenvalues! How to find the eigenvalues? $ det(A- \lambda I)= 0$ \begin{equation} det(A- \lambda I) = det\begin{bmatrix} 3-\lambda & 2 \\ 3 & -2-\lambda \end{bmatrix} \end{equation} This way we indeed find that 4 is an eigenvalue, and so is -3! You'll learn about the connection between eigenvalues and eigenmatrices in a second. https://georgemdallas.wordpress.com/2013/10/30/principal-component-analysis-4-dummies-eigenvectors-eigenvalues-and-dimension-reduction/ https://www.youtube.com/watch?v=ue3yoeZvt8E "toepassingen van de statistiek" sl 53-63 from the PCA chapter! ## 3. PCA: some notation ### 3.1 The data matrix Let's say we have P variables $X_1, X_2, \dots, X_p$ and $n$ observations $1,...,n$. Or data looks like this: \begin{bmatrix} X_{11} & X_{12} & X_{13} & \dots & X_{1p} \\ X_{21} & X_{22} & X_{23} & \dots & X_{2p} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ X_{n1} & X_{n2} & X_{n3} & \dots & X_{np} \end{bmatrix} For 2 variables, this is what our data could look like: ```python import numpy as np import matplotlib.pyplot as plt %matplotlib inline import numpy as np np.random.seed(123) X = np.random.normal(2, 1.5, 50) Y = np.random.normal(3, 0.6, 50) fig, ax = plt.subplots() ax.axhline(y=0, color='k') ax.axvline(x=0, color='k') plt.ylim(-5,5) plt.xlim(-7,7) plt.scatter(X, Y, s = 6) ``` ### 3.2 The mean & mean-corrected data The **mean** of the $j$-th variable: $\bar{X}_j = \dfrac{\sum_{i=1}^n X_{ij}}{n}= \bar X_{.j}$ To get to the **mean-corrected** data: substract the mean from each $X_{ij}$ $x_{ij} = X_{ij}-\bar X_{.j}$ Going back to our two variables example, this is how the data would be shifted: ```python fig, ax = plt.subplots() ax.axhline(y=0, color='k') ax.axvline(x=0, color='k') X_mean = X- np.mean(X) Y_mean = Y- np.mean(Y) plt.ylim(-5,5) plt.xlim(-7,7) plt.scatter(X_mean, Y_mean, s = 6) ``` ### 3.3 The variance & standardized data $s^2_j = \dfrac{\sum_{i=1}^n (X_{ij}-\bar X_{.j})^2}{n-1}= \dfrac{\sum_{i=1}^n x_{ij}^2}{n-1}$ To get to the **standardized** data: divide the mean-corrected data by the standard deviation $s_j$. $z_{ij} = \dfrac{x_{ij}}{s_{j}}$ Going back to the example with 2 variables, this is what standardized data would look like: ```python fig, ax = plt.subplots() ax.axhline(y=0, color='k') ax.axvline(x=0, color='k') X_mean = X- np.mean(X) Y_mean = Y- np.mean(Y) X_std= np.std(X) Y_std = np.std(Y) X_stdized = X_mean / X_std Y_stdized = Y_mean / Y_std plt.ylim(-5,5) plt.xlim(-7,7) plt.scatter(X_stdized, Y_stdized, s=6) ``` ### 3.4 The covariance The covariance for two variables $X_j$ and $X_k$: $s_{jk} = \dfrac{\sum_{i=1}^n (X_{ij}-\bar X_{.j})(X_{ij}-\bar X_{.k})}{n-1}= \dfrac{\sum_{i=1}^n x_{ij}x_{ik}}{n-1}$ Denote $\mathbf{S}$ the sample covariance matrix \begin{equation} \mathbf{S} = \begin{bmatrix} s^2_{1} & s_{12} & \dots & s_{1p} \\ s_{21} & s^2_{2} & \dots & s_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ s_{p1} & s_{p2} & \dots & s^2_{p} \end{bmatrix} \end{equation} When you do the same computation with standardized variables, you get the **correlation**. Remember that the correlation $r_{jk}$ always lies between -1 and 1. $r_{jk} = \dfrac{\sum_{i=1}^n z_{ij}z_{ik}}{n-1}$ Then, $\mathbf{R}$ is the correlation matrix \begin{equation} \mathbf{R} = \begin{bmatrix} 1 & r_{12} & \dots & r_{1p} \\ r_{21} & 1 & \dots & r_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ r_{p1} & r_{p2} & \dots & 1 \end{bmatrix} \end{equation} ## 4. How does PCA work? Matrices and eigendecomposition ### 4.1 Finding principal components $ \mathbf{X}= (X_1, X_2, \ldots, X_p)$ is a random variable. Then the principal components of $\mathbf{X}$, denoted by $PC_1, \ldots, PC_p$ satisfy these 3 conditions: - $(PC_1, PC_2, \ldots, PC_p)$ are mutually uncorrelated - $var(PC_1)\geq var(PC_2) \geq \ldots \geq var(PC_p)$ - $PC_j = c_{j1} X_1+c_{j2} X_2+\ldots+c_{jp} X_p$ Note that for $j=1,\ldots,p$ and $c_j = (c_{j1}, c_{j2}, \ldots, c_{jp})$' is a vector of constants satisfying $ \lVert{\mathbf{c_j} \rVert^2 = \mathbf{c'_j}\mathbf{c_j}} = \displaystyle\sum^p_{k=1} c^2_{kj}=1 $ The variance of $PC$ is then: $var(PC_j) = var( c_{j1} X_1+c_{j2} X_2+\ldots+c_{jp} X_p) \\ = c_{j1}^2 var(X_1) +c_{j2}^2 var(X_2) + \ldots + c_{jp}^2 var(X_p) + 2 \displaystyle\sum_k\sum_{l \neq k}c_{jk}c_{jl} cov(X_k, X_l) \\ = c_j' \Sigma c_j$ In words, this means that variances can easily be computed using the coefficients used while making the linear combinations. We can prove that $var(PC_1)\geq var(PC_2) \geq \ldots \geq var(PC_p)$ is actually given by the eigenvalues $\lambda_1\geq \lambda_2 \geq \ldots \geq \lambda_3$ and the eigenvectors are given by $c_j = (c_{j1}, c_{j2}, \ldots, c_{jp})$. From here on, we'll denote the eigenvectors by $e_j$ instead of $c_j$. # Sources http://www.bbk.ac.uk/ems/faculty/brooms/teaching/SMDA/SMDA-02.pdf https://stackoverflow.com/questions/13224362/principal-component-analysis-pca-in-python https://georgemdallas.wordpress.com/2013/10/30/principal-component-analysis-4-dummies-eigenvectors-eigenvalues-and-dimension-reduction/

33028a56cf1d2c9d1ad7039ac5d1024d9bdc5ab5

ipynb

Jupyter Notebook

Principal Component Analysis.ipynb

learn-co-students/ml-pca-staff

cb364402d9cfec4f3064942f9c3cc053b900ecf9

2018-05-27T21:48:21.000Z

2018-05-27T21:48:27.000Z

Principal Component Analysis.ipynb

learn-co-students/ml-pca-staff

cb364402d9cfec4f3064942f9c3cc053b900ecf9

Principal Component Analysis.ipynb

learn-co-students/ml-pca-staff

cb364402d9cfec4f3064942f9c3cc053b900ecf9

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Understanding the impact of timing on defaults > How do delays in recognising defaults impact the apparent profitability of Afterpay? - toc: true - badges: true - comments: true - categories: [Sympy,Finance,Afterpay] - image: images/2020-10-03-Afterpay-Customer-Defaults-Part-7/header.png ## The Context The thesis of this post is actually pretty simple. There is a delay between when customers make a transaction, and when *Afterpay* realises that they have defaulted. Because of this delay, combined with the rapid growth in the total value of transactions, defaults as a percentage of transaction value may be artificially reduced. > Important: Obviously I need a disclaimer. If you use anything I say as the basis for any decision, financial or otherwise, you are an idiot. ## The Model First off, let's load in a bunch of libraries. ```python %matplotlib inline import matplotlib.pyplot as plt import numpy as np from io import StringIO import pandas as pd import scipy.optimize plt.rcParams["figure.figsize"] = (10,10) from warnings import filterwarnings filterwarnings('ignore') ``` While reading through Afterpay's releases to the markets, I came across this chart, which appears on page 3 of [this](https://www.afterpaytouch.com/images/APT_ASX-Announcement_Q2-FY18_16-Jan-Final-2.pdf) release. Let's use this to build a simple quadratic model of the reported sales. ## Loading the data ```python #Underlying sales csv_data = StringIO('''anz_underlying_sales_value,date,month_count 0,FY15,0 37.3,FY16,12 561.2,FY17,24 2184.6,FY18,36 4314.1,FY19,48 6566.9,FY20,60''') df = pd.read_csv(csv_data, sep=",") ``` ## Fitting a curve Let's first fit quadratic: ```python def quadratic(t, a, b, c): y = a * t**2 + b * t + c return y xdata = df.month_count.values ydata = df.anz_underlying_sales_value.values popt, pcov = scipy.optimize.curve_fit(quadratic, xdata, ydata) print(popt) ``` [ 2.17012649 -17.61639881 -58.725 ] ```python x = np.linspace(0,60, 61) y = quadratic(x, *popt) plt.plot(xdata, ydata, 'o', label='data') plt.plot(x,y, label='fit') plt.title('ANZ Sales by preceding Financial Year ($M AUD)') plt.xlabel('Months after launch') plt.ylabel('ANZ Sales ($M AUD)') plt.legend(loc='best') plt.show() ``` ## Delays in reporting. So we found that we could model the annual reported sales as: $$2.170 t^2 - 17.61t - 58.725$$ The instantanious rate of sales, is given by : $$0.1808t^2 + 0.7021t -9.36$$. Don't wory how I arrived at this, I will show how in in the appendix of this post. ```python t = np.linspace(0,60, 61) sales = 0.1808*t**2 + 0.7021* t - 9.36 plt.plot(sales) plt.title('ANZ Sales by month ($M AUD)') plt.xlabel('Months after launch') plt.ylabel('ANZ Sales by month ($M AUD)') plt.show() ``` Now let's model a delay of 6 months between when the transaction happens, and when *Afterpay* finally realised there was a default. From this we can see there is a potentially significant difference between the true rate at which losses are occurring, and the rate at which we observe them occurring, at any point in time. ```python delay = 6 #months true_loss_rate = 0.01 losses_true = true_loss_rate*(0.1808*t**2 + 0.7021* t - 9.36) losses_observed = true_loss_rate*(0.1808*(t-delay)**2 + 0.7021* (t-delay) - 9.36) plt.plot(losses_observed,label='Observed') plt.plot(losses_true,label='True') plt.legend() plt.title('ANZ losses by month ($M AUD)') plt.xlabel('Months after launch') plt.ylabel('ANZ losses by month ($M AUD)') plt.show() ``` Now let's integrate by financial year. ```python def integrate_by_year(y): integrated = np.array([0,np.sum(y[0:12]),np.sum(y[12:24]),np.sum(y[24:36]),np.sum(y[36:48]),np.sum(y[48:60])]) return(integrated) observed_loss_rate = integrate_by_year(losses_true)/integrate_by_year(losses_observed) plt.plot(observed_loss_rate) plt.title('Ratio of true losses to observed losses') plt.xlabel('Years after launch') plt.ylabel('Ratio of true losses to observed losses') plt.ylim(1,2.5) plt.xlim(2,5) plt.grid() plt.show() ``` ## Conclusion In conclusion, we can clearly see the impact of a delay in recognising losses, in situations where there is rapid growth. Even after years of growth, with a 6 month delay in recognising losses, the true losses could be 30-40% higher than reported. # Appendix ### Finding an integral So we found that we could model the annual reported sales as: $$2.170 t^2 - 17.61t - 58.725$$ Let's call this function $f(t)$ We want to find the function $g(t)$, which is the underlying rate of sales, which I claimed was : $$0.1808t^2 + 0.7021t -9.36$$. This function, when integrated over 12 months, will give us the annual reported sales. To help us with the algebraic manipulation, we can use [Sympy](https://docs.sympy.org/latest/index.html). An alternative is to do the algebraic manipulation by hand, but this is probably faster and more scalable. ```python import sympy as sym sym.init_printing(use_unicode=True) a,b,c,d,t = sym.symbols('a b c d t') ``` So we are looking for a quadratic function, the definite integral of which is equal to $$2.170 t^2 - 17.61t - 58.725$$. Let's start by forming the definite integral. ```python expr = sym.simplify((a*t**3 + b*t**2 + c*t + d) - (a*(t-12)**3 + b*(t-12)**2 + c*(t-12) + d)) ``` ```python print(sym.collect(expr,t)) ``` 36*a*t**2 + 1728*a - 144*b + 12*c + t*(-432*a + 24*b) ```python fitted_quadratic = t**2 * 2.17012649 + t*-17.61639881 -58.725 ``` ### Solving for the coefficients Let's now form a set of simultaneous equations, and solve for each of the coefficients of $$t$$. ```python equations = [] for i in [2,1,0]: eq = sym.collect(expr,t).coeff(t, i) coeff = sym.collect(fitted_quadratic,t).coeff(t, i) equations.append(sym.Eq(eq,coeff)) result = sym.solve(equations,(a,b,c)) print(result) ``` {a: 0.0602812913888889, b: 0.351046627916667, c: -9.36169642500000} ### Finding the derivative Now all that's left to do, is to find the derivative of the indefinite integral.  ```python expr = result[a] * t**3 + result[b]*t**2 + result[c]*t print(sym.diff(expr, t)) ``` 0.180843874166667*t**2 + 0.702093255833333*t - 9.361696425 Voila!

bef5b31222796ebaec79d9b70dde2edf6ead1bdf

ipynb

Jupyter Notebook

_notebooks/2020-10-03-Afterpay-Customer-Defaults-Part-7.ipynb

CGCooke/Blog

ab1235939011d55674c0888dba4501ff7e4008c6

2020-10-29T06:32:23.000Z

2020-10-29T06:32:23.000Z

_notebooks/2020-10-03-Afterpay-Customer-Defaults-Part-7.ipynb

CGCooke/Blog

ab1235939011d55674c0888dba4501ff7e4008c6

2020-04-04T09:39:50.000Z

2022-03-25T12:30:56.000Z

_notebooks/2020-10-03-Afterpay-Customer-Defaults-Part-7.ipynb

CGCooke/Blog

ab1235939011d55674c0888dba4501ff7e4008c6

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python %matplotlib inline ``` # Visualize the hemodynamic response In this example, we describe how the hemodynamic response function was estimated in the previous model. We fit the same ridge model as in the previous example, and further describe the need to delay the features in time to account for the delayed BOLD response. Because of the temporal dynamics of neurovascular coupling, the recorded BOLD signal is delayed in time with respect to the stimulus. To account for this lag, we fit encoding models on delayed features. In this way, the linear regression model weighs each delayed feature separately and recovers the shape of the hemodynamic response function in each voxel separately. In turn, this method (also known as a Finite Impulse Response model, or FIR) maximizes the model prediction accuracy. With a repetition time of 2 seconds, we typically use 4 delays [1, 2, 3, 4] to cover the peak of the the hemodynamic response function. However, the optimal number of delays can vary depending on the experiment and the brain area of interest, so you should experiment with different delays. In this example, we show that a model without delays performs far worse than a model with delays. We also show how to visualize the estimated hemodynamic response function (HRF) from a model with delays. ```python ``` ## Path of the data directory ```python import os from voxelwise_tutorials.io import get_data_home directory = os.path.join(get_data_home(), "vim-5") print(directory) ``` ```python # modify to use another subject subject = "S01" ``` ## Load the data We first load the fMRI responses. ```python import numpy as np from voxelwise_tutorials.io import load_hdf5_array file_name = os.path.join(directory, "responses", f"{subject}_responses.hdf") Y_train = load_hdf5_array(file_name, key="Y_train") Y_test = load_hdf5_array(file_name, key="Y_test") print("(n_samples_train, n_voxels) =", Y_train.shape) print("(n_repeats, n_samples_test, n_voxels) =", Y_test.shape) ``` We average the test repeats, to remove the non-repeatable part of fMRI responses. ```python Y_test = Y_test.mean(0) print("(n_samples_test, n_voxels) =", Y_test.shape) ``` We fill potential NaN (not-a-number) values with zeros. ```python Y_train = np.nan_to_num(Y_train) Y_test = np.nan_to_num(Y_test) ``` Then, we load the semantic "wordnet" features. ```python feature_space = "wordnet" file_name = os.path.join(directory, "features", f"{feature_space}.hdf") X_train = load_hdf5_array(file_name, key="X_train") X_test = load_hdf5_array(file_name, key="X_test") print("(n_samples_train, n_features) =", X_train.shape) print("(n_samples_test, n_features) =", X_test.shape) ``` ## Define the cross-validation scheme We define the same leave-one-run-out cross-validation split as in the previous example. ```python from sklearn.model_selection import check_cv from voxelwise_tutorials.utils import generate_leave_one_run_out # indice of first sample of each run run_onsets = load_hdf5_array(file_name, key="run_onsets") print(run_onsets) ``` We define a cross-validation splitter, compatible with ``scikit-learn`` API. ```python n_samples_train = X_train.shape[0] cv = generate_leave_one_run_out(n_samples_train, run_onsets) cv = check_cv(cv) # copy the cross-validation splitter into a reusable list ``` ## Define the model We define the same model as in the previous example. See the previous example for more details about the model definition. ```python from sklearn.pipeline import make_pipeline from sklearn.preprocessing import StandardScaler from voxelwise_tutorials.delayer import Delayer from himalaya.kernel_ridge import KernelRidgeCV from himalaya.backend import set_backend backend = set_backend("torch_cuda", on_error="warn") X_train = X_train.astype("float32") X_test = X_test.astype("float32") alphas = np.logspace(1, 20, 20) pipeline = make_pipeline( StandardScaler(with_mean=True, with_std=False), Delayer(delays=[1, 2, 3, 4]), KernelRidgeCV( alphas=alphas, cv=cv, solver_params=dict(n_targets_batch=500, n_alphas_batch=5, n_targets_batch_refit=100)), ) ``` ```python from sklearn import set_config set_config(display='diagram') # requires scikit-learn 0.23 pipeline ``` ## Fit the model We fit on the train set, and score on the test set. ```python pipeline.fit(X_train, Y_train) scores = pipeline.score(X_test, Y_test) scores = backend.to_numpy(scores) print("(n_voxels,) =", scores.shape) ``` ## Intermission: understanding delays To have an intuitive understanding of what we accomplish by delaying the features before model fitting, we will simulate one voxel and a single feature. We will then create a ``Delayer`` object (which was used in the previous pipeline) and visualize its effect on our single feature. Let's start by simulating the data. ```python # number of total trs n_trs = 50 # repetition time for the simulated data TR = 2.0 rng = np.random.RandomState(42) y = rng.randn(n_trs) x = np.zeros(n_trs) # add some arbitrary value to our feature x[15:20] = .5 x += rng.randn(n_trs) * 0.1 # add some noise # create a delayer object and delay the features delayer = Delayer(delays=[0, 1, 2, 3, 4]) x_delayed = delayer.fit_transform(x[:, None]) ``` In the next cell we are plotting six lines. The subplot at the top shows the simulated BOLD response, while the other subplots show the simulated feature at different delays. The effect of the delayer is clear: it creates multiple copies of the original feature shifted forward in time by how many samples we requested (in this case, from 0 to 4 samples, which correspond to 0, 2, 4, 6, and 8 s in time with a 2 s TR). When these delayed features are used to fit a voxelwise encoding model, the brain response $y$ at time $t$ is simultaneously modeled by the feature $x$ at times $t-0, t-2, t-4, t-6, t-8$. In the remaining of this example we will see that this method improves model prediction accuracy and it allows to account for the underlying shape of the hemodynamic response function. ```python import matplotlib.pyplot as plt fig, axs = plt.subplots(6, 1, figsize=(8, 6.5), constrained_layout=True, sharex=True) times = np.arange(n_trs)*TR axs[0].plot(times, y, color="r") axs[0].set_title("BOLD response") for i, (ax, xx) in enumerate(zip(axs.flat[1:], x_delayed.T)): ax.plot(times, xx, color='k') ax.set_title("$x(t - {0:.0f})$ (feature delayed by {1} sample{2})".format( i*TR, i, "" if i == 1 else "s")) for ax in axs.flat: ax.axvline(40, color='gray') ax.set_yticks([]) _ = axs[-1].set_xlabel("Time [s]") plt.show() ``` ## Compare with a model without delays We define here another model without feature delays (i.e. no ``Delayer``). Because the BOLD signal is inherently slow due to the dynamics of neuro-vascular coupling, this model is unlikely to perform well. ```python pipeline_no_delay = make_pipeline( StandardScaler(with_mean=True, with_std=False), KernelRidgeCV( alphas=alphas, cv=cv, solver_params=dict(n_targets_batch=500, n_alphas_batch=5, n_targets_batch_refit=100)), ) pipeline_no_delay ``` We fit and score the model as the previous one. ```python pipeline_no_delay.fit(X_train, Y_train) scores_no_delay = pipeline_no_delay.score(X_test, Y_test) scores_no_delay = backend.to_numpy(scores_no_delay) print("(n_voxels,) =", scores_no_delay.shape) ``` Then, we plot the comparison of model prediction accuracies with a 2D histogram. All ~70k voxels are represented in this histogram, where the diagonal corresponds to identical prediction accuracy for both models. A distibution deviating from the diagonal means that one model has better prediction accuracy than the other. ```python from voxelwise_tutorials.viz import plot_hist2d ax = plot_hist2d(scores_no_delay, scores) ax.set( title='Generalization R2 scores', xlabel='model without delays', ylabel='model with delays', ) plt.show() ``` We see that the model with delays performs much better than the model without delays. This can be seen in voxels with scores above 0. The distribution of scores below zero is not very informative, since it corresponds to voxels with poor predictive performance anyway, and it only shows which model is overfitting the most. ## Visualize the HRF We just saw that delays are necessary to model BOLD responses. Here we show how the fitted ridge regression weights follow the hemodynamic response function (HRF). Fitting a kernel ridge regression results in a set of coefficients called the "dual" coefficients $w$. These coefficients differ from the "primal" coefficients $\beta$ obtained with a ridge regression, but the primal coefficients can be computed from the dual coefficients using the training features $X$: \begin{align}\beta = X^\top w\end{align} To better visualize the HRF, we will refit a model with more delays, but only on a selection of voxels to speed up the computations. ```python # pick the 10 best voxels voxel_selection = np.argsort(scores)[-10:] # define a pipeline with more delays pipeline_more_delays = make_pipeline( StandardScaler(with_mean=True, with_std=False), Delayer(delays=[0, 1, 2, 3, 4, 5, 6]), KernelRidgeCV( alphas=alphas, cv=cv, solver_params=dict(n_targets_batch=500, n_alphas_batch=5, n_targets_batch_refit=100)), ) pipeline_more_delays.fit(X_train, Y_train[:, voxel_selection]) # get the (primal) ridge regression coefficients primal_coef = pipeline_more_delays[-1].get_primal_coef() primal_coef = backend.to_numpy(primal_coef) # split the ridge coefficients per delays delayer = pipeline_more_delays.named_steps['delayer'] primal_coef_per_delay = delayer.reshape_by_delays(primal_coef, axis=0) print("(n_delays, n_features, n_voxels) =", primal_coef_per_delay.shape) # select the feature with the largest coefficients for each voxel feature_selection = np.argmax(np.sum(np.abs(primal_coef_per_delay), axis=0), axis=0) primal_coef_selection = primal_coef_per_delay[:, feature_selection, np.arange(len(voxel_selection))] plt.plot(delayer.delays, primal_coef_selection) plt.xlabel('Delays') plt.xticks(delayer.delays) plt.ylabel('Ridge coefficients') plt.title(f'Largest feature for the {len(voxel_selection)} best voxels') plt.axhline(0, color='k', linewidth=0.5) plt.show() ``` We see that the hemodynamic response function (HRF) is captured in the model weights. Note that in this dataset, the brain responses are recorded every two seconds.

f21ff95ce588509223b19c983a23c999eaadde8b

ipynb

Jupyter Notebook

tutorials/notebooks/movies_3T/03_plot_hemodynamic_response.ipynb

gallantlab/voxelwise_tutorials

3df639dd5fb957410f41b4a3b986c9f903f5333b

2021-09-08T22:22:26.000Z

2022-02-10T18:06:33.000Z

tutorials/notebooks/movies_3T/03_plot_hemodynamic_response.ipynb

gallantlab/voxelwise_tutorials

3df639dd5fb957410f41b4a3b986c9f903f5333b

2021-09-11T16:06:44.000Z

2021-12-16T23:39:40.000Z

tutorials/notebooks/movies_3T/03_plot_hemodynamic_response.ipynb

gallantlab/voxelwise_tutorials

3df639dd5fb957410f41b4a3b986c9f903f5333b

2021-09-13T19:11:00.000Z

2022-03-26T04:35:11.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python %reload_ext autoreload %aimport trochoid %autoreload 1 ``` ```python import math import numpy as np # %matplotlib notebook import matplotlib.pyplot as plt plt.style.use('seaborn-colorblind') plt.style.use('seaborn-whitegrid') plt.rcParams['figure.figsize'] = 800/72,800/72 plt.rcParams["font.size"] = 21 # plt.rcParams["legend.loc"] = 'upper right' import os import sys sys.path.append(os.getcwd())# 作業ティレクトリをパスに追加\n", from trochoid import * ``` ## トロコイド 定円の半径を $r_c$、動円の半径を $r_m$、回転角を $\theta$、描画点の半径を $r_d$ とする 外トロコイド: \begin{cases} x=(r_{c}+r_{m})\cos \theta -r_{d}\cos \left({\cfrac {r_{c}+r_{m}}{r_{m}}}\theta \right),\\y=(r_{c}+r_{m})\sin \theta -r_{d}\sin \left({\cfrac {r_{c}+r_{m}}{r_{m}}}\theta \right), \end{cases} 内トロコイド: \begin{cases} x=(r_{c}-r_{m})\cos \theta +r_{d}\cos \left({\cfrac {r_{c}-r_{m}}{r_{m}}}\theta \right),\\y=(r_{c}-r_{m})\sin \theta -r_{d}\sin \left({\cfrac {r_{c}-r_{m}}{r_{m}}}\theta \right), \end{cases} ## 一般形トロコイド 描画円がなぞる曲線を$\boldsymbol{p}$とし，接線方向ベクトルは$\boldsymbol{t}$,法線方向ベクトルは$\boldsymbol{n}$とする 円の転がってきたパスの長さ$s$は \begin{align} s &= \sum_0^n \sqrt{dx[i]^2 + dy[i]^2} \end{align} と表される 描画円は滑らずに転がるため \begin{align} r_m\theta &= s \end{align} したがって， \begin{align} \theta & = s/r_m \\ d\theta &= ds/r_m , \end{align} となる． 点$p[i]$における接線ベクトル$t[i]$,法線ベクトル$n[i]$は \begin{align} t[i] &= \frac{p[i+1]-p[i-1]}{2} \\ n[i] &= \frac{t[i+1]-t[i-1]}{2} \\ &= \frac{p[i+2]-p[i-2]}{4} \end{align} 描画円の中心の位置$p_m[i]$は単位法線ベクトル$n_u[i]$を用いて， \begin{align} p_m[i] &= p[i] + r_m n_u[i] \end{align} となる．ただし，$n_u[i]$は曲線の凹凸によって向きが反転する．これを避けるため，$n_u[i]$は改めて， \begin{align} n_u[i] &= R（\frac{\pi}{2}） t_u[i] \end{align} とする．ここで$R$はの回転行列である． 描画点の座標$p_d[i]$は \begin{align} p_d &= p_m[i] + r_d R(\theta[i]) e_x \end{align} となる．ここで$e_x$はx軸方向の単位ベクトルである ```python # trocoid along sin-curve fig = plt.figure() fig.add_subplot(1, 1, 1) sq_info = {} x=np.linspace(0,10,num=150) y=np.sin(x) plt.plot(x,y) x,y = trochoid(px=x,py=y,rm=0.1,rd=0.1) plt.plot(x,y) fig.axes[0].set_aspect('equal', 'box') # plt.savefig("masic.pdf") ``` ```python # trocid on path fig = plt.figure() fig.add_subplot(1, 1, 1) sq_info = {} x0,y0 = polygon(64, 1) plt.plot(x0,y0) x1,y1 = ctrochoid(rc=0.9, rm=0.99*1/3, rd=0.4,n=1024,outer=False) print(lcm(9,4)) # print(plot_trochoid(rc=0.95, rm=0.3, rd=0.4,outer=False,rmax=0.98,n=100)) plt.plot(x1,y1) # x,y=trochoid(rc=1.2, rm=1/3, rd=1/3,outer=False,n=100) # plt.plot(x,y) n = 7 # col_inverse = a[:, ::-1] rm = path_length(x1,y1)/(2*np.pi)/n x,y = trochoid(px=np.tile(x1,n),py=np.tile(y1,n),rm=rm,rd=0.02*rm) plt.plot(x,y) # x,y = ptrochoid(px=np.tile(x1,2),py=np.tile(y1,2),rm=0.1,rd=0.4) # plt.plot(x,y) # x2,y2 = trochoid(rc=0.9, rm=0.15/2, rd=1.2,n=1024) # plt.plot(x2,y2) # x,y = ptrochoid(px=np.tile(x2,12),py=np.tile(y2,12),rm=3,rd=0.4) # plt.plot(x,y) fig.axes[0].set_aspect('equal', 'box') # plt.savefig("trocids002.pdf") ``` ```python # trocid on path fig = plt.figure() fig.add_subplot(1, 1, 1) x0,y0 = polygon(64, 1) plt.plot(x0,y0) x1,y1 = ctrochoid(rc=1, rm=1/6, rd=0.8*1/6,n=1024,outer=False) plt.plot(x1,y1) m = 7 n = 11/m rm = path_length(x1,y1)/(2*np.pi)/n fig.axes[0].set_aspect('equal', 'box') plt.savefig("trocids002.pdf") ``` ```python def demo_trochoid(px, py, rm, rd, right=True, rmax=None, orient=0, *args, **kwargs): x = np.zeros(len(px)) y = np.zeros(len(py)) s = 0 theta = 0 rot = np.pi/2 if right is True: rot = -rot r_mat = np.matrix( [[np.cos(rot), -np.sin(rot)], [np.sin(rot), np.cos(rot)],] ) for i in range(len(px)): ds = 0 if i > 0: ds = np.linalg.norm(np.array([px[i]-px[i-1], py[i]-py[i-1]])) d_theta = ds/rm s = s+ds theta = theta + d_theta if (i - 1) < 0 : t = np.array([px[i+1]-px[i], py[i+1]-py[i]]) elif (i+1) >= len(px): t = np.array([px[i]-px[i-1], py[i]-py[i-1]]) else : t = np.array([px[i+1]-px[i-1], py[i+1]-py[i-1]])*0.5 t = t/(np.linalg.norm(t) + 1e-9) n = np.dot(r_mat, np.reshape(t, (2, 1))) p_m = np.array([[px[i]], [py[i]]]) + rm*n r_ort = np.matrix( [[np.cos(theta), -np.sin(theta)], [np.sin(theta), np.cos(theta)],] ) p_d = p_m + rd*np.dot(r_ort, np.array([[1], [0]])) x[i] = p_d.item(0) y[i] = p_d.item(1) if i %10 == 0: p_m=np.reshape(p_m,(1,-1)) plt.plot(*polygon(n=32,r=rm,cx=p_m.item(0),cy=p_m.item(1),color='gray',alpha=0.5) ) plt.plot([p_m.item(0),x[i]],[p_m.item(1),y[i]],'r-o') # info = {'rot': i} return (x,y) ``` ```python # trocoid along sin-curve fig = plt.figure() fig.add_subplot(1, 1, 1) sq_info = {} x=np.linspace(0,10,num=150) y=np.sin(x) plt.plot(x,y) x,y = demo_trochoid(px=x,py=y,rm=0.2,rd=0.3) plt.plot(x,y) fig.axes[0].set_aspect('equal', 'box') plt.savefig("torocoid_sin.pdf") ```

78d509b5bc4c1a7bfbd8deba09f8b024307690a9

ipynb

Jupyter Notebook

demo.ipynb

botamochi6277/trochoid-py

13efc06c86ed60e4b682d4b5c98d3ee6ad401a25

demo.ipynb

botamochi6277/trochoid-py

13efc06c86ed60e4b682d4b5c98d3ee6ad401a25

demo.ipynb

botamochi6277/trochoid-py

13efc06c86ed60e4b682d4b5c98d3ee6ad401a25

Qwen/Qwen-72B

1. YES 2. YES

__label__yue_Hant

# Mass Maps From Mass-Luminosity Inference Posterior In this notebook we start to explore the potential of using a mass-luminosity relation posterior to refine mass maps. Content: - [Math](#Math) - [Imports, Constants, Utils, Data](#Imports,-Constants,-Utils,-Data) - [Probability Functions](#Probability-Functions) - [Results](#Results) - [Discussion](#Discussion) ### Math Infering mass from mass-luminosity relation posterior ... \begin{align} P(M|L_{obs},z,\sigma_L^{obs}) &= \iint P(M|\alpha, S, L_{obs}, z)P(\alpha, S|L_{obs},z,\sigma_L^{obs})\ d\alpha dS\\ &\propto \iiint P(L_{obs}| L,\sigma_L^{obs})P(L|M,\alpha,S,z)P(M|z)P(\alpha, S|L_{obs},z,\sigma_L^{obs})\ dLd\alpha dS\\ &\approx \frac{P(M|z)}{n_{\alpha,S}}\sum_{\alpha,S \sim P(\alpha, S|L_{obs},z,\sigma_L^{obs})}\left( \frac{1}{n_L}\sum_{L\sim P(L|M,\alpha,S,z)}P(L_{obs}|L,\sigma_L^{obs})\right)\\ &= \frac{P(M|z)}{n_{\alpha,S}}\sum_{\alpha,S \sim P(\alpha, S|L_{obs},z,\sigma_L^{obs})}f(M;\alpha,S,z)\\ \end{align} Refine for individual halo ... \begin{align} P(M_k|L_{obs},z,\sigma_L^{obs}) &= \iint P(M_k|\alpha, S, L_{obs\ k}, z_k)P(\alpha, S|L_{obs},z,\sigma_L^{obs})\ d\alpha dS\\ &\propto \iiint P(L_{obs\ k}| L_k,\sigma_L^{obs})P(L_k|M_k,\alpha,S,z_k)P(M_k|z_k)P(\alpha, S|L_{obs},z,\sigma_L^{obs})\ dLd\alpha dS\\ &\approx \frac{P(M_k|z_k)}{n_{\alpha,S}}\sum_{\alpha,S \sim P(\alpha, S|L_{obs},z,\sigma_L^{obs})}\left( \frac{1}{n_L}\sum_{L\sim P(L_k|M_k,\alpha,S,z_k)}P(L_{obs\ k}|L_k,\sigma_L^{obs})\right)\\ &=\frac{P(M_k|z_k)}{n_{\alpha,S}}\sum_{\alpha,S \sim P(\alpha, S|L_{obs},z,\sigma_L^{obs})}f(M_k;\alpha,S,z_k)\\ \end{align} Can also factor it more conventionally for MCMC ... \begin{align} \underbrace{P(M_k|L_{obs},z,\sigma_L^{obs})}_{posterior} &\propto \underbrace{P(M_k|z_k)}_{prior}\underbrace{\iiint P(L_{obs\ k}| L_k,\sigma_L^{obs})P(L_k|M_k,\alpha,S,z_k)P(\alpha, S|L_{obs},z,\sigma_L^{obs})\ dLd\alpha dS}_{likelihood}\\ \end{align} In the code we have the following naming convention: - p1 for $P(M|z)$ - p2 for $P(\alpha, S|L_{obs},z,\sigma_L^{obs})$ - p3 for $P(L_k|M_k,\alpha,S,z_k)$ - p4 for $P(L_{obs\ k}|L_k, \sigma^{obs}_L)$ We use the terms **eval** and **samp** to help distinguish between evaluating a distribution and sampling from it. ### Imports, Constants, Utils, Data ```python %matplotlib inline import matplotlib.pyplot as plt from matplotlib import rc rc('text', usetex=True) from bigmali.grid import Grid from bigmali.prior import TinkerPrior from bigmali.hyperparameter import get import numpy as np from scipy.stats import lognorm from numpy.random import normal #globals that functions rely on grid = Grid() prior = TinkerPrior(grid) a_seed = get()[:-1] S_seed = get()[-1] mass_points = prior.fetch(grid.snap(0)).mass[2:-2] # cut edges tmp = np.loadtxt('/Users/user/Code/PanglossNotebooks/MassLuminosityProject/SummerResearch/mass_mapping.txt') z_data = tmp[:,0] lobs_data = tmp[:,1] mass_data = tmp[:,2] ra_data = tmp[:,3] dec_data = tmp[:,4] sigobs = 0.05 def fast_lognormal(mu, sigma, x): return (1/(x * sigma * np.sqrt(2 * np.pi))) * np.exp(- 0.5 * (np.log(x) - np.log(mu)) ** 2 / sigma ** 2) ``` ### Probability Functions ```python def p1_eval(zk): return prior.fetch(grid.snap(zk)).prob[2:-2] def p2_samp(nas=100): """ a is fixed on hyperseed, S is normal distribution centered at hyperseed. """ return normal(S_seed, S_seed / 10, size=nas) def p3_samp(mk, a, S, zk, nl=100): mu_lum = np.exp(a[0]) * ((mk / a[2]) ** a[1]) * ((1 + zk) ** (a[3])) return lognorm(S, scale=mu_lum).rvs(nl) def p4_eval(lobsk, lk, sigobs): return fast_lognormal(lk, sigobs, lobsk) def f(a, S, zk, lobsk, nl=100): ans = [] for mk in mass_points: tot = 0 for x in p3_samp(mk, a, S, zk, nl): tot += p4_eval(lobsk, x, sigobs) ans.append(tot / nl) return ans def mass_dist(ind=1, nas=10, nl=100): lobsk = lobs_data[ind] zk = z_data[ind] tot = np.zeros(len(mass_points)) for S in p2_samp(nas): tot += f(a_seed, S, zk, lobsk, nl) prop = p1_eval(zk) * tot / nas return prop / np.trapz(prop, x=mass_points) ``` ### Results ```python plt.subplot(3,3,1) dist = p1_eval(zk) plt.plot(mass_points, dist) plt.gca().set_xscale('log') plt.gca().set_yscale('log') plt.ylim([10**-25, 10]) plt.xlim([mass_points.min(), mass_points.max()]) plt.title('Prior') plt.xlabel(r'Mass $(M_\odot)$') plt.ylabel('Density') for ind in range(2,9): plt.subplot(3,3,ind) dist = mass_dist(ind) plt.plot(mass_points, dist, alpha=0.6, linewidth=2) plt.xlim([mass_points.min(), mass_points.max()]) plt.gca().set_xscale('log') plt.gca().set_yscale('log') plt.ylim([10**-25, 10]) plt.gca().axvline(mass_data[ind], color='red', linewidth=2, alpha=0.6) plt.title('Mass Distribution') plt.xlabel(r'Mass $(M_\odot)$') plt.ylabel('Density') # most massive ind = np.argmax(mass_data) plt.subplot(3,3,9) dist = mass_dist(ind) plt.plot(mass_points, dist, alpha=0.6, linewidth=2) plt.gca().set_xscale('log') plt.gca().set_yscale('log') plt.xlim([mass_points.min(), mass_points.max()]) plt.ylim([10**-25, 10]) plt.gca().axvline(mass_data[ind], color='red', linewidth=2, alpha=0.6) plt.title('Mass Distribution') plt.xlabel(r'Mass $(M_\odot)$') plt.ylabel('Density') # plt.tight_layout() plt.gcf().set_size_inches((10,6)) ``` ### Turning into Probabilistic Catalogue ```python index = range(2,9) + [np.argmax(mass_data)] plt.title('Simple Sketch of Field of View') plt.scatter(ra_data[index], dec_data[index] , s=np.log(mass_data[index]), alpha=0.6) plt.xlabel('ra') plt.ylabel('dec'); ``` Need to build graphic: - Make grid that will correspond to pixels - map ra-dec window to grid - snap objects onto grid and accumulate mass in each bin of the grid - plot the grayscale image # Discussion - While this is a simple toy model, the consistency between then predicted mass distribution and true mass is encouraging. - The noise in the mass distribution plots is interesting. The noise increases for masses that are further away than the truth. A similar effect may also exist in bigmali, could it lead to a failure mode? - In order to build probabilistic mass maps we will need to be able to sample from the mass distributions. One way to do this would be fitting a normal distribution and drawing from that distribution. This will also mitigate the influence of the noise for masses far from the true mass.

c9c84f045d7357dc237216cb26b7bb1889131558

ipynb

Jupyter Notebook

MassLuminosityProject/SummerResearch/MassMapsFromMassLuminosity_20170626.ipynb

davidthomas5412/PanglossNotebooks

719a3b9a5d0e121f0e9bc2a92a968abf7719790f

MassLuminosityProject/SummerResearch/MassMapsFromMassLuminosity_20170626.ipynb

davidthomas5412/PanglossNotebooks

719a3b9a5d0e121f0e9bc2a92a968abf7719790f

2016-12-13T02:05:57.000Z

2017-01-21T02:16:27.000Z

MassLuminosityProject/SummerResearch/MassMapsFromMassLuminosity_20170626.ipynb

davidthomas5412/PanglossNotebooks

719a3b9a5d0e121f0e9bc2a92a968abf7719790f

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python import numpy as np import sympy as sym x_1, x_2, x_3, x_4 = sym.symbols('x_1, x_2, x_3, x_4') y_1, y_2, y_3, y_4 = sym.symbols('y_1, y_2, y_3, y_4') XY = sym.Matrix([sym.symbols('x_1, x_2, x_3, x_4'), sym.symbols('y_1, y_2, y_3, y_4')]).transpose() xi, eta = sym.symbols('xi, eta') basis = sym.Matrix([xi, eta]) N1 = (1-xi)*(1-eta)/4 N2 = (1+xi)*(1-eta)/4 N3 = (1+xi)*(1+eta)/4 N4 = (1-xi)*(1+eta)/4 NN = sym.Matrix([N1, N2, N3, N4]) coordinate = (NN.transpose()*XY).transpose() NN_diff = NN.jacobian(basis) jacob = coordinate.jacobian(basis) jacob_det = sym.det(jacob) jacob_inv = sym.inv_quick(jacob) NN_diff_global = (NN_diff*jacob_inv) temp = NN_diff_global*NN_diff_global.transpose()/jacob_det sq3 = np.sqrt(1/3) ``` ```python NN ``` $\displaystyle \left[\begin{matrix}\frac{\left(1 - \eta\right) \left(1 - \xi\right)}{4}\\\frac{\left(1 - \eta\right) \left(\xi + 1\right)}{4}\\\frac{\left(\eta + 1\right) \left(\xi + 1\right)}{4}\\\frac{\left(1 - \xi\right) \left(\eta + 1\right)}{4}\end{matrix}\right]$ ```python coordinate ``` $\displaystyle \left[\begin{matrix}\frac{x_{1} \left(1 - \eta\right) \left(1 - \xi\right)}{4} + \frac{x_{2} \left(1 - \eta\right) \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\eta + 1\right) \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right) \left(\eta + 1\right)}{4}\\\frac{y_{1} \left(1 - \eta\right) \left(1 - \xi\right)}{4} + \frac{y_{2} \left(1 - \eta\right) \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\eta + 1\right) \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right) \left(\eta + 1\right)}{4}\end{matrix}\right]$ ```python jacob ``` $\displaystyle \left[\begin{matrix}- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4} & - \frac{x_{1} \left(1 - \xi\right)}{4} - \frac{x_{2} \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right)}{4}\\- \frac{y_{1} \left(1 - \eta\right)}{4} + \frac{y_{2} \left(1 - \eta\right)}{4} + \frac{y_{3} \left(\eta + 1\right)}{4} - \frac{y_{4} \left(\eta + 1\right)}{4} & - \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}\end{matrix}\right]$ ```python jacob.subs(([x_1, -0.7], [x_2, 0.2], [x_3, 0.8], [x_4, 0], [y_1, 0.3], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` $\displaystyle \left[\begin{matrix}0.425 - 0.025 \eta & 0.325 - 0.025 \xi\\0.075 \eta - 0.475 & 0.075 \xi + 0.325\end{matrix}\right]$ ```python jacob.subs(([xi, sq3], [eta, -sq3], [x_1, -0.8], [x_2, 0], [x_3, 0.8], [x_4, 0], [y_1, 0], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` $\displaystyle \left[\begin{matrix}0.5 & 0.5\\-0.5 & 0.5\end{matrix}\right]$ ```python jaco_inv = sym.inv_quick(jaco) jaco_inv ``` $\displaystyle \left[\begin{matrix}1.25 & -1.25\\1.25 & 1.25\end{matrix}\right]$ ```python NN_diff.subs(([xi, sq3], [eta, -sq3])) ``` $\displaystyle \left[\begin{matrix}-0.394337567297406 & -0.105662432702594\\0.394337567297406 & -0.394337567297406\\0.105662432702594 & 0.394337567297406\\-0.105662432702594 & 0.105662432702594\end{matrix}\right]$ ```python NN_diff ``` $\displaystyle \left[\begin{matrix}\frac{\eta}{4} - \frac{1}{4} & \frac{\xi}{4} - \frac{1}{4}\\\frac{1}{4} - \frac{\eta}{4} & - \frac{\xi}{4} - \frac{1}{4}\\\frac{\eta}{4} + \frac{1}{4} & \frac{\xi}{4} + \frac{1}{4}\\- \frac{\eta}{4} - \frac{1}{4} & \frac{1}{4} - \frac{\xi}{4}\end{matrix}\right]$ ```python jacob_det ``` $\displaystyle - \frac{\eta x_{1} y_{2}}{8} + \frac{\eta x_{1} y_{3}}{8} + \frac{\eta x_{2} y_{1}}{8} - \frac{\eta x_{2} y_{4}}{8} - \frac{\eta x_{3} y_{1}}{8} + \frac{\eta x_{3} y_{4}}{8} + \frac{\eta x_{4} y_{2}}{8} - \frac{\eta x_{4} y_{3}}{8} - \frac{x_{1} \xi y_{3}}{8} + \frac{x_{1} \xi y_{4}}{8} + \frac{x_{1} y_{2}}{8} - \frac{x_{1} y_{4}}{8} + \frac{x_{2} \xi y_{3}}{8} - \frac{x_{2} \xi y_{4}}{8} - \frac{x_{2} y_{1}}{8} + \frac{x_{2} y_{3}}{8} + \frac{x_{3} \xi y_{1}}{8} - \frac{x_{3} \xi y_{2}}{8} - \frac{x_{3} y_{2}}{8} + \frac{x_{3} y_{4}}{8} - \frac{x_{4} \xi y_{1}}{8} + \frac{x_{4} \xi y_{2}}{8} + \frac{x_{4} y_{1}}{8} - \frac{x_{4} y_{3}}{8}$ ```python jacob_inv ``` $\displaystyle \left[\begin{matrix}\frac{- \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}}{\left(- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}\right) - \left(- \frac{x_{1} \left(1 - \xi\right)}{4} - \frac{x_{2} \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \eta\right)}{4} + \frac{y_{2} \left(1 - \eta\right)}{4} + \frac{y_{3} \left(\eta + 1\right)}{4} - \frac{y_{4} \left(\eta + 1\right)}{4}\right)} & \frac{\frac{x_{1} \left(1 - \xi\right)}{4} + \frac{x_{2} \left(\xi + 1\right)}{4} - \frac{x_{3} \left(\xi + 1\right)}{4} - \frac{x_{4} \left(1 - \xi\right)}{4}}{\left(- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}\right) - \left(- \frac{x_{1} \left(1 - \xi\right)}{4} - \frac{x_{2} \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \eta\right)}{4} + \frac{y_{2} \left(1 - \eta\right)}{4} + \frac{y_{3} \left(\eta + 1\right)}{4} - \frac{y_{4} \left(\eta + 1\right)}{4}\right)}\\\frac{\frac{y_{1} \left(1 - \eta\right)}{4} - \frac{y_{2} \left(1 - \eta\right)}{4} - \frac{y_{3} \left(\eta + 1\right)}{4} + \frac{y_{4} \left(\eta + 1\right)}{4}}{\left(- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}\right) - \left(- \frac{x_{1} \left(1 - \xi\right)}{4} - \frac{x_{2} \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \eta\right)}{4} + \frac{y_{2} \left(1 - \eta\right)}{4} + \frac{y_{3} \left(\eta + 1\right)}{4} - \frac{y_{4} \left(\eta + 1\right)}{4}\right)} & \frac{- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4}}{\left(- \frac{x_{1} \left(1 - \eta\right)}{4} + \frac{x_{2} \left(1 - \eta\right)}{4} + \frac{x_{3} \left(\eta + 1\right)}{4} - \frac{x_{4} \left(\eta + 1\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \xi\right)}{4} - \frac{y_{2} \left(\xi + 1\right)}{4} + \frac{y_{3} \left(\xi + 1\right)}{4} + \frac{y_{4} \left(1 - \xi\right)}{4}\right) - \left(- \frac{x_{1} \left(1 - \xi\right)}{4} - \frac{x_{2} \left(\xi + 1\right)}{4} + \frac{x_{3} \left(\xi + 1\right)}{4} + \frac{x_{4} \left(1 - \xi\right)}{4}\right) \left(- \frac{y_{1} \left(1 - \eta\right)}{4} + \frac{y_{2} \left(1 - \eta\right)}{4} + \frac{y_{3} \left(\eta + 1\right)}{4} - \frac{y_{4} \left(\eta + 1\right)}{4}\right)}\end{matrix}\right]$ ```python D ``` $\displaystyle \left[\begin{matrix}1 & 1\\1 & 1\end{matrix}\right]$ ```python a = NN_diff_global.subs(([xi, -sq3], [eta, -sq3], [x_1, -0.8], [x_2, 0], [x_3, 0.8], [x_4, 0], [y_1, 0], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` ```python b = NN_diff_global.subs(([xi, sq3], [eta, -sq3], [x_1, -0.8], [x_2, 0], [x_3, 0.8], [x_4, 0], [y_1, 0], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` ```python c = NN_diff_global.subs(([xi, sq3], [eta, +sq3], [x_1, -0.8], [x_2, 0], [x_3, 0.8], [x_4, 0], [y_1, 0], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` ```python d = NN_diff_global.subs(([xi, -sq3], [eta, +sq3], [x_1, -0.8], [x_2, 0], [x_3, 0.8], [x_4, 0], [y_1, 0], [y_2, -0.8], [y_3, 0], [y_4, 0.8])) ``` ```python a ``` $\displaystyle \left[\begin{matrix}-0.985843918243516 & 0\\0.360843918243516 & -0.625\\0.264156081756484 & 0\\0.360843918243516 & 0.625\end{matrix}\right]$ ```python aa = np.array(a, dtype=np.float32) bb = np.array(b, dtype=np.float32) cc = np.array(c, dtype=np.float32) dd = np.array(d, dtype=np.float32) ``` ```python aa ``` array([[-0.9858439 , 0. ], [ 0.36084393, -0.625 ], [ 0.26415607, 0. ], [ 0.36084393, 0.625 ]], dtype=float32) ```python np.einsum('mi,ni->imn', aa, aa)*0.32 ``` array([[[ 0.31100422, -0.11383545, -0.08333333, -0.11383545], [-0.11383545, 0.04166667, 0.03050212, 0.04166667], [-0.08333333, 0.03050212, 0.0223291 , 0.03050212], [-0.11383545, 0.04166667, 0.03050212, 0.04166667]], [[ 0. , 0. , 0. , 0. ], [ 0. , 0.125 , 0. , -0.125 ], [ 0. , 0. , 0. , 0. ], [ 0. , -0.125 , 0. , 0.125 ]]], dtype=float32) ```python (np.einsum('mi,ni->imn', aa, aa)+np.einsum('mi,ni->imn', bb, bb)+\ np.einsum('mi,ni->imn', cc, cc)+np.einsum('mi,ni->imn', dd, dd))*0.32 ``` array([[[ 0.5833333 , -0.08333333, -0.41666663, -0.08333333], [-0.08333333, 0.08333334, -0.08333333, 0.08333334], [-0.41666663, -0.08333333, 0.5833333 , -0.08333333], [-0.08333333, 0.08333334, -0.08333333, 0.08333334]], [[ 0.08333334, -0.08333333, 0.08333334, -0.08333333], [-0.08333333, 0.5833333 , -0.08333333, -0.41666663], [ 0.08333334, -0.08333333, 0.08333334, -0.08333333], [-0.08333333, -0.41666663, -0.08333333, 0.5833333 ]]], dtype=float32) ```python ```

e11011c15250383c3c63214d0a31e762f294cf01

ipynb

Jupyter Notebook

equations4twoDimensionalElement.ipynb

AndrewWangJZ/pyfem

8e7df6aa69c1c761bb8ec67302847e30a83190b4

2022-03-10T17:22:53.000Z

2022-03-10T17:22:53.000Z

equations4twoDimensionalElement.ipynb

AndrewWangJZ/pyfem

8e7df6aa69c1c761bb8ec67302847e30a83190b4

equations4twoDimensionalElement.ipynb

AndrewWangJZ/pyfem

8e7df6aa69c1c761bb8ec67302847e30a83190b4

2022-03-10T12:47:34.000Z

2022-03-10T13:25:18.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__yue_Hant

# Descriptive Statistics. Working with dataset: - https://archive.ics.uci.edu/ml/datasets/Vertebral+Column ```python import numpy as np import pandas as pd import matplotlib.pyplot as plt ``` Vamos a ver aquí algunas medidas iniciales para el análisis de datos. El análisis inicial de los datos es muy importante para hacer estimaciones y sacar conclusiones. ```python names=["pelvic_incidence","pelvic_tilt","lumbar_lordosis_angle","sacral_slope","pelvic_radius","degree_spondylolisthesis","class"] data=pd.read_csv("../Datas/VertebralColumn/column_2C.dat",delimiter="\s+",names=names) ``` ```python data ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>pelvic_incidence</th> <th>pelvic_tilt</th> <th>lumbar_lordosis_angle</th> <th>sacral_slope</th> <th>pelvic_radius</th> <th>degree_spondylolisthesis</th> <th>class</th> </tr> </thead> <tbody> <tr> <th>0</th> <td>63.03</td> <td>22.55</td> <td>39.61</td> <td>40.48</td> <td>98.67</td> <td>-0.25</td> <td>AB</td> </tr> <tr> <th>1</th> <td>39.06</td> <td>10.06</td> <td>25.02</td> <td>29.00</td> <td>114.41</td> <td>4.56</td> <td>AB</td> </tr> <tr> <th>2</th> <td>68.83</td> <td>22.22</td> <td>50.09</td> <td>46.61</td> <td>105.99</td> <td>-3.53</td> <td>AB</td> </tr> <tr> <th>3</th> <td>69.30</td> <td>24.65</td> <td>44.31</td> <td>44.64</td> <td>101.87</td> <td>11.21</td> <td>AB</td> </tr> <tr> <th>4</th> <td>49.71</td> <td>9.65</td> <td>28.32</td> <td>40.06</td> <td>108.17</td> <td>7.92</td> <td>AB</td> </tr> <tr> <th>...</th> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> </tr> <tr> <th>305</th> <td>47.90</td> <td>13.62</td> <td>36.00</td> <td>34.29</td> <td>117.45</td> <td>-4.25</td> <td>NO</td> </tr> <tr> <th>306</th> <td>53.94</td> <td>20.72</td> <td>29.22</td> <td>33.22</td> <td>114.37</td> <td>-0.42</td> <td>NO</td> </tr> <tr> <th>307</th> <td>61.45</td> <td>22.69</td> <td>46.17</td> <td>38.75</td> <td>125.67</td> <td>-2.71</td> <td>NO</td> </tr> <tr> <th>308</th> <td>45.25</td> <td>8.69</td> <td>41.58</td> <td>36.56</td> <td>118.55</td> <td>0.21</td> <td>NO</td> </tr> <tr> <th>309</th> <td>33.84</td> <td>5.07</td> <td>36.64</td> <td>28.77</td> <td>123.95</td> <td>-0.20</td> <td>NO</td> </tr> </tbody> </table> <p>310 rows × 7 columns</p> </div> Podemos hacer primero una visión general de nuestros datos para entender cual es el compotamiento de las variables que tenemos en el estudio. Esto podemos hacerlo muy fácilmente con Python usando una librería de pandas y usando algunos de los algoritmos y funciones predefinidas. ```python print("El número de variables es N_var =",len(data.columns)) print("El número de casos o instancias (o individuos estudiados) es N =",len(data)) ``` El número de variables es N_var = 7 El número de casos o instancias (o individuos estudiados) es N = 310 Algunos estdísticos comunes y usuales. Por ejemplo podemos calcular la media, mediana y moda, así como también las desviaciones estándar y varianzas para cada una de las 7 variables. - Media (Promedio): \begin{equation} \bar x=\frac{1}{N}\sum_{i=0}^{N-1}x_i \end{equation} - Mediana (Cuartil 50): Con datos ordenados en orden ascendente: \begin{equation} median=\left\{\begin{matrix} x\left[\frac{N}{2}\right] & \mbox{si $N$ es par}\\ \frac{1}{2}\left(x\left[\frac{N-1}{2}\right]+x\left[\frac{N+1}{2}\right]\right) & \mbox{si $N$ es impar}\\ \end{matrix}\right. \end{equation} - Moda (dato con mayor frecuencia) - varianza: \begin{equation} s^2(x)=\mbox{Var}(x)=\frac{1}{N}\sum_{i=0}^{N-1}(x_i-\bar x)^2 \end{equation} - Desviación estándar: \begin{equation} \sigma_x=s(x)=\sqrt{\mbox{Var}(x)}=\sqrt{\frac{1}{N}\sum_{i=0}^{N-1}(x_i-\bar x)^2} \end{equation} ```python def average(var): return np.sum(var)/len(var) def median(var): var_ord=np.sort(var) N=len(var_ord) if N%2==0: median=var_ord[N//2] if N%2!=0: median=(var_ord[(N-1)//2]+var_ord[(N+1)//2])/2 return median def mode(var): dat,frec=np.unique(var,return_counts=True) if len(dat[frec==frec.max()])>1: print("Existen varias modas (la variable es multimodal)") if len(dat[frec==frec.max()])==1: print("Existe una moda (la variable es unimodal)") return dat[frec==frec.max()] ``` ```python def variance(var): return np.sum((var-average(var))**2)/len(var) def std(var): return np.sqrt(variance(var)) ``` ```python for i in data.columns[:-1]: print("Para la variable "+str(i),"el promedio (la media) es <{}> =".format(str(i)),average(data[i])) ``` Para la variable pelvic_incidence el promedio (la media) es <pelvic_incidence> = 60.49648387096773 Para la variable pelvic_tilt el promedio (la media) es <pelvic_tilt> = 17.542903225806448 Para la variable lumbar_lordosis_angle el promedio (la media) es <lumbar_lordosis_angle> = 51.93070967741936 Para la variable sacral_slope el promedio (la media) es <sacral_slope> = 42.953870967741935 Para la variable pelvic_radius el promedio (la media) es <pelvic_radius> = 117.92054838709676 Para la variable degree_spondylolisthesis el promedio (la media) es <degree_spondylolisthesis> = 26.296741935483873 ```python for i in data.columns[:-1]: print("Para la variable "+str(i),"la mediana es med({}) =".format(str(i)),median(data[i])) ``` Para la variable pelvic_incidence la mediana es med(pelvic_incidence) = 58.78 Para la variable pelvic_tilt la mediana es med(pelvic_tilt) = 16.42 Para la variable lumbar_lordosis_angle la mediana es med(lumbar_lordosis_angle) = 49.78 Para la variable sacral_slope la mediana es med(sacral_slope) = 42.44 Para la variable pelvic_radius la mediana es med(pelvic_radius) = 118.34 Para la variable degree_spondylolisthesis la mediana es med(degree_spondylolisthesis) = 12.07 ```python for i in data.columns[:-1]: print("Para la variable "+str(i),"la moda es moda({}) =".format(str(i)),mode(data[i])) ``` Existen varias modas (la variable es multimodal) Para la variable pelvic_incidence la moda es moda(pelvic_incidence) = [42.52 49.71 50.91 53.94 54.92 63.03 65.01 65.76 74.72] Existen varias modas (la variable es multimodal) Para la variable pelvic_tilt la moda es moda(pelvic_tilt) = [ 5.27 8.4 10.06 10.22 10.76 13.11 13.28 13.92 14.38 15.4 16.42 16.74 17.44 19.44 21.12 23.08 26.33 33.28 37.52] Existen varias modas (la variable es multimodal) Para la variable lumbar_lordosis_angle la moda es moda(lumbar_lordosis_angle) = [35. 42. 47. 52. 58.] Existe una moda (la variable es unimodal) Para la variable sacral_slope la moda es moda(sacral_slope) = [56.31] Existen varias modas (la variable es multimodal) Para la variable pelvic_radius la moda es moda(pelvic_radius) = [110.71 116.56 116.59 116.8 117.98 119.32 129.39] Existen varias modas (la variable es multimodal) Para la variable degree_spondylolisthesis la moda es moda(degree_spondylolisthesis) = [-4.08 -2.01 1.01 3.09 4.96] ```python for i in data.columns[:-1]: print("Para la variable "+str(i),"la varianza es Var({}) =".format(str(i)),variance(data[i])) ``` Para la variable pelvic_incidence la varianza es Var(pelvic_incidence) = 296.12512021748176 Para la variable pelvic_tilt la varianza es Var(pelvic_tilt) = 99.83976124869929 Para la variable lumbar_lordosis_angle la varianza es Var(lumbar_lordosis_angle) = 343.13176723829343 Para la variable sacral_slope la varianza es Var(sacral_slope) = 179.5889740478668 Para la variable pelvic_radius la varianza es Var(pelvic_radius) = 176.7871277637877 Para la variable degree_spondylolisthesis la varianza es Var(degree_spondylolisthesis) = 1406.119148417274 ```python for i in data.columns[:-1]: print("Para la variable "+str(i),"la desviación estándar es STD({}) =".format(str(i)),std(data[i])) ``` Para la variable pelvic_incidence la desviación estándar es STD(pelvic_incidence) = 17.208286382364797 Para la variable pelvic_tilt la desviación estándar es STD(pelvic_tilt) = 9.991984850303732 Para la variable lumbar_lordosis_angle la desviación estándar es STD(lumbar_lordosis_angle) = 18.523816216921755 Para la variable sacral_slope la desviación estándar es STD(sacral_slope) = 13.401081077579779 Para la variable pelvic_radius la desviación estándar es STD(pelvic_radius) = 13.296132060256761 Para la variable degree_spondylolisthesis la desviación estándar es STD(degree_spondylolisthesis) = 37.49825527164263 También podemos usar una función de Pandas que reune algunos estadísticos básicos: ```python vertebral_stats=data.describe().T ``` ```python vertebral_stats ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>count</th> <th>mean</th> <th>std</th> <th>min</th> <th>25%</th> <th>50%</th> <th>75%</th> <th>max</th> </tr> </thead> <tbody> <tr> <th>pelvic_incidence</th> <td>310.0</td> <td>60.496484</td> <td>17.236109</td> <td>26.15</td> <td>46.4325</td> <td>58.690</td> <td>72.8800</td> <td>129.83</td> </tr> <tr> <th>pelvic_tilt</th> <td>310.0</td> <td>17.542903</td> <td>10.008140</td> <td>-6.55</td> <td>10.6675</td> <td>16.360</td> <td>22.1200</td> <td>49.43</td> </tr> <tr> <th>lumbar_lordosis_angle</th> <td>310.0</td> <td>51.930710</td> <td>18.553766</td> <td>14.00</td> <td>37.0000</td> <td>49.565</td> <td>63.0000</td> <td>125.74</td> </tr> <tr> <th>sacral_slope</th> <td>310.0</td> <td>42.953871</td> <td>13.422748</td> <td>13.37</td> <td>33.3475</td> <td>42.405</td> <td>52.6925</td> <td>121.43</td> </tr> <tr> <th>pelvic_radius</th> <td>310.0</td> <td>117.920548</td> <td>13.317629</td> <td>70.08</td> <td>110.7100</td> <td>118.265</td> <td>125.4675</td> <td>163.07</td> </tr> <tr> <th>degree_spondylolisthesis</th> <td>310.0</td> <td>26.296742</td> <td>37.558883</td> <td>-11.06</td> <td>1.6000</td> <td>11.765</td> <td>41.2850</td> <td>418.54</td> </tr> </tbody> </table> </div> también podemos usar las funciones que están predefinidas en Python para estas tareas: ```python import scipy.stats as st # La mediana puede ser también calculada, así como el coeficiente de asimetría y curtosis respectivamente print(data.columns) print("media =",np.mean(data.loc[:, data.columns != 'class'],axis=0)) print("median =",np.median(data.loc[:, data.columns != 'class'],axis=0)) print("varianza =",np.var(data.loc[:, data.columns != 'class'],axis=0)) print("desviación estándar =",np.std(data.loc[:, data.columns != 'class'],axis=0)) print("skewness =",st.skew(data.loc[:, data.columns != 'class'],axis=0)) print("kurtosis =",st.kurtosis(data.loc[:, data.columns != 'class'],axis=0)) ``` Index(['pelvic_incidence', 'pelvic_tilt', 'lumbar_lordosis_angle', 'sacral_slope', 'pelvic_radius', 'degree_spondylolisthesis', 'class'], dtype='object') media = pelvic_incidence 60.496484 pelvic_tilt 17.542903 lumbar_lordosis_angle 51.930710 sacral_slope 42.953871 pelvic_radius 117.920548 degree_spondylolisthesis 26.296742 dtype: float64 median = [ 58.69 16.36 49.565 42.405 118.265 11.765] varianza = pelvic_incidence 296.125120 pelvic_tilt 99.839761 lumbar_lordosis_angle 343.131767 sacral_slope 179.588974 pelvic_radius 176.787128 degree_spondylolisthesis 1406.119148 dtype: float64 desviación estándar = pelvic_incidence 17.208286 pelvic_tilt 9.991985 lumbar_lordosis_angle 18.523816 sacral_slope 13.401081 pelvic_radius 13.296132 degree_spondylolisthesis 37.498255 dtype: float64 skewness = [ 0.51786409 0.67329913 0.5964695 0.78883655 -0.17607403 4.29696644] kurtosis = [ 0.2006561 0.64597047 0.13977901 2.94050906 0.90035703 37.4374569 ] ```python vertebral_stats["median"]=np.median(data.loc[:, data.columns != 'class'],axis=0) vertebral_stats["skewness"]=st.skew(data.loc[:, data.columns != 'class'],axis=0) vertebral_stats["kurtosis"]=st.kurtosis(data.loc[:, data.columns != 'class'],axis=0) ``` ```python vertebral_stats ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>count</th> <th>mean</th> <th>std</th> <th>min</th> <th>25%</th> <th>50%</th> <th>75%</th> <th>max</th> <th>median</th> <th>skewness</th> <th>kurtosis</th> </tr> </thead> <tbody> <tr> <th>pelvic_incidence</th> <td>310.0</td> <td>60.496484</td> <td>17.236109</td> <td>26.15</td> <td>46.4325</td> <td>58.690</td> <td>72.8800</td> <td>129.83</td> <td>58.690</td> <td>0.517864</td> <td>0.200656</td> </tr> <tr> <th>pelvic_tilt</th> <td>310.0</td> <td>17.542903</td> <td>10.008140</td> <td>-6.55</td> <td>10.6675</td> <td>16.360</td> <td>22.1200</td> <td>49.43</td> <td>16.360</td> <td>0.673299</td> <td>0.645970</td> </tr> <tr> <th>lumbar_lordosis_angle</th> <td>310.0</td> <td>51.930710</td> <td>18.553766</td> <td>14.00</td> <td>37.0000</td> <td>49.565</td> <td>63.0000</td> <td>125.74</td> <td>49.565</td> <td>0.596469</td> <td>0.139779</td> </tr> <tr> <th>sacral_slope</th> <td>310.0</td> <td>42.953871</td> <td>13.422748</td> <td>13.37</td> <td>33.3475</td> <td>42.405</td> <td>52.6925</td> <td>121.43</td> <td>42.405</td> <td>0.788837</td> <td>2.940509</td> </tr> <tr> <th>pelvic_radius</th> <td>310.0</td> <td>117.920548</td> <td>13.317629</td> <td>70.08</td> <td>110.7100</td> <td>118.265</td> <td>125.4675</td> <td>163.07</td> <td>118.265</td> <td>-0.176074</td> <td>0.900357</td> </tr> <tr> <th>degree_spondylolisthesis</th> <td>310.0</td> <td>26.296742</td> <td>37.558883</td> <td>-11.06</td> <td>1.6000</td> <td>11.765</td> <td>41.2850</td> <td>418.54</td> <td>11.765</td> <td>4.296966</td> <td>37.437457</td> </tr> </tbody> </table> </div> ```python ``` ```python ``` Ahora entendamos el problema en el que nos embarcamos. Resulta que los datos pueden contener outliers, o en otras palabras, hay datos que parecen estar fuera del rango de valores del conjunto total. Veamos esto con otro conjunto de datos: https://archive.ics.uci.edu/ml/datasets/ionosphere ```python #names=["pelvic_incidence","pelvic_tilt","lumbar_lordosis_angle","sacral_slope","pelvic_radius","degree_spondylolisthesis","class"] data=pd.read_csv("../Datas/Ionosphere/ionosphere.data",delimiter=",",names=["Atr"+str(i) for i in range(35)]) ``` ```python data ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>Atr0</th> <th>Atr1</th> <th>Atr2</th> <th>Atr3</th> <th>Atr4</th> <th>Atr5</th> <th>Atr6</th> <th>Atr7</th> <th>Atr8</th> <th>Atr9</th> <th>...</th> <th>Atr25</th> <th>Atr26</th> <th>Atr27</th> <th>Atr28</th> <th>Atr29</th> <th>Atr30</th> <th>Atr31</th> <th>Atr32</th> <th>Atr33</th> <th>Atr34</th> </tr> </thead> <tbody> <tr> <th>0</th> <td>1</td> <td>0</td> <td>0.99539</td> <td>-0.05889</td> <td>0.85243</td> <td>0.02306</td> <td>0.83398</td> <td>-0.37708</td> <td>1.00000</td> <td>0.03760</td> <td>...</td> <td>-0.51171</td> <td>0.41078</td> <td>-0.46168</td> <td>0.21266</td> <td>-0.34090</td> <td>0.42267</td> <td>-0.54487</td> <td>0.18641</td> <td>-0.45300</td> <td>g</td> </tr> <tr> <th>1</th> <td>1</td> <td>0</td> <td>1.00000</td> <td>-0.18829</td> <td>0.93035</td> <td>-0.36156</td> <td>-0.10868</td> <td>-0.93597</td> <td>1.00000</td> <td>-0.04549</td> <td>...</td> <td>-0.26569</td> <td>-0.20468</td> <td>-0.18401</td> <td>-0.19040</td> <td>-0.11593</td> <td>-0.16626</td> <td>-0.06288</td> <td>-0.13738</td> <td>-0.02447</td> <td>b</td> </tr> <tr> <th>2</th> <td>1</td> <td>0</td> <td>1.00000</td> <td>-0.03365</td> <td>1.00000</td> <td>0.00485</td> <td>1.00000</td> <td>-0.12062</td> <td>0.88965</td> <td>0.01198</td> <td>...</td> <td>-0.40220</td> <td>0.58984</td> <td>-0.22145</td> <td>0.43100</td> <td>-0.17365</td> <td>0.60436</td> <td>-0.24180</td> <td>0.56045</td> <td>-0.38238</td> <td>g</td> </tr> <tr> <th>3</th> <td>1</td> <td>0</td> <td>1.00000</td> <td>-0.45161</td> <td>1.00000</td> <td>1.00000</td> <td>0.71216</td> <td>-1.00000</td> <td>0.00000</td> <td>0.00000</td> <td>...</td> <td>0.90695</td> <td>0.51613</td> <td>1.00000</td> <td>1.00000</td> <td>-0.20099</td> <td>0.25682</td> <td>1.00000</td> <td>-0.32382</td> <td>1.00000</td> <td>b</td> </tr> <tr> <th>4</th> <td>1</td> <td>0</td> <td>1.00000</td> <td>-0.02401</td> <td>0.94140</td> <td>0.06531</td> <td>0.92106</td> <td>-0.23255</td> <td>0.77152</td> <td>-0.16399</td> <td>...</td> <td>-0.65158</td> <td>0.13290</td> <td>-0.53206</td> <td>0.02431</td> <td>-0.62197</td> <td>-0.05707</td> <td>-0.59573</td> <td>-0.04608</td> <td>-0.65697</td> <td>g</td> </tr> <tr> <th>...</th> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> <td>...</td> </tr> <tr> <th>346</th> <td>1</td> <td>0</td> <td>0.83508</td> <td>0.08298</td> <td>0.73739</td> <td>-0.14706</td> <td>0.84349</td> <td>-0.05567</td> <td>0.90441</td> <td>-0.04622</td> <td>...</td> <td>-0.04202</td> <td>0.83479</td> <td>0.00123</td> <td>1.00000</td> <td>0.12815</td> <td>0.86660</td> <td>-0.10714</td> <td>0.90546</td> <td>-0.04307</td> <td>g</td> </tr> <tr> <th>347</th> <td>1</td> <td>0</td> <td>0.95113</td> <td>0.00419</td> <td>0.95183</td> <td>-0.02723</td> <td>0.93438</td> <td>-0.01920</td> <td>0.94590</td> <td>0.01606</td> <td>...</td> <td>0.01361</td> <td>0.93522</td> <td>0.04925</td> <td>0.93159</td> <td>0.08168</td> <td>0.94066</td> <td>-0.00035</td> <td>0.91483</td> <td>0.04712</td> <td>g</td> </tr> <tr> <th>348</th> <td>1</td> <td>0</td> <td>0.94701</td> <td>-0.00034</td> <td>0.93207</td> <td>-0.03227</td> <td>0.95177</td> <td>-0.03431</td> <td>0.95584</td> <td>0.02446</td> <td>...</td> <td>0.03193</td> <td>0.92489</td> <td>0.02542</td> <td>0.92120</td> <td>0.02242</td> <td>0.92459</td> <td>0.00442</td> <td>0.92697</td> <td>-0.00577</td> <td>g</td> </tr> <tr> <th>349</th> <td>1</td> <td>0</td> <td>0.90608</td> <td>-0.01657</td> <td>0.98122</td> <td>-0.01989</td> <td>0.95691</td> <td>-0.03646</td> <td>0.85746</td> <td>0.00110</td> <td>...</td> <td>-0.02099</td> <td>0.89147</td> <td>-0.07760</td> <td>0.82983</td> <td>-0.17238</td> <td>0.96022</td> <td>-0.03757</td> <td>0.87403</td> <td>-0.16243</td> <td>g</td> </tr> <tr> <th>350</th> <td>1</td> <td>0</td> <td>0.84710</td> <td>0.13533</td> <td>0.73638</td> <td>-0.06151</td> <td>0.87873</td> <td>0.08260</td> <td>0.88928</td> <td>-0.09139</td> <td>...</td> <td>-0.15114</td> <td>0.81147</td> <td>-0.04822</td> <td>0.78207</td> <td>-0.00703</td> <td>0.75747</td> <td>-0.06678</td> <td>0.85764</td> <td>-0.06151</td> <td>g</td> </tr> </tbody> </table> <p>351 rows × 35 columns</p> </div> ```python len(data.columns) ``` 35 ```python plt.figure(figsize=(30,6)) data.iloc[:,2:].boxplot(grid=True,fontsize=15) plt.legend(fontsize=16) #plt.xticks(range(1,7),np.array(["pelvic \n incidence","pelvic \n tilt","lumbar \n lordosis \n angle","sacral \n slope","pelvic \n radius","degree \n spondylolisthesis"]),fontsize=16) plt.yticks(fontsize=16) plt.savefig("../../figures/ionosphere_boxplot1.png",bbox_inches ="tight") plt.show() ``` ```python ionosphere_stats=data.describe().T ``` ```python ionosphere_stats["LII"]=ionosphere_stats["25%"]-((ionosphere_stats["75%"]-ionosphere_stats["25%"])*1.5) ionosphere_stats["LIS"]=ionosphere_stats["75%"]+((ionosphere_stats["75%"]-ionosphere_stats["25%"])*1.5) ionosphere_stats["LEI"]=ionosphere_stats["25%"]-((ionosphere_stats["75%"]-ionosphere_stats["25%"])*3.0) ionosphere_stats["LES"]=ionosphere_stats["75%"]+((ionosphere_stats["75%"]-ionosphere_stats["25%"])*3.0) ``` ```python ionosphere_stats ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>count</th> <th>mean</th> <th>std</th> <th>min</th> <th>25%</th> <th>50%</th> <th>75%</th> <th>max</th> <th>LII</th> <th>LIS</th> <th>LEI</th> <th>LES</th> </tr> </thead> <tbody> <tr> <th>Atr0</th> <td>351.0</td> <td>0.891738</td> <td>0.311155</td> <td>0.0</td> <td>1.000000</td> <td>1.00000</td> <td>1.000000</td> <td>1.0</td> <td>1.000000</td> <td>1.000000</td> <td>1.000000</td> <td>1.000000</td> </tr> <tr> <th>Atr1</th> <td>351.0</td> <td>0.000000</td> <td>0.000000</td> <td>0.0</td> <td>0.000000</td> <td>0.00000</td> <td>0.000000</td> <td>0.0</td> <td>0.000000</td> <td>0.000000</td> <td>0.000000</td> <td>0.000000</td> </tr> <tr> <th>Atr2</th> <td>351.0</td> <td>0.641342</td> <td>0.497708</td> <td>-1.0</td> <td>0.472135</td> <td>0.87111</td> <td>1.000000</td> <td>1.0</td> <td>-0.319663</td> <td>1.791797</td> <td>-1.111460</td> <td>2.583595</td> </tr> <tr> <th>Atr3</th> <td>351.0</td> <td>0.044372</td> <td>0.441435</td> <td>-1.0</td> <td>-0.064735</td> <td>0.01631</td> <td>0.194185</td> <td>1.0</td> <td>-0.453115</td> <td>0.582565</td> <td>-0.841495</td> <td>0.970945</td> </tr> <tr> <th>Atr4</th> <td>351.0</td> <td>0.601068</td> <td>0.519862</td> <td>-1.0</td> <td>0.412660</td> <td>0.80920</td> <td>1.000000</td> <td>1.0</td> <td>-0.468350</td> <td>1.881010</td> <td>-1.349360</td> <td>2.762020</td> </tr> <tr> <th>Atr5</th> <td>351.0</td> <td>0.115889</td> <td>0.460810</td> <td>-1.0</td> <td>-0.024795</td> <td>0.02280</td> <td>0.334655</td> <td>1.0</td> <td>-0.563970</td> <td>0.873830</td> <td>-1.103145</td> <td>1.413005</td> </tr> <tr> <th>Atr6</th> <td>351.0</td> <td>0.550095</td> <td>0.492654</td> <td>-1.0</td> <td>0.211310</td> <td>0.72873</td> <td>0.969240</td> <td>1.0</td> <td>-0.925585</td> <td>2.106135</td> <td>-2.062480</td> <td>3.243030</td> </tr> <tr> <th>Atr7</th> <td>351.0</td> <td>0.119360</td> <td>0.520750</td> <td>-1.0</td> <td>-0.054840</td> <td>0.01471</td> <td>0.445675</td> <td>1.0</td> <td>-0.805613</td> <td>1.196448</td> <td>-1.556385</td> <td>1.947220</td> </tr> <tr> <th>Atr8</th> <td>351.0</td> <td>0.511848</td> <td>0.507066</td> <td>-1.0</td> <td>0.087110</td> <td>0.68421</td> <td>0.953240</td> <td>1.0</td> <td>-1.212085</td> <td>2.252435</td> <td>-2.511280</td> <td>3.551630</td> </tr> <tr> <th>Atr9</th> <td>351.0</td> <td>0.181345</td> <td>0.483851</td> <td>-1.0</td> <td>-0.048075</td> <td>0.01829</td> <td>0.534195</td> <td>1.0</td> <td>-0.921480</td> <td>1.407600</td> <td>-1.794885</td> <td>2.281005</td> </tr> <tr> <th>Atr10</th> <td>351.0</td> <td>0.476183</td> <td>0.563496</td> <td>-1.0</td> <td>0.021120</td> <td>0.66798</td> <td>0.957895</td> <td>1.0</td> <td>-1.384043</td> <td>2.363058</td> <td>-2.789205</td> <td>3.768220</td> </tr> <tr> <th>Atr11</th> <td>351.0</td> <td>0.155040</td> <td>0.494817</td> <td>-1.0</td> <td>-0.065265</td> <td>0.02825</td> <td>0.482375</td> <td>1.0</td> <td>-0.886725</td> <td>1.303835</td> <td>-1.708185</td> <td>2.125295</td> </tr> <tr> <th>Atr12</th> <td>351.0</td> <td>0.400801</td> <td>0.622186</td> <td>-1.0</td> <td>0.000000</td> <td>0.64407</td> <td>0.955505</td> <td>1.0</td> <td>-1.433258</td> <td>2.388763</td> <td>-2.866515</td> <td>3.822020</td> </tr> <tr> <th>Atr13</th> <td>351.0</td> <td>0.093414</td> <td>0.494873</td> <td>-1.0</td> <td>-0.073725</td> <td>0.03027</td> <td>0.374860</td> <td>1.0</td> <td>-0.746602</td> <td>1.047737</td> <td>-1.419480</td> <td>1.720615</td> </tr> <tr> <th>Atr14</th> <td>351.0</td> <td>0.344159</td> <td>0.652828</td> <td>-1.0</td> <td>0.000000</td> <td>0.60194</td> <td>0.919330</td> <td>1.0</td> <td>-1.378995</td> <td>2.298325</td> <td>-2.757990</td> <td>3.677320</td> </tr> <tr> <th>Atr15</th> <td>351.0</td> <td>0.071132</td> <td>0.458371</td> <td>-1.0</td> <td>-0.081705</td> <td>0.00000</td> <td>0.308975</td> <td>1.0</td> <td>-0.667725</td> <td>0.894995</td> <td>-1.253745</td> <td>1.481015</td> </tr> <tr> <th>Atr16</th> <td>351.0</td> <td>0.381949</td> <td>0.618020</td> <td>-1.0</td> <td>0.000000</td> <td>0.59091</td> <td>0.935705</td> <td>1.0</td> <td>-1.403558</td> <td>2.339263</td> <td>-2.807115</td> <td>3.742820</td> </tr> <tr> <th>Atr17</th> <td>351.0</td> <td>-0.003617</td> <td>0.496762</td> <td>-1.0</td> <td>-0.225690</td> <td>0.00000</td> <td>0.195285</td> <td>1.0</td> <td>-0.857152</td> <td>0.826747</td> <td>-1.488615</td> <td>1.458210</td> </tr> <tr> <th>Atr18</th> <td>351.0</td> <td>0.359390</td> <td>0.626267</td> <td>-1.0</td> <td>0.000000</td> <td>0.57619</td> <td>0.899265</td> <td>1.0</td> <td>-1.348898</td> <td>2.248163</td> <td>-2.697795</td> <td>3.597060</td> </tr> <tr> <th>Atr19</th> <td>351.0</td> <td>-0.024025</td> <td>0.519076</td> <td>-1.0</td> <td>-0.234670</td> <td>0.00000</td> <td>0.134370</td> <td>1.0</td> <td>-0.788230</td> <td>0.687930</td> <td>-1.341790</td> <td>1.241490</td> </tr> <tr> <th>Atr20</th> <td>351.0</td> <td>0.336695</td> <td>0.609828</td> <td>-1.0</td> <td>0.000000</td> <td>0.49909</td> <td>0.894865</td> <td>1.0</td> <td>-1.342297</td> <td>2.237163</td> <td>-2.684595</td> <td>3.579460</td> </tr> <tr> <th>Atr21</th> <td>351.0</td> <td>0.008296</td> <td>0.518166</td> <td>-1.0</td> <td>-0.243870</td> <td>0.00000</td> <td>0.188760</td> <td>1.0</td> <td>-0.892815</td> <td>0.837705</td> <td>-1.541760</td> <td>1.486650</td> </tr> <tr> <th>Atr22</th> <td>351.0</td> <td>0.362475</td> <td>0.603767</td> <td>-1.0</td> <td>0.000000</td> <td>0.53176</td> <td>0.911235</td> <td>1.0</td> <td>-1.366853</td> <td>2.278087</td> <td>-2.733705</td> <td>3.644940</td> </tr> <tr> <th>Atr23</th> <td>351.0</td> <td>-0.057406</td> <td>0.527456</td> <td>-1.0</td> <td>-0.366885</td> <td>0.00000</td> <td>0.164630</td> <td>1.0</td> <td>-1.164157</td> <td>0.961902</td> <td>-1.961430</td> <td>1.759175</td> </tr> <tr> <th>Atr24</th> <td>351.0</td> <td>0.396135</td> <td>0.578451</td> <td>-1.0</td> <td>0.000000</td> <td>0.55389</td> <td>0.905240</td> <td>1.0</td> <td>-1.357860</td> <td>2.263100</td> <td>-2.715720</td> <td>3.620960</td> </tr> <tr> <th>Atr25</th> <td>351.0</td> <td>-0.071187</td> <td>0.508495</td> <td>-1.0</td> <td>-0.332390</td> <td>-0.01505</td> <td>0.156765</td> <td>1.0</td> <td>-1.066123</td> <td>0.890497</td> <td>-1.799855</td> <td>1.624230</td> </tr> <tr> <th>Atr26</th> <td>351.0</td> <td>0.541641</td> <td>0.516205</td> <td>-1.0</td> <td>0.286435</td> <td>0.70824</td> <td>0.999945</td> <td>1.0</td> <td>-0.783830</td> <td>2.070210</td> <td>-1.854095</td> <td>3.140475</td> </tr> <tr> <th>Atr27</th> <td>351.0</td> <td>-0.069538</td> <td>0.550025</td> <td>-1.0</td> <td>-0.443165</td> <td>-0.01769</td> <td>0.153535</td> <td>1.0</td> <td>-1.338215</td> <td>1.048585</td> <td>-2.233265</td> <td>1.943635</td> </tr> <tr> <th>Atr28</th> <td>351.0</td> <td>0.378445</td> <td>0.575886</td> <td>-1.0</td> <td>0.000000</td> <td>0.49664</td> <td>0.883465</td> <td>1.0</td> <td>-1.325197</td> <td>2.208662</td> <td>-2.650395</td> <td>3.533860</td> </tr> <tr> <th>Atr29</th> <td>351.0</td> <td>-0.027907</td> <td>0.507974</td> <td>-1.0</td> <td>-0.236885</td> <td>0.00000</td> <td>0.154075</td> <td>1.0</td> <td>-0.823325</td> <td>0.740515</td> <td>-1.409765</td> <td>1.326955</td> </tr> <tr> <th>Atr30</th> <td>351.0</td> <td>0.352514</td> <td>0.571483</td> <td>-1.0</td> <td>0.000000</td> <td>0.44277</td> <td>0.857620</td> <td>1.0</td> <td>-1.286430</td> <td>2.144050</td> <td>-2.572860</td> <td>3.430480</td> </tr> <tr> <th>Atr31</th> <td>351.0</td> <td>-0.003794</td> <td>0.513574</td> <td>-1.0</td> <td>-0.242595</td> <td>0.00000</td> <td>0.200120</td> <td>1.0</td> <td>-0.906668</td> <td>0.864193</td> <td>-1.570740</td> <td>1.528265</td> </tr> <tr> <th>Atr32</th> <td>351.0</td> <td>0.349364</td> <td>0.522663</td> <td>-1.0</td> <td>0.000000</td> <td>0.40956</td> <td>0.813765</td> <td>1.0</td> <td>-1.220648</td> <td>2.034413</td> <td>-2.441295</td> <td>3.255060</td> </tr> <tr> <th>Atr33</th> <td>351.0</td> <td>0.014480</td> <td>0.468337</td> <td>-1.0</td> <td>-0.165350</td> <td>0.00000</td> <td>0.171660</td> <td>1.0</td> <td>-0.670865</td> <td>0.677175</td> <td>-1.176380</td> <td>1.182690</td> </tr> </tbody> </table> </div> ```python ionosphere_stats2=pd.DataFrame() ionosphere_stats2.index=ionosphere_stats.index atipicos,extremos=[],[] for i in ionosphere_stats2.index: LII=ionosphere_stats[ionosphere_stats.index==i]["LII"][0] LIS=ionosphere_stats[ionosphere_stats.index==i]["LIS"][0] LEI=ionosphere_stats[ionosphere_stats.index==i]["LEI"][0] LES=ionosphere_stats[ionosphere_stats.index==i]["LES"][0] atipicos.append(len(data[((data[i]<=LII)&(data[i]>=LIS))|((data[i]>=LIS)&(data[i]<=LES))])) extremos.append(len(data[((data[i]<=LEI)|(data[i]>=LES))])) ionosphere_stats2["Valores atipicos"]=atipicos ionosphere_stats2["Valores extremos"]=extremos ionosphere_stats2["Valores atipicos (%)"]=np.round(ionosphere_stats2["Valores atipicos"]/len(data)*100,2) ionosphere_stats2["Valores extremos (%)"]=np.round(ionosphere_stats2["Valores extremos"]/len(data)*100,2) ionosphere_stats2.loc['Total'] = ionosphere_stats2.sum(axis=0) ``` ```python ionosphere_stats2[24:] ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>Valores atipicos</th> <th>Valores extremos</th> <th>Valores atipicos (%)</th> <th>Valores extremos (%)</th> </tr> </thead> <tbody> <tr> <th>Atr24</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr25</th> <td>20.0</td> <td>0.0</td> <td>5.70</td> <td>0.0</td> </tr> <tr> <th>Atr26</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr27</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr28</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr29</th> <td>31.0</td> <td>0.0</td> <td>8.83</td> <td>0.0</td> </tr> <tr> <th>Atr30</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr31</th> <td>33.0</td> <td>0.0</td> <td>9.40</td> <td>0.0</td> </tr> <tr> <th>Atr32</th> <td>0.0</td> <td>0.0</td> <td>0.00</td> <td>0.0</td> </tr> <tr> <th>Atr33</th> <td>40.0</td> <td>0.0</td> <td>11.40</td> <td>0.0</td> </tr> <tr> <th>Total</th> <td>971.0</td> <td>755.0</td> <td>276.64</td> <td>215.1</td> </tr> </tbody> </table> </div> Tomemos por ejemplo los datos que tienen valores atípicos en mayor cantidad, por ejemplo, los datos con `Atr33`, y veamos como cambia la distribución cual eliminamos los valores atípicos. ```python def filter_outlier(col): LII=ionosphere_stats[ionosphere_stats.index==col]["LII"][0] LIS=ionosphere_stats[ionosphere_stats.index==col]["LIS"][0] LEI=ionosphere_stats[ionosphere_stats.index==col]["LEI"][0] LES=ionosphere_stats[ionosphere_stats.index==col]["LES"][0] return data[~(((data[col]<=LII)&(data[col]>=LIS))|((data[col]>=LIS)&(data[col]<=LES)))][col] ``` ```python plt.figure(figsize=(16,6)) plt.subplot(121) plt.hist(filter_outlier("Atr33"),label="Atr33 (Outliers removed) ({} rows)".format(len(filter_outlier("Atr33"))),alpha=0.4,color='b') plt.hist(data["Atr33"],label="Atr33 (All data) ({} rows)".format(len(data)),alpha=0.4,color='r') plt.vlines(np.mean(filter_outlier("Atr33")),0,120,label="Outliers removed mean's (<Atr33>={})".format(np.round(np.mean(filter_outlier("Atr33")),4)),alpha=0.99,color='b',linestyle="--",linewidth=4) plt.vlines(np.mean(data["Atr33"]),0,120,label="All data mean's (<Atr33>={})".format(np.round(np.mean(data["Atr33"]),4)),alpha=0.99,color='r',linestyle="--",linewidth=4) plt.legend(title="Absolute distribution",fontsize=11) plt.ylabel("Counts",fontsize=14) plt.xticks(fontsize=14) plt.yticks(fontsize=14) plt.subplot(122) plt.hist(filter_outlier("Atr33"),label="Atr33 (Outliers removed) ({} rows)".format(len(filter_outlier("Atr33"))),alpha=0.4,cumulative=True,color='b') plt.hist(data["Atr33"],label="Atr33 (All data) ({} rows)".format(len(data)),alpha=0.4,cumulative=True,color='r') plt.vlines(np.median(filter_outlier("Atr33")),0,350,label="Outliers removed median's (<Atr33>={})".format(np.round(np.mean(filter_outlier("Atr33")),4)),alpha=0.99,color='b',linestyle="--",linewidth=6) plt.vlines(np.median(data["Atr33"]),0,350,label="All data median's (<Atr33>={})".format(np.round(np.mean(data["Atr33"]),4)),alpha=0.99,color='r',linestyle="--",linewidth=3) plt.legend(title="Cumulative distribution",fontsize=11) plt.ylabel("Counts",fontsize=14) plt.xticks(fontsize=14) plt.yticks(fontsize=14) plt.savefig("../../figures/ionosphere_hist2.png",bbox_inches ="tight") plt.show() ``` Entonces como podemos ver cláramente la distribución de los datos cambian cuando eliminamos de nuestros datos los valores atípicos y por tanto también cambian las medidas de tendencia central, en particular la diferencia porcentual para el cálculo de la media en los datos originales y los datos cuando han sido removidos los valores atípicos es de: ```python np.round(abs(np.mean(filter_outlier("Atr33"))-np.mean(data["Atr33"]))/np.mean(data["Atr33"])*100,2) ``` 794.06 ```python np.mean(filter_outlier("Atr33")) ``` -0.10050096463022508 ```python np.mean(data["Atr33"]) ``` 0.014480113960113956 ```python ``` ```python ``` ```python ``` ```python ``` ```python ``` ```python ``` ```python ``` ```python ``` ```python ```

eff9a206b7016237a647fea29a2d094c0409c2a3

ipynb

Jupyter Notebook

Notebooks/descriptive_statistics.ipynb

sierraporta/Data_Mining_Excersices

3790466d1d8314d83178b61035fc6c28b567ab59

Notebooks/descriptive_statistics.ipynb

sierraporta/Data_Mining_Excersices

3790466d1d8314d83178b61035fc6c28b567ab59

Notebooks/descriptive_statistics.ipynb

sierraporta/Data_Mining_Excersices

3790466d1d8314d83178b61035fc6c28b567ab59

Qwen/Qwen-72B

1. YES 2. YES

__label__kor_Hang

# Models, Data, Learning Problems In this lab we start our first data analysis on a concrete problem. We are using Fisher's famous <a href="https://en.wikipedia.org/wiki/Iris_flower_data_set">Iris data set</a>. The goal is to classify flowers from the Iris family into one of three species, that look as follows: <table> <tr> <td> </td> <td> </td> <td> </td> </tr> <tr> <td>Iris Setosa</td> <td>Iris Versicolor</td> <td>Iris Virginica</td> </tr> </table> Our data set contains 50 flowers from each class, thus 150 in total. There are four features, the length and width of the petal (dt. Kronblatt) and sepal (dt. Kelchblatt) in centimeters. Your goal is to go through the notebook, understand premade code and text as well as filling blanks and exercises left for you. You may also edit the notebook as you wish. A good way to learn is to add comments (lines starting with #) or modifying the code and see what changes. The data set is distributed with sci-kit learn, the only thing we have to do is to important a function and call it. ```python from sklearn.datasets import load_iris data = load_iris() X = data.data y = data.target print(type(X)) print(X.shape) print(f"First three rows of data\n{X[:3]}") print(f"First three labels: {y[:3]}") ``` <class 'numpy.ndarray'> (150, 4) First three rows of data [[5.1 3.5 1.4 0.2] [4.9 3. 1.4 0.2] [4.7 3.2 1.3 0.2]] First three labels: [0 0 0] Not only do we get the input matrix $X \in \mathbb{R}^{150 \times 4}$ and target $y \in \mathbb{R}^{150}$, but also meta information such as what the class labels $0, 1, 2$ stand for and what the features (i.e. columns of $X$) correspond to. ```python print(data.target_names) print(data.feature_names) ``` ['setosa' 'versicolor' 'virginica'] ['sepal length (cm)', 'sepal width (cm)', 'petal length (cm)', 'petal width (cm)'] As a first step we focus our analysis on the first two variables, the sepal length and sepal width. Since we obtain a representation of the data in two dimensions, we are able to plot it. ```python X_2 = X[:, :2] y_2 = y ``` ```python # Configures Jupyter to show graphics in the notebook %matplotlib inline from matplotlib import pyplot as plt # standard import # We write a function so we can reuse it later. def generate_scatter_plot(X, y): class_names = data.target_names class_colors = ['blue','yellow','green'] fig = plt.figure(figsize=(12, 6)) # increase size of plot for i, class_color in enumerate(class_colors): # plot the points only of this class label plt.scatter(X[y == i, 0], X[y == i, 1], c=class_color, label=class_names[i]) plt.xlabel(data.feature_names[0]) # label the axis plt.ylabel(data.feature_names[1]) plt.legend(loc="best") # with legend generate_scatter_plot(X_2, y) ``` We see that we could discriminate the iris setosa linearly from other two species. The linear function could even have a slope of about $1$. Let us substitute the first feature with the difference of the two features. ```python import numpy as np x_new = X[:, 0] - X[:, 1] X_new = np.column_stack((x_new, X[:, 1])) print(X_new.shape) generate_scatter_plot(X_new, y) plt.xlabel("sepal length - sepal width") ``` Remember that our main goal is to find a model, $$ y_\theta: X \rightarrow Y $$ such that $y_\theta(x)$ models the knowledge we got from our training data plus the inductive bias. The plot gives the decision rule (or part of): <center>"If sepal length - sepal width $\leq$ 2.2 $\rightarrow$ Classify iris setosa"</center> <b>Exercise 1:</b> Implement the naive decision rule as given above. If the condition for iris setosa is not fulfilled, classify the result as 'iris versicolor'. ```python def naive_decision_rule(x): # X matrix with 4 dimensional columns # returns the expected class label for each row of X (0 = setosa, 1 = versicolor, 2 = virginica) # FILL IN x_new = x[0] - x[1] pred_class = 0 if (x_new <= 2.2) else 1 return pred_class ``` The following function takes a decision rule (or model) and a matrix of data points to generate the predictions for this matrix. ```python def predict(model, X): """Builds prediction on a matrix X given a model for each data point in a row. Returns a flat vector of predictions. """ return np.apply_along_axis(model, axis=1, arr=X) y_pred = predict(naive_decision_rule, X) print(y_pred[:50]) # print first 50 predictions print(y_pred) # print all 150 ``` [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0] [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1] The predictions of the first 50 numbers should be zero and one for all others. Now we have to judge the quality of our model, we do this by using the zero-loss function of the lecture. <b>Exercise 2:</b> Implement the zero loss function as defined in the lecture, $$ \begin{align} l(x_i, y_i; \theta) &= l_{0/1}(y_\theta(x_i), y_i) = \begin{cases} 0, & \mbox{ if } y_\theta(x_i) = y_i \\ 1, & \mbox{ otherwise } \end{cases} \\ l(X, y; \theta) &= \sum_i{ l(x_i, y_i; \theta). } \end{align} $$ In lay-man terms one counts how often the label predicted differed from the observed label. ```python def zero_one_loss(y_pred, y_true,loss): # FILL IN #print (y_pred) #print (y_true) for i in range(len(y_pred)): if y_pred[i] - y_true[i] != 0: loss= loss+1 return loss ``` ```python print(f"The 0-1-loss of the naive decision rule is {zero_one_loss(y_pred, y,loss=0)} (should be 50)") ``` The 0-1-loss of the naive decision rule is 50 (should be 50) <b>Exercise 3:</b> Improve the decision rule to have a maximum number of missclassifications of $10$. As an informal constraint use "Occams Razor" as an inductive bias, i.e. as simplest as possible. <b>Discussion topic:</b> Why could a complex model with zero missclassifications perform worse in reality (we got out and measure new flowers) than a simple model with more misclassifications? ```python import numpy as np x_new1 = X[:, 2]- X[:, 3] X_new1 = np.column_stack((x_new1, X[:, 3])) print(X_new.shape) generate_scatter_plot(X_new1, y) plt.xlabel("P length - P width") ``` ```python # Place for your analysis. def my_decision_rule(x): # Take 'petal length (cm)', 'petal width (cm)' into account # 'petal width (cm)' at Y axis x_new = x[2] - x[3] if (x_new <= 1.8): pred_class = 0 elif ((x_new >= 1.8) & (x_new <= 3.6)) & (x[3] <= 1.6) : pred_class = 1 else: pred_class = 2 return pred_class pass ``` ```python # Evaluation script y_pred = predict(my_decision_rule, X) print(y_pred) loss = zero_one_loss(y_pred, y,loss=0) print(f"Your loss {loss}.") if loss <= 10: print("You have made it!") else: print("Uhm, try again. Maybe you have flipped some class?") ``` [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2] Your loss 4. You have made it! ```python ```

767802580f170bbd2f134fa0331849daf2be5c93

ipynb

Jupyter Notebook

Lab2/.ipynb_checkpoints/Models, Data, Learning Problems-checkpoint.ipynb

pratik98/Machine-LearningSummer2020

ab1ab87c2bd3c9ffb42a88dfb1b93891ed8aa746

Lab2/.ipynb_checkpoints/Models, Data, Learning Problems-checkpoint.ipynb

pratik98/Machine-LearningSummer2020

ab1ab87c2bd3c9ffb42a88dfb1b93891ed8aa746

Lab2/.ipynb_checkpoints/Models, Data, Learning Problems-checkpoint.ipynb

pratik98/Machine-LearningSummer2020

ab1ab87c2bd3c9ffb42a88dfb1b93891ed8aa746

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python import pyzx as zx import sympy from fractions import Fraction ``` ```python gamma = sympy.Symbol('gamma') ``` ```python g=zx.graph.GraphSym() v= g.add_vertex(zx.VertexType.Z, qubit=0, row=1, phase=gamma) w= g.add_vertex(zx.VertexType.Z, qubit=1, row=1, phase=1) x= g.add_vertex(zx.VertexType.Z, qubit=2, row=1, phase=1) g.add_edge(g.edge(v,w),edgetype=zx.EdgeType.SIMPLE) ``` ```python zx.draw_matplotlib(g) ``` ```python g.phase(0) ``` ```python e = zx.editor.edit(g) ``` ```python gamma = sympy.Symbol('gamma') ``` ```python type(Fraction(numerator=2, denominator=1) * gamma) ``` ```python ```

ab0ae4fb568902f150646992c1dfbda41a1962cd

ipynb

Jupyter Notebook

scratchpads/symbolic_diagrams.ipynb

tobsto/pyzx

9d92f9163bec91315e60423703fb7a7ad0d96fc8

scratchpads/symbolic_diagrams.ipynb

tobsto/pyzx

9d92f9163bec91315e60423703fb7a7ad0d96fc8

scratchpads/symbolic_diagrams.ipynb

tobsto/pyzx

9d92f9163bec91315e60423703fb7a7ad0d96fc8

Qwen/Qwen-72B

1. YES 2. YES

__label__kor_Hang

Create a reduced basis for a simple sinusoid model ``` %matplotlib inline import numpy as np from misc import * import matplotlib.pyplot as plt from lalapps import pulsarpputils as pppu ``` Create the signal model: \begin{equation} h(t) = \frac{h_0}{2}\left[\frac{1}{2}F_+ (1+\cos{}^2\iota)\cos{\phi_0} + F_{\times}\cos{\iota} \sin{\phi_0 }\right]. \end{equation} ``` def signalmodel(t, h0, cosi, phi0, psi, FA, FB): """ The real part of a heterodyned CW signal model, where the frequency and frequency derivatives represents offsets from the values used in the initial heterodyne. """ ctwopsi = np.cos(2.*psi) stwopsi = np.sin(2.*psi) fps = FA*ctwopsi + FB*stwopsi fcs = FB*ctwopsi - FA*stwopsi phase = phi0 ht = 0.5*h0*(0.5*fps*(1.+cosi**2)*np.cos(phase) + fcs*cosi*np.sin(phase)) return ht ``` Initialise the model time series and other constant parameters, including antenna patterns that don't including the polarisation angle. ``` # a time series t0 = 0 tend = 86400.*10 N = (tend-t0)/60. ts = np.linspace(t0, tend, N) dt = ts[1]-ts[0] ra = 0. # right ascension at zero radians dec = 0. # declination at zero radians det = 'H1' # LIGO Hanford detector FAs = np.zeros(len(ts)) FBs = np.zeros(len(ts)) # create antenna response (without polsaition angle) for i, t in enumerate(ts): FA, FB = pppu.antenna_response(t, ra, dec, 0.0, det) FAs[i] = FA FBs[i] = FB ``` Create a training set of 2000 waveforms with random frequency and frequency derivatives within a narrow range. ``` # number of training waveforms TS_size = 2000 phi0s = np.random.rand(TS_size)*(2.*np.pi) psis = np.random.rand(TS_size)*(np.pi/2.)-(np.pi/4.) cosis = np.random.rand(TS_size)*2. - 1. # allocate memory and create training set TS = np.zeros(TS_size*len(ts)).reshape(TS_size, len(ts)) # store training space in TS_size X len(ts) array h0 = 1. for i in range(TS_size): TS[i] = signalmodel(ts, h0, cosis[i], phi0s[i], psis[i], FAs, FBs) # normalize TS[i] /= np.sqrt(abs(dot_product(dt, TS[i], TS[i]))) ``` Allocate memory for reduced basis vectors. ``` # Allocate storage for projection coefficients of training space waveforms onto the reduced basis elements proj_coefficients = np.zeros(TS_size*TS_size).reshape(TS_size, TS_size) # Allocate matrix to store the projection of training space waveforms onto the reduced basis projections = np.zeros(TS_size*len(ts)).reshape(TS_size, len(ts)) rb_errors = [] #### Begin greedy: see Field et al. arXiv:1308.3565v2 #### tolerance = 1e-12 # set maximum RB projection error sigma = 1 # (2) of Algorithm 1. (projection error at 0th iteration) rb_errors.append(sigma) ``` Run greedy algorithm for creating the reduced basis ``` RB_matrix = [TS[0]] # (3) of Algorithm 1. (seed greedy algorithm (arbitrary)) iter = 0 while sigma >= tolerance: # (5) of Algorithm 1. # project the whole training set onto the reduced basis set projections = project_onto_basis(dt, RB_matrix, TS, projections, proj_coefficients, iter) residual = TS - projections # Find projection errors projection_errors = [dot_product(dt, residual[i], residual[i]) for i in range(len(residual))] sigma = abs(max(projection_errors)) # (7) of Algorithm 1. (Find largest projection error) # break out if sigma is less than tolerance, so another basis is not added to the set # (this can be required is the waveform only requires a couple of basis vectors, and it # stops a further basis containing large amounts of numerical noise being added) if sigma < tolerance: break print sigma, iter index = np.argmax(projection_errors) # Find Training-space index of waveform with largest proj. error rb_errors.append(sigma) #Gram-Schmidt to get the next basis and normalize next_basis = TS[index] - projections[index] # (9) of Algorithm 1. (Gram-Schmidt) next_basis /= np.sqrt(abs(dot_product(dt, next_basis, next_basis))) #(10) of Alg 1. (normalize) RB_matrix.append(next_basis) # (11) of Algorithm 1. (append reduced basis set) iter += 1 ``` 0.999999972831 0 Check that this basis does give the expected residuals for a new set of random waveforms generated from the same parameter range. ``` #### Error check #### TS_rand_size = 4000 TS_rand = np.zeros(TS_rand_size*len(ts)).reshape(TS_rand_size, len(ts)) # Allocate random training space phi0s_rand = np.random.rand(TS_rand_size)*(2.*np.pi) psis_rand = np.random.rand(TS_rand_size)*(np.pi/2.)-(np.pi/4.) cosis_rand = np.random.rand(TS_rand_size)*2. - 1. for i in range(TS_rand_size): TS_rand[i] = signalmodel(ts, h0, cosis_rand[i], phi0s_rand[i], psis_rand[i], FAs, FBs) # normalize TS_rand[i] /= np.sqrt(abs(dot_product(dt, TS_rand[i], TS_rand[i]))) ### find projection errors ### iter = 0 proj_rand = np.zeros(len(ts)) proj_error = [] for h in TS_rand: while iter < len(RB_matrix): proj_coefficients_rand = dot_product(dt, RB_matrix[iter], h) proj_rand += proj_coefficients_rand*RB_matrix[iter] iter += 1 residual = h - proj_rand projection_errors = abs(dot_product(dt, residual, residual)) proj_error.append(projection_errors) proj_rand = np.zeros(len(ts)) iter = 0 plt.scatter(np.linspace(0, len(proj_error), len(proj_error)), np.log10(proj_error)) plt.ylabel('log10 projection error') plt.show() ``` Now we will create the empirical interpolant and find time time stamp 'nodes' at which it matches the full model. ``` # put basis into complex form e = np.array(RB_matrix) indices = [] ts_nodes = [] V = np.zeros((len(e), len(e))) ``` ``` from scipy.linalg import inv # seed EIM algorithm indices.append( int(np.argmax( np.abs(e[0]) )) ) # (2) of Algorithm 2 ts_nodes.append(ts[indices]) # (3) of Algorithm 2 for i in range(1, len(e)): #(4) of Algorithm 2 #build empirical interpolant for e_iter for j in range(len(indices)): # Part of (5) of Algorithm 2: making V_{ij} for k in range(len(indices)): # Part of (5) of Algorithm 2: making V_{ij} V[k][j] = e[j][indices[k]] # Part of (5) of Algorithm 2: making V_{ij} invV = inv(V[0:len(indices), 0:len(indices)]) # Part of (5) of Algorithm 2: making V_{ij} B = B_matrix(invV, e) # Part of (5) of Algorithm 2: making B_j(f) interpolant = emp_interp(B, e[i], indices) # Part of (5) of Algorithm 2: making the empirical interpolant of e res = interpolant - e[i] # 6 of Algorithm 2 index = int(np.argmax(np.abs(res))) # 7 of Algorithm 2 print "ts_{%i} = %f"%(i, ts[index]) indices.append(index) # 8 of Algorithm 2 ts_nodes.append( ts[index] ) # 9 of Algorithm 2 # make B matrix with all the indices for j in range(len(indices)): for k in range(len(indices)): V[k][j] = e[j][indices[k]] invV = inv(V[0:len(indices), 0:len(indices)]) B = B_matrix(invV, e) ``` ts_{1} = 83585.804570 Compare the interpolant with the full signal model ``` h_for_comparison = signalmodel(ts, h0, cosis[0], phi0s[0], psis[0], FAs, FBs) interpolant_for_comparison = np.inner(B.T, h_for_comparison[indices]) plt.plot(ts, h_for_comparison-interpolant_for_comparison, 'b') plt.xlabel('time (s)') plt.show() ``` ``` print len(ts_nodes) print len(ts) ``` 2 14400 ``` H_size = 2000 H = np.zeros(H_size*len(ts)).reshape(H_size, len(ts)) # Allocate random training space phi0s_rand = np.random.rand(H_size)*(2.*np.pi) psis_rand = np.random.rand(H_size)*(np.pi/2.)-(np.pi/4.) cosis_rand = np.random.rand(H_size)*2. - 1. # create set of test waveforms for i in range(H_size): H[i] = signalmodel(ts, h0, cosis[i], phi0s[i], psis[i], FAs, FBs) # find errors between full waveform and interpolants list_of_errors = [] for i in range(H_size): interpolant = np.inner(B.T, H[i][indices]) interpolant /= np.sqrt(np.vdot(interpolant, interpolant)) #normalize H[i] /= np.sqrt(np.vdot(H[i], H[i]) ) #normalize error = abs(np.vdot(H[i] - interpolant, H[i] - interpolant )) list_of_errors.append(error) print error plt.scatter(np.linspace(0, H_size, H_size), np.log10(list_of_errors)) plt.ylabel('log10 interpolation error') plt.show() ``` Now let's find the weights for the "reduced order quadrature", which are calculated as \begin{equation} w_j = \sum_{i=1}^N d_i B_{j,i} \end{equation} where $N$ is the full number of time steps, and $B$ is the matrix produced for the empirical interpolant. First we'll just create some fake data consisting of Gaussian noise. ``` data = np.random.randn(len(ts)) # create weights w = np.inner(B, data.T) ``` Now, compare calculating $\sum_{i=1}^N d_i h_i$ with the full set of time stamps, with the using the interpolant $\sum_{i=1}^M w_i h(F_i)$, where in this case $h$ is only calculated at the interpolant nodes. ``` d_dot_h = np.vdot(data, signalmodel(ts, 1., cosis_rand[12], phi0s_rand[65], psis_rand[101], FAs, FBs)) FA_nodes = FAs[indices] FB_nodes = FBs[indices] hsig = signalmodel(ts_nodes, 1., cosis_rand[12], phi0s_rand[65], psis_rand[101], FA_nodes, FB_nodes) ROQ = np.dot(w, hsig) print "regular inner product = %.15e"%d_dot_h print "ROQ = %.15e"%ROQ ``` regular inner product = -8.576471621009784e+00 ROQ = -8.576471621009867e+00 Now test the speed-up ``` import time t1 = time.time() for i in range(50000): np.dot(data, signalmodel(ts, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FAs, FBs)) # regular inner product e1 = time.time() t2 = time.time() for i in range(50000): np.dot(w, signalmodel(ts_nodes, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FA_nodes, FB_nodes)) # ROQ inner product e2 = time.time() print "regular inner product took %f s"%((e1-t1)/50000.) print "ROQ took %f s"%((e2-t2)/50000.) print "speedup = %f"%((e1-t1) / (e2-t2)) ``` regular inner product took 0.000124 s ROQ took 0.000018 s speedup = 6.927863 Now we want to see if we can do the same thing to compute the inner product of the model $\sum_{i=1}^N h_ih_i$. For this we need to use our interpolant \begin{equation} \mathcal{I}[h](t) = \sum_{j=1}^M B_j(t)h(T_j) \end{equation} where $M$ is the number of nodes in the intepolant time stamps $T$, and $B$ is the matrix produced for the interpolant. Given this we have \begin{equation} \langle\, h | h\,\rangle \approx \langle\, \mathcal{I}[h] | \mathcal{I}[h]\,\rangle = \sum_{i=1}^N \left( \sum_{j=1}^M B_j(t_i)h(T_j)\right)^2. \end{equation} We want to try and separate out the longer sum over $N$, so that it can be performed in pre-processing. This part can be separated out as an array with components given by $\bar{B}_{mn} = \sum_{i=1}^N B_m(t_i) B_n(t_i)$, where $m$ and $n$ run from 1 to $M$. Then to get the "reduced order quadrature" we need to do \begin{equation} \langle\, \mathcal{I}[h] | \mathcal{I}[h]\,\rangle = \vec{H} \bar{B} \vec{H}^{T}, \end{equation} where $\vec{H} = [h(T_1), h(T_2), \ldots]$. Note that this is an order $M^2$ operation, which is quick for $M^2 \ll N$, but otherwise does not save computational time. ``` # create new weights w2 = np.zeros((B.shape[0], B.shape[0])) for i in range(B.shape[0]): for j in range(B.shape[0]): w2[i,j] = np.sum(B[i]*B[j]) ``` ``` sigfull = signalmodel(ts, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FAs, FBs) sigred = signalmodel(ts_nodes, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FA_nodes, FB_nodes) h_dot_h = np.vdot(sigfull, sigfull) ROQh = np.dot(np.dot(sigred, w2), sigred) print "regular inner product = %.15e"%h_dot_h print "ROQ = %.15e"%ROQh ``` regular inner product = 4.557619458915702e+02 ROQ = 4.557619458915702e+02 ``` t1 = time.time() for i in range(50000): sigfullnew = signalmodel(ts, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FAs, FBs) np.dot(sigfullnew, sigfullnew) # regular inner product e1 = time.time() t2 = time.time() for i in range(50000): sigrednew = signalmodel(ts_nodes, 1., cosis_rand[0], phi0s_rand[0], psis_rand[0], FA_nodes, FB_nodes) np.dot(np.dot(sigrednew, w2), sigrednew) # ROQ inner product e2 = time.time() print "regular inner product took %f s"%((e1-t1)/50000.) print "ROQ took %f s"%((e2-t2)/50000.) print "speedup = %f"%((e1-t1) / (e2-t2)) ``` regular inner product took 0.000124 s ROQ took 0.000018 s speedup = 6.796928 The other way to get this inner product is to analytically integrate the model using the known form of the antenna pattern (e.g. equations 10-13 of [Jaranowski, Krolak & Schutz (1998)](http://adsabs.harvard.edu/abs/1998PhRvD..58f3001J)), such that \begin{align} \langle\,h(t)|h(t)\,\rangle =& \frac{1}{\Delta t}\int_{t_1}^{t_2} h^2 {\rm d}t \\ =& \frac{h_0^2}{2^2\Delta t}\Bigg[ \left(\frac{(1+\cos{}^2\iota)}{2}\cos{\phi_0}\right)^2 \left(\int_{t_1}^{t_2} F_+^2 {\rm d}t\right) + \left(\cos{\iota}\sin{\phi_0}\right)^2\left(\int_{t_1}^{t_2} F_{\times}^2 {\rm d}t\right) +\\ & 2\frac{(1+\cos{}^2\iota)}{2}\cos{\iota}\cos{\phi_0}\sin{\phi_0}\left(\int_{t_1}^{t_2} F_+ F_{\times} {\rm d}t\right) \Bigg]. \end{align} These integral would be cheap to produce once at the start of the run. Things would become more complex if the model has a varying phase evolution due to searching over frequency parameters. In that case it might be possible to use the stationary phase approx _Note_: write down the above integrals in full. Also note that the above is all just for the real part of a heterodyned signal. The imaginary part is given by \begin{equation} h(t) = \frac{h_0}{2}\left[\frac{1}{2}F_+ (1+\cos{}^2\iota)\sin{\phi_0} - F_{\times}\cos{\iota} \cos{\phi_0 }\right]. \end{equation} I should redo all the above, but just getting the reduced basis, interpolant and ROQ for the antenna patterns (i.e. over $\psi$ [or in fact not even over that as it just gives pre-factors, so really it's just the two $a$ and $b$ functions of the antenna pattern that need orthogonalising!]), as the other things are all just pre-factors that are very quickly computed.

cf113704f31a6e081450678be604e7f6a67ad44d

ipynb

Jupyter Notebook

ROQ/Reduced basis for CW signal model (no frequency range).ipynb

mattpitkin/random_scripts

8fcfc1d25d8ca7ef66778b7b30be564962e3add3

ROQ/Reduced basis for CW signal model (no frequency range).ipynb

mattpitkin/random_scripts

8fcfc1d25d8ca7ef66778b7b30be564962e3add3

ROQ/Reduced basis for CW signal model (no frequency range).ipynb

mattpitkin/random_scripts

8fcfc1d25d8ca7ef66778b7b30be564962e3add3

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

EE 502 P: Analytical Methods for Electrical Engineering # Homework 1: Python Setup ## Due October 10, 2021 by 11:59 PM ### <span style="color: red">Mayank Kumar</span> Copyright &copy; 2021, University of Washington <hr> **Instructions**: Please use this notebook as a template. Answer all questions using well formatted Markdown with embedded LaTeX equations, executable Jupyter cells, or both. Submit your homework solutions as an `.ipynb` file via Canvas. <span style="color: red'"> Although you may discuss the homework with others, you must turn in your own, original work. </span> **Things to remember:** - Use complete sentences. Equations should appear in text as grammatical elements. - Comment your code. - Label your axes. Title your plots. Use legends where appropriate. - Before submitting a notebook, choose Kernel -> Restart and Run All to make sure your notebook runs when the cells are evaluated in order. Note : Late homework will be accepted up to one week after the due date and will be worth 50% of its full credit score. ### 0. Warmup (Do not turn in) - Get Jupyter running on your computer, or learn to use Google Colab's Jupyter environment. - Make sure you can click through the Lecture 1 notes on Python. Try changing some of the cells to see the effects. - If you haven't done any Python, follow one of the links in Lecture 1 to a tutorial and work through it. - If you haven't done any Numpy or Sympy, read through the linked documentation and tutorials for those too. ### 1. Complex Numbers Write a function `rand_complex(n)` that returns a list of `n` random complex numbers uniformly distributed in the unit circle (i.e., the magnitudes of the numbers are all between 0 and 1). Give the function a docstring. Demonstrate the function by making a list of 25 complex numbers. ```python def rand_complex(n): """ n : number of complex number to be generated function during call imports random and numpy libraries for processing. This function prints the complex number uniformly distributed in the unit circle. """ import random # import "random" library to generate random numbers. import numpy as np # import numpy to process mathematical expression a = [ ] # declare an empty list for i in range(n): x = random.uniform(-1,1) # generate random number between -1 and 1. y = random.uniform(-(np.sqrt(1-(x*x))),np.sqrt(1-(x*x))) # generating random value of y between -max and max. # max implies corresponding point on the circle for with y will lie on the circle. a.append(complex(x,y)) # append complex number x+yj to the list a. # print("Complex %d: %f + %fj" %(i+1,x, y)) # print("Value %d: %f" %((i+1),np.sqrt(x*x+y*y))) return a #function call for 25 complex numbers rand_complex(25) ``` [(-0.23291727633158343-0.03223906965648793j), (0.1075264193171328-0.6287554693178661j), (-0.9648352685957642-0.09836768328893047j), (0.9729332666351493-0.07563889894437151j), (-0.31264579365485834-0.6952447113453855j), (-0.7543046289338111-0.2007322497775722j), (0.35390768875343515+0.24066078970197824j), (0.40023028389808935-0.4322478564776867j), (0.7575837355779895-0.42474223841029735j), (-0.017343727286792454-0.9318922926373809j), (0.9893226815073348-0.07429562306514148j), (-0.7557002838583444+0.5007781393244392j), (0.14229428659162435-0.05580365383869268j), (0.34332631754502363+0.7610185062340256j), (-0.6663127004466431-0.43778123350460907j), (0.698490378322481+0.33122161883243073j), (0.3875019475298742+0.5520329887602473j), (-0.7285191732411849+0.20066390378294496j), (-0.6876991691513978-0.10572637939038865j), (0.2895621483951627-0.14222573877072409j), (0.559677483878626+0.18654677359454308j), (0.8954820520485438-0.2273070618668761j), (0.4062147904057407+0.3113418259043812j), (0.5540089171896212-0.42022748042357205j), (-0.8270129193619533-0.27250082404388287j)] ### 2. Hashes Write a function `to_hash(L) `that takes a list of complex numbers `L` and returns an array of hashes of equal length, where each hash is of the form `{ "re": a, "im": b }`. Give the function a docstring and test it by converting a list of 25 numbers generated by your `rand_complex` function. ```python def to_hash(L): """ L: It is a list of complex numbers. With a list as input, the function returns an array of hashes in form {"re": a, "im": b} """ b = [] #declare an empty list for i in range(len(L)): x1 = L[i].real #extract real part of the (i)th value of list L y1 = L[i].imag #extract imaginary part of the (i)th value of list L b.append({"re": x1, "im": y1}) #append the values in hash form. return b #function call using the function from Question 1. to_hash(rand_complex(25)) #extra line of code to verify the outcome. #d = to_hash(rand_complex(25)) #print(d[1]) #d[1]["re"] ``` ### 3. Matrices Write a function `lower_traingular(n)` that returns an $n \times n$ numpy matrix with zeros on the upper diagonal, and ones on the diagonal and lower diagonal. For example, `lower_triangular(3)` would return ```python array([[1, 0, 0], [1, 1, 0], [1, 1, 1]]) ``` ```python import numpy as np # variable np is used inside the function to generate array. #function defination starts here. def lower_triangular(n): """ function takes integer n as input and returns lower triangular matrix of dimension nxn. import module "numpy as np" to use this function without any errors. """ c = np.ones(n*n) # create an array of nxn elements with value = 1. c = c.reshape (n,n) # Reshape the array to a matrix of dimension nxn. for i in range (n): for j in range (n): if j > i: # checks for the elements where we have to change it to zero. c[i][j] = 0 # Replace the elements in upper triangle other than diagonal elements with 0. return c print("lower triangular matrix :") lower_triangular(3) ``` lower triangular matrix : array([[1., 0., 0.], [1., 1., 0.], [1., 1., 1.]]) ### 4. Numpy Write a function `convolve(M,K)` that takes an $n \times m$ matrix $M$ and a $3 \times 3$ matrix $K$ (called the kernel) and returns their convolution as in [this diagram](https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTYo2_VuAlQhfeEGJHva3WUlnSJLeE0ApYyjw&usqp=CAU). Please do not use any predefined convolution functions from numpy or scipy. Write your own. If the matrix $M$ is too small, your function should return a exception. You can read more about convolution in [this post](https://setosa.io/ev/image-kernels/). The matrix returned will have two fewer rows and two fewer columns than $M$. Test your function by making a $100 \times 100$ matrix of zeros and ones that as an image look like the letter X and convolve it with the kernel $$ K = \frac{1}{16} \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{pmatrix} $$ Use `imshow` to display both images using subplots. ```python # import numpy as np #import numpy library into variable np import numpy as np import matplotlib.pyplot as plt %matplotlib inline #define kernel K = np.array ([[1,2,1], [2,4,2], [1,2,1]]) K = K/16 # divide all elements of matrix by 16. #create image for testing as per the question m = 100 n = 100 im = np.ones(m*n) #created an array of length mn. im = im.reshape(m,n) #reshaped the array in matrix of dimension mxn. #creating image in form X. for i in range (m): for j in range (n): if ((i==j) or (i+j == n-1)): #select all the elements where manipulation is required. im[i][j] = 0 #replace 1 with 0 at all the locations selected above. fig = plt.figure(figsize = (8,8)) #define figure size fig.add_subplot(1,2,1) #add a subplot to the figure plt.imshow(im) #show input image which was created earlier. plt.axis('off') #axis are turned off plt.title("Original") #adding title to the image #function defination starts here def convolve (M,K): """ function take two inputs. M: it is the original image on which convolution should be performed. K: it is the kernel which will be applied to the input matrix. it returns matrix with (m-2) x (n-2) """ im_out = np.zeros(M.shape[0]* M.shape[1]).reshape(M.shape[0], M.shape[1]) #declare a matrix with same dimension as M if (M.shape[0] < 3 or M.shape[1] < 3): #raising exception for the matrices with either if dimensions < 3. raise Exception("Convolution of matrix M can't be calculated. Check input matrix for dimension.") else: for i in range(1, M.shape[0]-1): for j in range(1, M.shape[1]-1): im_out[i][j] = np.sum(K * M[(i-1):(i+2),(j-1):(j+2)]) #performing sum on matrix multiplication with kernal and current pixels of the image. if (im_out[i][j]) > 255: #checking for the pixels which get more than 255. im_out[i][j] = 255 elif (im_out[i][j] < 0): #checking for the pixels which get less than 0. im_out[i][j] = 0 im_out = im_out[1:(M.shape[0]-1),1:(M.shape[1]-1)] #removed the blank pixels where convolution was not possible. return im_out out_image = convolve(im,K) #function call, output is saved for further usage #print(len(out_image[0])) fig.add_subplot(1,2,2) #add a subplot to the figure plt.imshow(out_image) #Show the output image to the location plt.axis('off') #axis is turned off as it is not required plt.title("Output") #adding title to the output image ``` ### 5. Symbolic Manipulation Use sympy to specify and solve the following equations for $x$. - $x^2 + 2x - 1 = 0$ - $a x^2 + bx + c = 0$ Also, evaluate the following integrals using sympy - $\int x^2 dx$ - $\int x e^{6x} dx$ - $\int (3t+5)\cos(\frac{t}{4}) dt$ ```python #Importing related libraries import math from sympy import * init_printing(use_latex='mathjax') #solving First equation i.e., x^2 + 2x - 1 = 0 x = symbols("x") expr_1 = (x**2) + (2*x) - 1 result_1 = solve(expr_1,x) print("solution of equation x^2 + 2x - 1 = 0 is :") result_1 ``` ```python #solving second equation i.e., ax^2 + bx + c = 0 a,b,c = symbols("a b c") x = symbols("x") expr_2 = (a*(x**2)) + b*x + c result_2 = solve(expr_2,x) print("solution of equation ax^2 + bx + c = 0 is : " ) result_2 ``` solution of equation ax^2 + bx + c = 0 is : $\displaystyle \left[ \frac{- b + \sqrt{- 4 a c + b^{2}}}{2 a}, \ - \frac{b + \sqrt{- 4 a c + b^{2}}}{2 a}\right]$ ```python #evaluating integral 1 x = symbols("x") expr_3 = x**2 integrate(expr_3,x) ``` ```python #evaluating integral 2 x = symbols("x") expr_4 = x * exp(6*x) integrate(expr_4) ``` ```python #evaluating integral 3 t = symbols("t") expr_5 = ((3*t) + 5) * cos(t/4) integrate(expr_5) ``` ### 6. Typesetting Use LaTeX to typeset the following equations. ## 17 Equations that changed the world ### by Ian Stewart #### Typesetting starts Now --- **1. Pythagoras theorm:** \begin{align} a^2 + b^2 = c^2 \end{align} ___ **2. Logrithms:** \begin{align} \log xy = \log x +\log y \end{align} ___ **3. Calculus:** \begin{align} \frac{df}{dt} = \lim_{h \rightarrow 0} \frac {f(t+h) - f(t)}{f(h)} \end{align} I have made some modification in the formula written above. as it was not conveying intended meaning. ___ **4. Law of Gravity:** \begin{align} F = G\frac{m_1 m_2}{r^2} \end{align} ___ **5. Square root of minus one** \begin{align} i^2 = -1 \end{align} ___ **6. The Euler's formula for polyhedra** \begin{align} V - E + F = 2 \end{align} ___ **7. Normal Distribution** \begin{align} \phi(x) = \frac {1}{\sqrt{2\pi\rho}} e^\frac{(x-\mu)^2}{2\rho^2} \end{align} ___ **8. Wave Equation** \begin{align} \frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2} \end{align} ___ **9. Fourier Transform** \begin{align} f(w) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i x w} dx \end{align} ___ **10. Navier-Stokes Equation** \begin{align} \rho(\frac{\partial v}{\partial t} + v.\nabla v ) = - \nabla p + \nabla.T + f \end{align} ___ **11. Maxwell's Equation** \begin{align} \nabla . E = 0 \hspace{50 pt}\nabla.H = 0 \\ \nabla \times E = -\frac{1}{c} \frac{\partial H}{\partial t} \hspace{30 pt}\nabla \times H = \frac{1}{c} \frac{\partial E}{\partial t} \end{align} ___ **12. Second Law of Thermodynamics** \begin{align} dS \geq 0 \end{align} ___ **13. Relativity** \begin{align} E = mc^2 \end{align} ___ **14. Schrodinger's Equation** \begin{align} i\hbar \frac{\partial}{\partial t} \Psi = H \Psi \end{align} ___ **15. Information Theory** \begin{align} H = -\sum p(x)\log p(x) \end{align} ___ **16. Chaos Theory** \begin{align} x_{t + 1} = k x_t(1 - x_t) \end{align} ___ **17.Black-Scholes Equation** \begin{align} \frac{1}{2}\sigma S^2 \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} + \frac{\partial V}{\partial t} - r V = 0 \end{align} ___ ```python ```

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Jupyter Notebook

Basics/HW_01_Python_EEP502_Mayank Kumar.ipynb

krmayankb/Analytical_Methods

a1bf58e1b9056949f5aa0fb25070e2d0ffbf5c4f

Basics/HW_01_Python_EEP502_Mayank Kumar.ipynb

krmayankb/Analytical_Methods

a1bf58e1b9056949f5aa0fb25070e2d0ffbf5c4f

Basics/HW_01_Python_EEP502_Mayank Kumar.ipynb

krmayankb/Analytical_Methods

a1bf58e1b9056949f5aa0fb25070e2d0ffbf5c4f

Qwen/Qwen-72B

1. YES 2. YES

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# Selección óptima de portafolios II Entonces, tenemos que: - La LAC describe las posibles selecciones de riesgo-rendimiento entre un activo libre de riesgo y un activo riesgoso. - Su pendiente es igual al radio de Sharpe del activo riesgoso. - La asignación óptima de capital para cualquier inversionista es el punto tangente de la curva de indiferencia del inversionista con la LAC. Para todo lo anterior, supusimos que ya teníamos el portafolio óptimo (activo riesgoso). En la clase pasada aprendimos a hallar este portafolio óptimo si el conjunto de activos riesgosos estaba conformado únicamente por dos activos: $$w_{1,EMV}=\frac{(E[r_1]-r_f)\sigma_2^2-(E[r_2]-r_f)\sigma_{12}}{(E[r_2]-r_f)\sigma_1^2+(E[r_1]-r_f)\sigma_2^2-((E[r_1]-r_f)+(E[r_2]-r_f))\sigma_{12}}.$$ - Sin embargo, la complejidad del problema crece considerablemente con el número de variables, y la solución analítica deja de ser viable cuando mencionamos que un portafolio bien diversificado consta aproximadamente de 50-60 activos. - En esos casos, este problema se soluciona con rutinas numéricas que hagan la optimización por nosotros, porque son una solución viable y escalable a más variables. **Objetivos:** - ¿Cuál es el portafolio óptimo de activos riesgosos cuando tenemos más de dos activos? - ¿Cómo construir la frontera de mínima varianza cuando tenemos más de dos activos? *Referencia:* - Notas del curso "Portfolio Selection and Risk Management", Rice University, disponible en Coursera. ___ ## 1. Maximizando el radio de Sharpe ### ¿Qué pasa si tenemos más de dos activos riesgosos? En realidad es algo muy similar a lo que teníamos con dos activos. - Para dos activos, construir la frontera de mínima varianza es trivial: todas las posibles combinaciones. - Con más de dos activos, recordar la definición: la frontera de mínima varianza es el lugar geométrico de los portafolios que proveen el mínimo riesgo para un nivel de rendimiento dado. <font color=blue> Ver en el tablero.</font> Analíticamente: - $n$ activos, - caracterizados por $(\sigma_i,E[r_i])$, - cada uno con peso $w_i$, con $i=1,2,\dots,n$. Entonces, buscamos los pesos tales que \begin{align} \min_{w_1,\dots,w_n} & \quad \sum_{i=1}^{n}w_i^2\sigma_i^2+\sum_{i=1}^{n}\sum_{j=1,j\neq i}^{n}w_iw_j\sigma_{ij}\\ \text{s.a.} & \quad \sum_{i=1}^{n}w_i=1, w_i\geq0\\ & \quad \sum_{i=1}^{n}w_iE[r_i]=\bar{\mu}, \end{align} donde $\bar{\mu}$ corresponde a un nivel de rendimiento objetivo. **Obviamente, tendríamos que resolver este problema para muchos niveles de rendimiento objetivo.** - <font color=blue> Explicar relación con gráfica.</font> - <font color=green> Recordar clase 10.</font> Lo anterior se puede escribir vectorialmente como: \begin{align} \min_{\boldsymbol{w}} & \quad \boldsymbol{w}^T\Sigma\boldsymbol{w}\\ \text{s.a.} & \quad \boldsymbol{1}^T\boldsymbol{w}=1, \boldsymbol{w}\geq0\\ & \quad E[\boldsymbol{r}^T]\boldsymbol{w}=\bar{\mu}, \end{align} donde: - $\boldsymbol{w}=\left[w_1,\dots,w_n\right]^T$ es el vector de pesos, - $\boldsymbol{1}=\left[1,\dots,1\right]^T$ es un vector de unos, - $E[\boldsymbol{r}]=\left[E[r_1],\dots,E[r_n]\right]^T$ es el vector de rendimientos esperados, y - $\Sigma=\left[\begin{array}{cccc}\sigma_{1}^2 & \sigma_{12} & \dots & \sigma_{1n} \\ \sigma_{21} & \sigma_{2}^2 & \dots & \sigma_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \sigma_{n1} & \sigma_{n2} & \dots & \sigma_{n}^2\end{array}\right]$ es la matriz de varianza-covarianza. **Esta última forma es la que comúnmente usamos al programar, por ser eficiente y escalable a problemas de N variables.** ### Entonces, ¿para cuántos niveles de rendimiento objetivo tendríamos que resolver el anterior problema con el fin de graficar la frontera de mínima varianza? - Observar que el problema puede volverse muy pesado a medida que incrementamos el número de activos en nuestro portafolio... - Una tarea bastante compleja. ### Sucede que, en realidad, sólo necesitamos conocer dos portafolios que estén sobre la *frontera de mínima varianza*. - Si logramos encontrar dos portafolios sobre la frontera, entonces podemos a la vez encontrar todas las posibles combinaciones de estos dos portafolios para trazar la frontera de mínima varianza. - Ver el caso de dos activos. ### ¿Qué portafolios usar? Hasta ahora, hemos estudiando profundamente como hallar dos portafolios muy importantes que de hecho yacen sobre la frontera de mínima varianza: 1. Portafolio de EMV: máximo SR. 2. Portafolio de mínima varianza: básicamente, el mismo problema anterior, sin la restricción de rendimiento objetivo. Luego, tomar todas las posibles combinaciones de dichos portafolios usando las fórmulas para dos activos de medias y varianzas: - w: peso para el portafolio EMV, - 1-w: peso para le portafolio de mínima varianza. ## 2. Ejemplo ilustrativo. Retomamos el ejemplo de mercados de acciones en los países integrantes del $G5$: EU, RU, Francia, Alemania y Japón. ```python # Importamos pandas y numpy import pandas as pd import numpy as np ``` ```python # Resumen en base anual de rendimientos esperados y volatilidades annual_ret_summ = pd.DataFrame(columns=['EU', 'RU', 'Francia', 'Alemania', 'Japon'], index=['Media', 'Volatilidad']) annual_ret_summ.loc['Media'] = np.array([0.1355, 0.1589, 0.1519, 0.1435, 0.1497]) annual_ret_summ.loc['Volatilidad'] = np.array([0.1535, 0.2430, 0.2324, 0.2038, 0.2298]) annual_ret_summ.round(4) ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>EU</th> <th>RU</th> <th>Francia</th> <th>Alemania</th> <th>Japon</th> </tr> </thead> <tbody> <tr> <th>Media</th> <td>0.1355</td> <td>0.1589</td> <td>0.1519</td> <td>0.1435</td> <td>0.1497</td> </tr> <tr> <th>Volatilidad</th> <td>0.1535</td> <td>0.243</td> <td>0.2324</td> <td>0.2038</td> <td>0.2298</td> </tr> </tbody> </table> </div> ```python # Matriz de correlación corr = pd.DataFrame(data= np.array([[1.0000, 0.5003, 0.4398, 0.3681, 0.2663], [0.5003, 1.0000, 0.5420, 0.4265, 0.3581], [0.4398, 0.5420, 1.0000, 0.6032, 0.3923], [0.3681, 0.4265, 0.6032, 1.0000, 0.3663], [0.2663, 0.3581, 0.3923, 0.3663, 1.0000]]), columns=annual_ret_summ.columns, index=annual_ret_summ.columns) corr.round(4) ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>EU</th> <th>RU</th> <th>Francia</th> <th>Alemania</th> <th>Japon</th> </tr> </thead> <tbody> <tr> <th>EU</th> <td>1.0000</td> <td>0.5003</td> <td>0.4398</td> <td>0.3681</td> <td>0.2663</td> </tr> <tr> <th>RU</th> <td>0.5003</td> <td>1.0000</td> <td>0.5420</td> <td>0.4265</td> <td>0.3581</td> </tr> <tr> <th>Francia</th> <td>0.4398</td> <td>0.5420</td> <td>1.0000</td> <td>0.6032</td> <td>0.3923</td> </tr> <tr> <th>Alemania</th> <td>0.3681</td> <td>0.4265</td> <td>0.6032</td> <td>1.0000</td> <td>0.3663</td> </tr> <tr> <th>Japon</th> <td>0.2663</td> <td>0.3581</td> <td>0.3923</td> <td>0.3663</td> <td>1.0000</td> </tr> </tbody> </table> </div> ```python # Tasa libre de riesgo rf = 0.05 ``` Esta vez, supondremos que tenemos disponibles todos los mercados de acciones y el activo libre de riesgo. #### 1. Construir la frontera de mínima varianza ##### 1.1. Encontrar portafolio de mínima varianza ```python # Importamos funcion minimize del modulo optimize de scipy from scipy.optimize import minimize ``` ```python ## Construcción de parámetros # 1. Sigma: matriz de varianza-covarianza Sigma = S.dot(corr).dot(S) S = np.diag(annual_ret_summ.loc['Volatilidad']) Sigma = S.dot(corr).dot(S).astype(float) # 2. Eind: rendimientos esperados activos individuales Eind = annual_ret_summ.loc['Media'].values.astype(float) ``` ```python # Función objetivo def varianza(w, Sigma): return w.T.dot(Sigma).dot(w) ``` ___ ### Explicación funciones lambda ```python def f(x): return x**2 ``` ```python f(1), f(10), f(13) ``` (1, 100, 169) ```python g = lambda x: x**2 ``` ```python g(1), g(10), g(13) ``` (1, 100, 169) ___ ```python # Número de activos n = len(Eind) # Dato inicial w0 = np.ones(n) / n # Cotas de las variables bnds = ((0, 1),) * n # Restricciones cons = {"type": "eq", "fun": lambda w: w.sum() - 1} ``` ```python # Portafolio de mínima varianza minvar = minimize(fun=varianza, x0=w0, args=(Sigma,), bounds=bnds, constraints=cons, tol=1e-10) ``` ```python minvar ``` fun: 0.018616443574850344 jac: array([0.03723289, 0.03879431, 0.03847277, 0.03723289, 0.03723289]) message: 'Optimization terminated successfully.' nfev: 70 nit: 10 njev: 10 status: 0 success: True x: array([6.20463011e-01, 8.67361738e-19, 5.42101086e-19, 2.03475313e-01, 1.76061676e-01]) ```python # Pesos, rendimiento, riesgo y razón de Sharpe del portafolio de mínima varianza w_minvar = minvar.x e_minvar = Eind.dot(w_minvar) s_minvar = (w_minvar.T.dot(Sigma).dot(w_minvar))**0.5 rs_minvar = (e_minvar - rf) / s_minvar e_minvar, s_minvar, rs_minvar ``` (0.1396278783022745, 0.13644208872210342, 0.6568931855391253) ##### 1.2. Encontrar portafolio EMV ```python # Función objetivo def menos_rs(w, Eind, rf, Sigma): ep = Eind.dot(w) sp = (w.T.dot(Sigma).dot(w))**0.5 rs = (ep - rf) / sp return -rs ``` ```python # Número de activos n = len(Eind) # Dato inicial w0 = np.ones(n) / n # Cotas de las variables bnds = ((0, 1),) * n # Restricciones cons = {"type": "eq", "fun": lambda w: w.sum() - 1} ``` ```python # Portafolio EMV emv = minimize(fun=menos_rs, x0=w0, args=(Eind, rf, Sigma), bounds=bnds, constraints=cons, tol=1e-10) ``` ```python emv ``` fun: -0.6644375126752685 jac: array([-0.36087245, -0.36087371, -0.36087482, -0.36087404, -0.36087622]) message: 'Optimization terminated successfully.' nfev: 56 nit: 8 njev: 8 status: 0 success: True x: array([0.50729071, 0.07475346, 0.02413765, 0.18995791, 0.20386028]) ```python # Pesos, rendimiento, riesgo y razón de Sharpe del portafolio EMV w_emv = emv.x e_emv = Eind.dot(w_emv) s_emv = (w_emv.T.dot(Sigma).dot(w_emv))**0.5 rs_emv = (e_emv - rf) / s_emv e_emv, s_emv, rs_emv ``` (0.14205956750708812, 0.13855263399626944, 0.6644375126752685) ```python e_minvar, s_minvar, rs_minvar ``` (0.1396278783022745, 0.13644208872210342, 0.6568931855391253) ##### 1.3. Construir frontera de mínima varianza También debemos encontrar la covarianza (o correlación) entre estos dos portafolios: ```python # Covarianza entre los portafolios cov = w_emv.T.dot(Sigma).dot(w_minvar) #cov = w_minvar.T.dot(Sigma).dot(w_emv) cov ``` 0.018689768271811187 ```python # Correlación entre los portafolios corr = cov / (s_emv * s_minvar) corr ``` 0.988645903271872 ```python w_minvar ``` array([6.20463011e-01, 8.67361738e-19, 5.42101086e-19, 2.03475313e-01, 1.76061676e-01]) ```python w_emv ``` array([0.50729071, 0.07475346, 0.02413765, 0.18995791, 0.20386028]) ```python # Vector de w w = np.linspace(0, 1, 101) ``` ```python # DataFrame de portafolios: # 1. Índice: i # 2. Columnas 1-2: w, 1-w # 3. Columnas 3-4: E[r], sigma # 4. Columna 5: Sharpe ratio f_mv = pd.DataFrame({ "w": w, "1-w": 1 - w, "Media": w * e_emv + (1 - w) * e_minvar, "Vol": ((w * s_emv)**2 + ((1 - w) * s_minvar)**2 + 2 * w * (1 - w) * cov)**0.5 }) f_mv["rs"] = (f_mv["Media"] - rf) / f_mv["Vol"] f_mv.head() ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>w</th> <th>1-w</th> <th>Media</th> <th>Vol</th> <th>rs</th> </tr> </thead> <tbody> <tr> <th>0</th> <td>0.00</td> <td>1.00</td> <td>0.139628</td> <td>0.136442</td> <td>0.656893</td> </tr> <tr> <th>1</th> <td>0.01</td> <td>0.99</td> <td>0.139652</td> <td>0.136448</td> <td>0.657045</td> </tr> <tr> <th>2</th> <td>0.02</td> <td>0.98</td> <td>0.139677</td> <td>0.136453</td> <td>0.657195</td> </tr> <tr> <th>3</th> <td>0.03</td> <td>0.97</td> <td>0.139701</td> <td>0.136460</td> <td>0.657343</td> </tr> <tr> <th>4</th> <td>0.04</td> <td>0.96</td> <td>0.139725</td> <td>0.136466</td> <td>0.657490</td> </tr> </tbody> </table> </div> ```python # Importar librerías de gráficos from matplotlib import pyplot as plt ``` ```python # Gráfica de dispersión de puntos coloreando # de acuerdo a SR, los activos individuales # y los portafolios hallados # Frontera plt.scatter(f_mv["Vol"], f_mv["Media"], c=f_mv["rs"], label="Front. MV") # Activos ind for i in range(n): plt.plot(annual_ret_summ.iloc[1, i], annual_ret_summ.iloc[0, i], "o", ms=10, label=annual_ret_summ.columns[i]) # Port. óptimos plt.plot(s_minvar, e_minvar, '*g', ms=10, label="P. Min. Var") plt.plot(s_emv, e_emv, '*b', ms=10, label="P. EMV") # Etiquetas de los ejes plt.xlabel("Volatilidad $\sigma$") plt.ylabel("Rendimiento esperado $E[r]$") plt.colorbar() # Leyenda plt.legend(loc="upper left", bbox_to_anchor=(1.3, 1)) ``` **A partir de lo anterior, solo restaría construir la LAC y elegir la distribución de capital de acuerdo a las preferencias (aversión al riesgo).** ___ ```python # Vector de wp variando entre 0 y 1.5 con n pasos wp = np.linspace(0, 1.5, 101) ``` ```python # DataFrame de CAL: # 1. Índice: i # 2. Columnas 1-2: wp, wrf # 3. Columnas 3-4: E[r], sigma # 4. Columna 5: Sharpe ratio lac = pd.DataFrame({"wp": wp, "wrf": 1 - wp, "Media": wp * e_emv + (1 - wp) * rf, "Vol": wp * s_emv}) lac["RS"] = (lac["Media"] - rf) / lac["Vol"] ``` ```python lac.head(10) ``` <div> <style scoped> .dataframe tbody tr th:only-of-type { vertical-align: middle; } .dataframe tbody tr th { vertical-align: top; } .dataframe thead th { text-align: right; } </style> <table border="1" class="dataframe"> <thead> <tr style="text-align: right;"> <th></th> <th>wp</th> <th>wrf</th> <th>Media</th> <th>Vol</th> <th>RS</th> </tr> </thead> <tbody> <tr> <th>0</th> <td>0.000</td> <td>1.000</td> <td>0.050000</td> <td>0.000000</td> <td>NaN</td> </tr> <tr> <th>1</th> <td>0.015</td> <td>0.985</td> <td>0.051381</td> <td>0.002078</td> <td>0.664438</td> </tr> <tr> <th>2</th> <td>0.030</td> <td>0.970</td> <td>0.052762</td> <td>0.004157</td> <td>0.664438</td> </tr> <tr> <th>3</th> <td>0.045</td> <td>0.955</td> <td>0.054143</td> <td>0.006235</td> <td>0.664438</td> </tr> <tr> <th>4</th> <td>0.060</td> <td>0.940</td> <td>0.055524</td> <td>0.008313</td> <td>0.664438</td> </tr> <tr> <th>5</th> <td>0.075</td> <td>0.925</td> <td>0.056904</td> <td>0.010391</td> <td>0.664438</td> </tr> <tr> <th>6</th> <td>0.090</td> <td>0.910</td> <td>0.058285</td> <td>0.012470</td> <td>0.664438</td> </tr> <tr> <th>7</th> <td>0.105</td> <td>0.895</td> <td>0.059666</td> <td>0.014548</td> <td>0.664438</td> </tr> <tr> <th>8</th> <td>0.120</td> <td>0.880</td> <td>0.061047</td> <td>0.016626</td> <td>0.664438</td> </tr> <tr> <th>9</th> <td>0.135</td> <td>0.865</td> <td>0.062428</td> <td>0.018705</td> <td>0.664438</td> </tr> </tbody> </table> </div> ```python plt.scatter(f_mv["Vol"], f_mv["Media"], c=f_mv["rs"], label="Front. MV") # Activos ind for i in range(n): plt.plot(annual_ret_summ.iloc[1, i], annual_ret_summ.iloc[0, i], "o", ms=10, label=annual_ret_summ.columns[i]) # Port. óptimos plt.plot(s_minvar, e_minvar, '*g', ms=10, label="P. Min. Var") plt.plot(s_emv, e_emv, '*b', ms=10, label="P. EMV") # LAC plt.plot(lac["Vol"], lac["Media"], "--k", lw=3, label="LAC") # Etiquetas de los ejes plt.xlabel("Volatilidad $\sigma$") plt.ylabel("Rendimiento esperado $E[r]$") plt.colorbar() # Leyenda plt.legend(loc="upper left", bbox_to_anchor=(1.3, 1)) plt.axis([0.13, 0.15, 0.135, 0.145]) ``` ```python # Para gamma=5 g = 5 w_opt = (e_emv - rf) / (g * s_emv**2) w_opt, 1 - w_opt ``` (0.9591120623418227, 0.04088793765817733) ```python # Ponderaciones finales w_opt * w_emv, 1 - w_opt ``` (array([0.48654864, 0.07169694, 0.02315071, 0.18219092, 0.19552485]), 0.04088793765817733) ## 3. Comentarios finales ### 3.1. Restricciones adicionales Los inversionistas pueden tener restricciones adicionales: 1. Restricciones en posiciones cortas. 2. Pueden requerir un rendimiento mínimo. 3. Inversión socialmente responsable: prescinden de inversiones en negocios o paises considerados éticamente o políticamente indeseables. Todo lo anterior se puede incluir como restricciones en el problema de optimización, y puede ser llevado a cabo a costa de un cociente de Sharpe menor. ### 3.2. Críticas a la optimización media varianza 1. Solo importan medias y varianzas: recordar que la varianza subestima el riesgo en algunos casos. 2. Preferencias media-varianza tratan las ganancias y pérdidas simétricamente: el sentimiento de insatisfacción de una perdida es mayor al sentimiento de satisfacción de una ganancia (aversión a pérdidas). 3. La aversión al riesgo es constante: la actitud frente al riesgo puede cambiar, por ejemplo con el estado de la economía. 4. Horizonte corto (un periodo). 5. Basura entra - basura sale: la optimización media varianza es supremamente sensible a las entradas: estimaciones de rendimientos esperados y varianzas. ___ # Anuncios parroquiales ## 1. Quiz la próxima clase (clases 12, 13, y 14). ## 2. Revisar archivo Tarea 6. ## 3. [Nota interesante](http://yetanothermathprogrammingconsultant.blogspot.com/2016/08/portfolio-optimization-maximize-sharpe.html) <footer id="attribution" style="float:right; color:#808080; background:#fff;"> Created with Jupyter by Esteban Jiménez Rodríguez. </footer>

ef4a4284245f8350e3dbb12ffdd0567ea76fab56

ipynb

Jupyter Notebook

Modulo3/Clase14_SeleccionOptimaPortII.ipynb

if722399/porinvo2021

19ee9421b806f711c71b2affd1633bbfba40a9eb

Modulo3/Clase14_SeleccionOptimaPortII.ipynb

if722399/porinvo2021

19ee9421b806f711c71b2affd1633bbfba40a9eb

Modulo3/Clase14_SeleccionOptimaPortII.ipynb

if722399/porinvo2021

19ee9421b806f711c71b2affd1633bbfba40a9eb

Qwen/Qwen-72B

1. YES 2. YES

__label__spa_Latn

# Week 2 worksheet 3: Gaussian elimination This notebook was created by Charlotte Desvages. Gaussian elimination is a direct method to solve linear systems of the form $Ax = b$, with $A \in \mathbb{R}^{n\times n}$ and $b \in \mathbb{R}^n$, to find the unknown $x \in \mathbb{R}^n$. This week, we put what we have seen so far into practice, and program different steps of the Gaussian elimination algorithm: forward substitution, backward substitution, and elementary row operations. The best way to learn programming is to write code. Don't hesitate to edit the code in the example cells, or add your own code, to test your understanding. You will find practice exercises throughout the notebook, denoted by 🚩 ***Exercise $x$:***. ## How workshops will work: 1. Read the Exercise 2. Write tests and fail these tests 3. Write the solution 4. Pass the tests ### Displaying solutions Solutions will be released after the workshops, as a new `.txt` file in the same GitHub repository. After pulling the file to your workspace, run the following cell to create clickable buttons under each exercise, which will allow you to reveal the solutions. ```python %run scripts/create_widgets.py W02-W3 ``` --- ### 📚 Book sections - **ASC**: sections 4.1, ***4.2***, 4.7, 4.8 - **PCP**: sections 2.3, 2.4, 5.6 🚩 Section **4.2** of **ASC** is **mandatory reading** this week, particularly when working through sections 3 and 4 of this notebook. You probably have seen Gaussian elimination in your first year Linear Algebra course, so this should be familiar already -- but this will be a good refresher. --- ## 1. NumPy's `np.linalg` Numpy has a **sub-module** called `linalg`, which contains many useful functions for linear algebra and matrix operations. If we imported Numpy as `np`, for example, then to use the functions in `linalg`, you will need to prefix them with `np.linalg.`. Some of the functions provided by the `np.linalg` submodule are: ```python import numpy as np # Create a random 3x3 matrix and a vector of three 1s A = np.random.random([3, 3]) b = np.ones(3) print(np.linalg.eigvals(A)) # Eigenvalues of a matrix: note the complex values here, j=sqrt(-1) eig_val_A, eig_vec_A = np.linalg.eig(A) # Eigenvalues and right eigenvectors print("Eigenvalues: ", eig_val_A) print("Eigenvectors: ", eig_vec_A) print('\nInverse and determinant:') print("A^(-1) =", np.linalg.inv(A)) # Inverse of a matrix print("det(A) =", np.linalg.det(A)) # Determinant of a matrix print('\nSolution of Ax = b:') print("x =", np.linalg.solve(A, b)) # Solve Ax = b for x ``` --- **📚 Learn more:** * [numpy.linalg](https://docs.scipy.org/doc/numpy/reference/routines.linalg.html) * **ASC**: section 4.8 --- 🚩 ***Exercise 1:*** Create two variables `M` and `y`, assigned with Numpy arrays representing the matrix $M$ and vector $y$ defined as (you can reuse the code from the linear algebraic equations example) $$ M = \begin{pmatrix} 9 & 3 & 0 \\ -2 & -2 & 1 \\ 0 & -1 & 1 \end{pmatrix}, \qquad y = \begin{pmatrix} 0.4 \\ -3 \\ 0.3 \end{pmatrix}. $$ Then, solve the system $Mx = y$ for $x$, using `np.linalg.solve()`. *For checking:* the result should be `[-3.16666667 9.63333333 9.93333333]`. ```python # Tests assert x[0] == -57/18, "x[0] should be -57/18" ``` ```python %run scripts/show_solutions.py W02-W3_ex1 ``` --- ## 2. Diagonal matrices Diagonal matrices have elements off the leading diagonal equal to zero. Elements on the leading diagonal of a diagonal matrix may or may not be equal to zero. A diagonal matrix $A$ is invertible iff none of its diagonal elements are equal to zero. ### 2.1. Solving diagonal systems When $A$ is a diagonal matrix, the linear system $Ax = b$ can be written as $$ \begin{pmatrix} a_{11} & & & \\ & a_{22} & & \\ & & \ddots & \\ & & & a_{nn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} \qquad \Leftrightarrow \qquad \begin{cases} a_{11} x_1 &= b_1 \\ a_{22} x_2 &= b_2 \\ &\vdots \\ a_{nn} x_n &= b_n \end{cases}, $$ where $a_{ii}, b_{i}, i = 1, \dots, n$ are known. The matrix $A$ is invertible (and therefore the system $Ax = b$ has precisely one solution) iff all $a_{ii} \neq 0$. --- 🚩 ***Exercise 2:*** Write a function `linsolve_diag()` which solves the linear system $A x = b$, returning $x$ in the output variable `x` (as a NumPy array), without using `np.linalg.solve()`. Here the input `A` should be assumed to be an invertible **diagonal** square matrix, and `b` a column vector. *Hints:* - Use the `.shape` attribute of NumPy arrays to determine the size of the input matrix and vector. - The solution may be computed using a `for` loop. - There is also an efficient way to do this via a NumPy function which extracts the diagonal elements of a matrix. *For checking:* the solution to the given example is $[20, 10]$. ```python import numpy as np def linsolve_diag(A, b): ''' Solves the diagonal system Ax = b for x, assuming A is invertible. ''' # your code here x = np.ones(2) return x # Use the function on the following example A = np.array([[2, 0], [0, 0.5]]) b = np.array([40, 5]) x = linsolve_diag(A, b) print(x) # Test for the expected solution assert x[0] == 20 assert x[1] == 10 ``` ```python %run scripts/show_solutions.py W02-W3_ex2 ``` --- 🚩 ***Exercise 3:*** Use your `linsolve_diag` function to solve the linear system \begin{equation} \left( \begin{array}{ccc} 3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 10 \end{array} \right) x = \left( \begin{array}{c} 3 \\ 1 \\ 1 \end{array} \right), \end{equation} for $x$. ```python ``` ```python %run scripts/show_solutions.py W02-W3_ex3 ``` ### 2.2. Measuring computation time The `time()` function in Python's `time` module allows Python to read the current time from your computer's clock. We can therefore use it to time how long it takes a section of code to run, as follows: ```python import time t0 = time.time() # Code to time t = time.time() - t0 print(f"Elapsed time: {t:.6f} seconds") ``` and the resulting time is stored in the variable `t`, as the time elapsed between the first and the second measurement. --- **📚 Learn more:** - [The `time` module](https://docs.python.org/3/library/time.html) - Python documentation - [`time.time()`](https://docs.python.org/3/library/time.html#time.time) - Python documentation - **PCP**: section 5.6, which discusses measuring computation time and efficiency, and provides examples using a different Python module called [`timeit`](https://docs.python.org/3/library/timeit.html) --- 🚩 ***Exercise 4:*** The following code generates a randomised invertible diagonal square matrix $A$ with dimension $N$, stored in the variable `A`, and a right-hand-side vector $b$, stored in the variable `b`, for a given value of `N`. Use `time.time()` to time how long it takes the `np.linalg.solve()` function to solve $A x = b$ for $x$. Compare this against the time it takes your `linsolve_diag()` function from Exercise 2 to solve for $x$, for different values of `N`. Display the measured times in a way that is convenient to read (you can use an f-string, for instance; see the Week 1 workshop task). *Hint:* limit `N` to less than $\sim 1,000$ to avoid using excessive memory. ```python import time # Create a randomised invertible diagonal matrix A and vector b N = 500 A = np.diag(np.random.random([N])) + np.eye(N) b = np.random.random([N]) # your code here # remember to add tests ``` ```python %run scripts/show_solutions.py W02-W3_ex4 ``` --- ## 3. Forward and backward substitution Gaussian elimination can be performed in 2 steps: forward substitution and backward substitution. In your previous courses on linear algebra, you probably have performed this by hand on small systems ($4\times 4$ or so). We can *implement* (program) the procedure in Python to be able to solve systems of any size much more quickly. ### 3.1. Lower triangular systems: forward substitution **Lower triangular matrices** have elements above the leading diagonal equal to zero. Elements on or below the leading diagonal may or may not be equal to zero. Linear systems involving lower triangular invertible square matrices can be solved via **forward substitution**. For example for the linear system \begin{equation} \left( \begin{array}{ccc} 2 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 1 & 2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 4 \\ 1 \\ 4 \end{array} \right), \end{equation} applying the matrix multiplication gives \begin{equation} \left( \begin{array}{c} 2 x_1 \\ -x_1 + x_2 \\ -x_1 + x_2 + 2 x_3 \end{array} \right) = \left( \begin{array}{c} 4 \\ 1 \\ 4 \end{array} \right), \end{equation} where, for instance, $-x_1 + x_2 + 2 x_3$ is the 3rd element of the vector $Ax$. Comparing the first elements gives $x_1$. Since $x_1$ is now known, comparing the second elements gives $x_2$. Since $x_1$ and $x_2$ are now known, comparing the third elements gives $x_3$. In other words, $x_1$ is trivial to compute, and is then *substituted* into the next equation, which means that $x_2$ is now trivial to compute, etc. The substitutions cascade *forward*. #### Forward substitution in Python The function `linsolve_lt()` below solves the linear system $A x = b$ using forward substitution, returning $x$ in the output variable `x`. Here the input `A` should be assumed to be an invertible **lower triangular** square matrix, and `b` a column vector. ```python def linsolve_lt(A, b): ''' Solves the lower triangular system Ax = b. ''' N = b.shape[0] x = np.zeros(N) for i in range(N): x[i] = (b[i] - A[i, :i] @ x[:i]) / A[i, i] return x # Solving the system in the example above A = np.array([[2, 0, 0], [-1, 1, 0], [-1, 1, 2]], dtype=float) b = np.array([4, 1, 4], dtype=float) x = linsolve_lt(A, b) print(x) ``` --- 🚩 ***Exercise 5:*** Examine the function `linsolve_lt()` carefully to understand how and why it works. Add code comments in the function definition to explain each step. *Hint:* - pen and paper will be useful here! Write (or sketch) what line 8 achieves depending on the value of `row`. For instance, what happens at the first iteration of the loop (when `row` is `0`)? at the second iteration (when `row` is `1`)? etc. - @ or np.matmul() indicates matrix/matrix-vector products ```python ``` ```python %run scripts/show_solutions.py W02-W3_ex5 ``` --- ### 3.2. Upper triangular systems: backward substitution **Upper triangular matrices** have elements below the leading diagonal equal to zero. Elements on or above the leading diagonal may or may not be equal to zero. Linear systems involving upper triangular invertible square matrices can be solved via **backward substitution**. Backward substitution is similar to forward substitution, but starts from the last row, and substitutions cascade backward until the first row. #### Backward substitution in Python 🚩 ***Exercise 6:*** Write a function `linsolve_ut()` which solves the linear system $A x = b$ using backward substitution, returning $x$ in the output variable `x`. Here the input `A` should be assumed to be an invertible **upper triangular** square matrix, and `b` a column vector. You can start from `linsolve_lt()` above and adapt it to use backward substitution. *For checking:* The solution to the given example is $[-1, 2]$. ```python def linsolve_ut(A, b): ''' Solves the upper triangular system Ax = b. ''' # your code here return x # Testing with an example A = np.array([[1, 1], [0, 0.5]]) b = np.array([1, 1]) x = linsolve_ut(A, b) print(x) ``` ```python %run scripts/show_solutions.py W02-W3_ex6 ``` --- 🚩 ***Exercise 7:*** The following code generates an invertible upper triangular square matrix $A$ with dimension $N$, stored in the variable `A`, and a right-hand-side vector $b$, stored in the variable `b`, for a given value of `N`. Use `time.time()` to time how long it takes the `np.linalg.solve()` function to solve $A x = b$ for $x$. Compare this against the time it takes your `linsolve_ut()` function to solve for $x$, for different values of `N`. *Hint:* Limit `N` to less than $\sim 1,000$ to avoid using excessive memory. ```python import time # Create a randomised invertible upper triangular matrix A and vector b N = 800 A = np.triu(np.random.random([N])) + np.eye(N) b = np.random.random([N]) # your code here ``` ```python %run scripts/show_solutions.py W02-W3_ex7 ``` ## 4. Gaussian elimination We now know how to solve lower and upper triangular systems. Now, consider a system which is not triangular -- for instance: $$ \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}. $$ We can build the *augmented matrix* by adding $b$ as an extra column in $A$: $$ \begin{pmatrix} 1 & 1 & 1 & 2 \\ 2 & 1 & -1 & 1 \\ 1 & 1 & 2 & 0 \end{pmatrix}. $$ The goal is now to **reduce** this augmented matrix into **reduced row echelon form** (RREF), i.e. $$ \begin{pmatrix} 1 & 0 & 0 & x_1 \\ 0 & 1 & 0 & x_2 \\ 0 & 0 & 1 & x_3 \end{pmatrix}, $$ and the final column is then the solution of the original linear problem. We do this by applying **elementary row operations** to the augmented matrix, to create zeros under each diagonal element: \begin{align*} \left( \begin{array}{cccc} 1 & 1 & 1 & 2 \\ 2 & 1 & -1 & 1 \\ 1 & 1 & 2 & 0 \end{array} \right) \underset{R_2 - 2 R_1}{\rightarrow} & \left( \begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & -1 & -3 & -3 \\ 1 & 1 & 2 & 0 \end{array} \right) \nonumber \\ \underset{R_3 - R_1}{\rightarrow} & \left( \begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & -1 & -3 & -3 \\ 0 & 0 & 1 & -2 \end{array} \right) \nonumber \\ \end{align*} This is equivalent to the linear equations: \begin{align*} 1 x_1 + 1 x_2 + 1 x_3 & = 2, \nonumber \\ 0 x_1 - 1 x_2 - 3 x_3 & = -3, \nonumber \\ 0 x_1 + 0 x_2 + 1 x_3 & = -2. \end{align*} We could keep going, and apply further elementary row operations to the augmented matrix... but this system is now **upper triangular**, and therefore we can solve it using **backward substitution**! Time to tie it all together -- remember that you can check section **4.2** in **ASC** for help for these final problems. ### 4.1. Elementary row operations 🚩 ***Exercise 8:*** Write a function `row_op()` which applies the elementary row operation \begin{equation} \left( \textrm{Row} j \right) \rightarrow \beta \times \left( \textrm{Row } j \right) + \alpha \times \left( \textrm{Row } i \right), \end{equation} where $\alpha, \beta \in \mathbb{R}$. *For checking:* The solution to the given example is $[[0, 4], [1, 2]]$. *Hint:* Input arguments of functions can be modified if they are e.g. lists or NumPy arrays (remember section 4 of the Week 3 tutorial), so you can apply this operation to `A` itself, and thus change it. As long as you don't redefine `A` from scratch inside the function, you don't even need to return it, as it will be changed in place. Since you don't return it (i.e. you return `None`), there is no result to assign to a new variable -- so simply calling your function (as in the example), without assigning the output to `A` for instance, will still work. Here is a simpler example to illustrate this: ```python def change_in_place(x): ''' Change the first element of x in-place. ''' x[0] = 12345 # note that we don't return anything here! # Test our function z = np.array([10, 15, 20, 25]) print(z) change_in_place(z) # this changes z itself, no "output" to store in a variable here print(z) ``` ```python def row_op(A, alpha, i, beta, j): ''' Applies row operation beta*A_j + alpha*A_i to A_j, the jth row of the matrix A. Changes A in place. ''' # your code here # Testing with an example A = np.array([[2, 0], [1, 2]]) alpha, beta = 2, -1 i, j = 1, 0 # If you don't return A, it will be changed in-place when the function is executed print(A) row_op(A, alpha, i, beta, j) # this changes A print(A) ``` ```python %run scripts/show_solutions.py W02-W3_ex8 ``` ### 4.2. Row echelon form 🚩 ***Exercise 9 (challenging):*** Write a function `REF()` which takes as inputs `A` and `b`, a square invertible matrix and a vector, and returns `C` and `d`, which are respectively `A` and `b` transformed by successive elementary row operations, so that `C` is upper triangular (and the system $Cx = d$ is equivalent to $Ax = b$). Your function should first build the augmented matrix $( A | b )$, and use elementary row operations as in the example above to reduce it to row echelon form. Finally, it should split the final augmented matrix into a square matrix `C` and a vector `d`. Use your function `row_op()` to perform the row operations: **you do not need to re-define it here**, you can simply *call* it -- i.e. use the command `row_op(..)` with appropriate input arguments inside your function `REF()`. You will have to calculate $\alpha$ and $\beta$ for each row operation. For instance, in the example above, the first row operation performed is $R_2 \to R_2 - 2R_1$, therefore we have $i=1$, $j=2$, $\alpha = -2$, and $\beta = 1$. How can you know that these values of $\alpha$ and $\beta$ will ensure that the element in the second row, first column becomes 0? (*hint: you should see similarities with your forward substitution algorithm.*) *Hint:* think about how you would do this on paper. You will need to create zeros under the diagonal element in each column (one after another), and you will need a separate row operation for each row (in a given column) to make the leading element zero. You will need 2 nested loops. *For checking:* `C` and `d` should be as in the example above. ```python def REF(A, b): ''' Reduces the augmented matrix (A|b) into row echelon form, returns (C|d). ''' # your code here return C, d # Testing with an example A = np.array([[1, 1, 1], [2, 1, -1], [1, 1, 2]], dtype=float) b = np.array([2, 1, 0], dtype=float) C, d = REF(A, b) print(C) print(d) ``` ```python %run scripts/show_solutions.py W02-W3_ex9 ``` ### 4.3. Completing Gaussian elimination We have done all the hard work now, all that is left is to put it all together. 🚩 ***Exercise 10:*** Write a function `gauss()` which, given an invertible matrix `A` and a column vector `b`, solves the system $Ax = b$ and returns the result as `x`. This function should make use of your previous functions `REF()` and `linsolve_ut()`. (Again, no need to define them again here, just call them.) *For checking:* the result here should be $[-5, 9, -2]$. *For further checking:* given an arbitrary `A` and `b`, how can you check that `x` is indeed the solution to $Ax = b$ (to machine precision)? ```python def gauss(A, b): ''' Solve the linear system Ax = b, given a square invertible matrix A and a vector b, using Gaussian elimination. ''' # your code here return x # Test the function A = np.array([[1, 1, 1], [2, 1, -1], [1, 1, 2]], dtype=float) b = np.array([2, 1, 0], dtype=float) x = gauss(A, b) print(x) ``` ```python %run scripts/show_solutions.py W02-W3_ex9 ```

031b6b6b2db1e915d9224a4d190df4e81a503f8f

ipynb

Jupyter Notebook

Workshops/W02-W3_NMfCE_Gaussian_elimination.ipynb

DrFriedrich/nmfce-2021-22

2ccee5a97b24bd5c1e80e531957240ffb7163897

Workshops/W02-W3_NMfCE_Gaussian_elimination.ipynb

DrFriedrich/nmfce-2021-22

2ccee5a97b24bd5c1e80e531957240ffb7163897

Workshops/W02-W3_NMfCE_Gaussian_elimination.ipynb

DrFriedrich/nmfce-2021-22

2ccee5a97b24bd5c1e80e531957240ffb7163897

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python '''import numpy as np import math import sympy import sympy as sp from sympy import Eq, IndexedBase, symbols, Idx, Indexed, Sum, S, N from sympy.functions.special.tensor_functions import KroneckerDelta from sympy.vector import Vector, CoordSys3D, AxisOrienter, BodyOrienter, Del, curl, divergence, gradient, is_conservative, is_solenoidal, scalar_potential, Point, scalar_potential_difference, Del, express, matrix_to_vector import matplotlib.pyplot as plt from sympy.physics.vector import ReferenceFrame import matplotlib as mpl from mpl_toolkits.mplot3d import Axes3D from IPython.display import display from IPython.display import display_latex from sympy import latex from sympy import Array, Matrix, transpose, zeros, diff, Function, Derivative, cos, sin, sqrt, solve, linsolve, acos, atan, asin from sympy import symbols from sympy.plotting import plot''' from numpy import log, log2 from scipy.linalg import logm, expm from spatialmath import SO2, SE2 from spatialmath.base import rotx, rot2, det, simplify, skew, vex, trot2, transl2, trplot2 from spatialmath.base import transl, trplot import spatialmath.base.symbolic as sym ``` ```python np.set_printoptions(formatter={'float': lambda x: "{0: 0.4f}".format(x)}) ``` Given $^A\xi_B$, A is the reference frame B is the target frame $^AP = ^A\xi_B \cdot ^BP$ We use the operator ⊕ to indicate composition of relative poses $^AP = (^A\xi_B \oplus ^B\xi_C) \bullet ^CP$ ```python R = rot2(0.2) d = det(R) d ``` $\displaystyle 1.0$ ```python theta = sym.symbol('theta') R = rot2(theta) R ``` array([[cos(theta), -sin(theta)], [sin(theta), cos(theta)]], dtype=object) ```python simplify(R * R) ``` $\displaystyle \left[\begin{matrix}\cos^{2}{\left(\theta \right)} & \sin^{2}{\left(\theta \right)}\\\sin^{2}{\left(\theta \right)} & \cos^{2}{\left(\theta \right)}\end{matrix}\right]$ ```python simplify(det(R)) ``` $\displaystyle 1$ Matrix Exponential ```python R = rot2(0.3) R ``` array([[ 0.9553, -0.2955], [ 0.2955, 0.9553]]) ```python R = rot2(0.3) S = logm(R) ``` ```python k = skew(2) ``` ```python vex(S) ``` array([ 0.3000]) ```python expm(S) ``` array([[ 0.9553, -0.2955], [ 0.2955, 0.9553]]) $R = e^{[\theta]_x} \exists SO(2)$ ```python R = rot2(0.3) R = expm(skew(0.3)) R ``` array([[ 0.9553, -0.2955], [ 0.2955, 0.9553]]) Pose in 2-Dimensions Homogeneous Transformation Matrix $^AP = (^AR_B)$ ```python T1 = transl2(1, 2) * trot2(30, 'deg') T1 ``` array([[ 0.8660, -0.0000, 0.0000], [ 0.0000, 0.8660, 0.0000], [ 0.0000, 0.0000, 1.0000]]) ```python trplot( transl(1,2,3), frame='A', rviz=True, width=1, dims=[0, 10, 0, 10, 0, 10]) ``` ```python trplot2( transl2(1,2) * trot2(30, 'deg'), frame='A', rviz=True, width=1, dims=[0, 5, 0, 5]) ``` ```python transl2(1,2) * trot2(30, 'deg') ``` array([[ 0.8660, -0.0000, 0.0000], [ 0.0000, 0.8660, 0.0000], [ 0.0000, 0.0000, 1.0000]]) ```python ```

f1364f9bfba7c6242df4854ee74b9b3b8f8ade39

ipynb

Jupyter Notebook

.ipynb_checkpoints/Robotics-checkpoint.ipynb

Valentine-Efagene/Jupyter-Notebooks

91a1d98354a270d214316eba21e4a435b3e17f5d

.ipynb_checkpoints/Robotics-checkpoint.ipynb

Valentine-Efagene/Jupyter-Notebooks

91a1d98354a270d214316eba21e4a435b3e17f5d

.ipynb_checkpoints/Robotics-checkpoint.ipynb

Valentine-Efagene/Jupyter-Notebooks

91a1d98354a270d214316eba21e4a435b3e17f5d

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

| |Pierre Proulx, ing, professeur| |:---|:---| |Département de génie chimique et de génie biotechnologique |** GCH200-Phénomènes d'échanges I **| ### Section10.4, chauffage par effet Brinkman: source de chauffage causé par la viscosité > >> Ici on traitera le problème de façon légèrement différente de Transport Phenomena, en effet, pour traiter ce problème on peut aller chercher les équations générales de l'annexe B.9. >>Regardons donc l'équation B.9-1 qui est utilisée puisqu'on suppose que le système a une courbure négligeable et peut être traité en coordonnées cartésiennes. Tous les termes de gauche sont nuls car: >>> On veut le résultat en état de régime, donc la dérivée en fonction du temps est nulle. >>> Il n'y a pas de vitesse dans la direction x ni dans la direction y, et il n'y a pas de gradient de température dans la direction z. >>> Le seul terme dans les dérivées de la température qui reste à droite est celui dans la direction x, et le terme de source $\Phi$, donc: >>>> ### $k \frac {\partial^2 T}{\partial x^2}+\mu \Phi_v = 0$ >> Le terme de source visqueuse est obtenu par l'annexe B7 > qui peut sembler très complexe (en effet) mais tous les termes s'annulent sauf celui du gradient de $v_z$ en fonction de x donc: >>>> ### $\Phi_v =\big(\frac {\partial v_z}{\partial x} \big)^2 $ >> L'équation à résoudre sera donc directement: >>>> ### $k \frac {\partial^2 T}{\partial x^2}+ \mu \big(\frac {\partial v_z}{\partial x} \big)^2= 0$ > Et puisque le profil de vitesse vous a déjà été donné sur la figure 10.4-2, on pourra résoudre ```python # # Pierre Proulx # # Préparation de l'affichage et des outils de calcul symbolique # import sympy as sp from IPython.display import * sp.init_printing(use_latex=True) %matplotlib inline ``` ```python # Paramètres, variables et fonctions # mu,k,b,Tb,T0,vb,x=sp.symbols('mu k b T_b T_0 v_b x') T=sp.Function('T')(x) v=sp.Function('v')(x) ``` ```python v=vb*x/b eq=k*T.diff(x,x)+mu*v.diff(x)**2 display(eq) ``` ```python T=sp.dsolve(eq) display('Le profil de température, solution générale avec les constantes inconnues',T) ``` ```python condition_1=sp.Eq(T.rhs.subs(x,0),T0) condition_2=sp.Eq(T.rhs.subs(x,b),Tb) display('Les conditions aux limites',condition_1, condition_2) constantes=sp.solve([condition_1,condition_2],sp.symbols('C1,C2')) display('Les constantes après solution',constantes) T=T.subs(constantes) display('Le profil de température final',T) ``` ```python ### Tracage T=T.rhs dico={'T_0':0,'T_b':10,'k':0.1,'mu':0.01,'b':0.01,'v_b':10} T1=T.subs(dico) dico={'T_0':0,'T_b':10,'k':0.1,'mu':0.05,'b':0.01,'v_b':10} T2=T.subs(dico) dico={'T_0':0,'T_b':10,'k':0.1,'mu':0.1,'b':0.01,'v_b':10} T3=T.subs(dico) dico={'T_0':0,'T_b':10,'k':0.1,'mu':0.15,'b':0.01,'v_b':10} T4=T.subs(dico) import matplotlib.pyplot as plt plt.rcParams['figure.figsize']=10,8 p=sp.plot((T1,(x,0,0.01)),(T2,(x,0,0.01)),(T3,(x,0,0.01)),(T4,(x,0,0.01)), legend=True,ylabel='T',xlabel='x',show=False) p[0].label='mu=0.01' p[1].label='.05' p[2].label='.1' p[3].label='.15' p[0].line_color='blue' p[1].line_color='red' p[2].line_color='green' p[3].line_color='black' p.show() ```

9fc8807afd7b991f7f0e53625a5b3166f1535334

ipynb

Jupyter Notebook

Chap-10-Section-10-4.ipynb

pierreproulx/GCH200

66786aa96ceb2124b96c93ee3d928a295f8e9a03

2018-02-26T16:29:58.000Z

2018-02-26T16:29:58.000Z

Chap-10-Section-10-4.ipynb

pierreproulx/GCH200

66786aa96ceb2124b96c93ee3d928a295f8e9a03

Chap-10-Section-10-4.ipynb

pierreproulx/GCH200

66786aa96ceb2124b96c93ee3d928a295f8e9a03

2018-02-27T15:04:33.000Z

2021-06-03T16:38:07.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__fra_Latn

```julia using MomentClosure, Latexify, OrdinaryDiffEq, Catalyst ``` ┌ Info: Precompiling MomentClosure [01a1b25a-ecf0-48c5-ae58-55bfd5393600] └ @ Base loading.jl:1278 $$ G \stackrel{c_1}{\rightarrow} G+P, \\ G^* \stackrel{c_2}{\rightarrow} G^*+P, \\ P \stackrel{c_3}{\rightarrow} 0 \\ G+P \underset{c_5}{\stackrel{c_4}{\rightleftharpoons}} G^* $$ On/off gene states are merged into a Bernoulli variable $g(t)$ which can be either $1$ ($G$) or $0$ ($G^*$). The number of proteins in the system is given by $p(t)$. ### Using `ReactionSystemMod` ```julia @parameters t, c₁, c₂, c₃, c₄, c₅ @variables p(t), g(t) vars = [g, p] ps = [c₁, c₂, c₃, c₄, c₅] S = [0 0 0 -1 1; 1 1 -1 -1 1] as = [c₁*g, # G -> G + P c₂*(1-g), # G* -> G* + P c₃*p, # P -> 0 c₄*g*p, # G+P -> G* c₅*(1-g)] # G* -> G+P # system size Ω can be included as an additional parameter if needed binary_vars = [1] rn = ReactionSystemMod(t, vars, ps, as, S) ``` ReactionSystemMod(t, Term{Real,Nothing}[g(t), p(t)], Sym{ModelingToolkit.Parameter{Real},Nothing}[c₁, c₂, c₃, c₄, c₅], SymbolicUtils.Mul{Real,Int64,Dict{Any,Number},Nothing}[c₁*g(t), c₂*(1 - g(t)), c₃*p(t), c₄*g(t)*p(t), c₅*(1 - g(t))], [0 0 … -1 1; 1 1 … -1 1]) ### Using Catalyst.jl `ReactionSystem` * $\rightarrow$ indicates a reaction that follows the law of mass action (need to indicate only the reaction coefficient, full propensity function is constructed automatically) * $\Rightarrow$ indicates a reaction that does not follow the law of mass action (need to define the full propensity function) ```julia rn = @reaction_network begin (c₁), g → g+p (c₂*(1-g)), 0 ⇒ p (c₃), p → 0 (c₄), g+p → 0 (c₅*(1-g)), 0 ⇒ g+p end c₁ c₂ c₃ c₄ c₅ ``` \begin{align} \require{mhchem} \ce{ g &->[c{_1}] g + p}\\ \ce{ \varnothing &<=>[c{_2} \left( 1 - g\left( t \right) \right)][c{_3}] p}\\ \ce{ g + p &<=>[c{_4}][c{_5} \left( 1 - g\left( t \right) \right)] \varnothing} \end{align} identical stoichiometry matrix and propensity functions are recovered ```julia propensities(rn, combinatoric_ratelaw=false) ``` 5-element Array{SymbolicUtils.Mul{Real,Int64,Dict{Any,Number},Nothing},1}: c₁*g(t) c₂*(1 - g(t)) c₃*p(t) c₄*g(t)*p(t) c₅*(1 - g(t)) ```julia get_S_mat(rn) ``` 2×5 Array{Int64,2}: 0 0 0 -1 1 1 1 -1 -1 1 ### Moment equations Generate raw moment equations up to 3rd order. The argument `combinatoric_ratelaw = false` indicates whether binomial coefficients are included when constructing the propensity functions for the reactions that follow the law of mass action (does not play a role in this specific scenarion) Equivalently, central moment equations can be generated using `generate_central_moment_eqs(rn, 3, 5, combinatoric_ratelaw=false)` ```julia raw_eqs = generate_raw_moment_eqs(rn, 3, combinatoric_ratelaw=false) latexify(raw_eqs) ``` \begin{align*} \frac{d\mu{_{10}}}{dt} =& c{_5} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{01}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} - c{_2} \mu{_{10}} - c{_3} \mu{_{01}} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{20}}}{dt} =& c{_5} + c{_4} \mu{_{11}} + c{_5} \mu{_{10}} - 2 c{_4} \mu{_{21}} - 2 c{_5} \mu{_{20}} \\ \frac{d\mu{_{11}}}{dt} =& c{_5} + c{_1} \mu{_{20}} + c{_2} \mu{_{10}} + c{_4} \mu{_{11}} + c{_5} \mu{_{01}} - c{_2} \mu{_{20}} - c{_3} \mu{_{11}} - c{_4} \mu{_{12}} - c{_4} \mu{_{21}} - c{_5} \mu{_{11}} - c{_5} \mu{_{20}} \\ \frac{d\mu{_{02}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + c{_3} \mu{_{01}} + c{_4} \mu{_{11}} + 2 c{_1} \mu{_{11}} + 2 c{_2} \mu{_{01}} + 2 c{_5} \mu{_{01}} - c{_2} \mu{_{10}} - 2 c{_2} \mu{_{11}} - 2 c{_3} \mu{_{02}} - 2 c{_4} \mu{_{12}} - c{_5} \mu{_{10}} - 2 c{_5} \mu{_{11}} \\ \frac{d\mu{_{30}}}{dt} =& c{_5} + 3 c{_4} \mu{_{21}} + 2 c{_5} \mu{_{10}} - c{_4} \mu{_{11}} - 3 c{_4} \mu{_{31}} - 3 c{_5} \mu{_{30}} \\ \frac{d\mu{_{21}}}{dt} =& c{_5} + c{_1} \mu{_{30}} + c{_2} \mu{_{20}} + c{_4} \mu{_{12}} + c{_5} \mu{_{01}} + c{_5} \mu{_{10}} + c{_5} \mu{_{11}} + 2 c{_4} \mu{_{21}} - c{_2} \mu{_{30}} - c{_3} \mu{_{21}} - c{_4} \mu{_{11}} - 2 c{_4} \mu{_{22}} - c{_4} \mu{_{31}} - c{_5} \mu{_{20}} - 2 c{_5} \mu{_{21}} - c{_5} \mu{_{30}} \\ \frac{d\mu{_{12}}}{dt} =& c{_5} + c{_1} \mu{_{20}} + c{_2} \mu{_{10}} + c{_3} \mu{_{11}} + c{_4} \mu{_{21}} + c{_5} \mu{_{02}} + 2 c{_1} \mu{_{21}} + 2 c{_2} \mu{_{11}} + 2 c{_4} \mu{_{12}} + 2 c{_5} \mu{_{01}} - c{_2} \mu{_{20}} - 2 c{_2} \mu{_{21}} - 2 c{_3} \mu{_{12}} - c{_4} \mu{_{11}} - c{_4} \mu{_{13}} - 2 c{_4} \mu{_{22}} - c{_5} \mu{_{12}} - c{_5} \mu{_{20}} - 2 c{_5} \mu{_{21}} \\ \frac{d\mu{_{03}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + 3 c{_1} \mu{_{11}} + 3 c{_1} \mu{_{12}} + 3 c{_2} \mu{_{01}} + 3 c{_2} \mu{_{02}} + 3 c{_3} \mu{_{02}} + 3 c{_4} \mu{_{12}} + 3 c{_5} \mu{_{01}} + 3 c{_5} \mu{_{02}} - c{_2} \mu{_{10}} - 3 c{_2} \mu{_{11}} - 3 c{_2} \mu{_{12}} - c{_3} \mu{_{01}} - 3 c{_3} \mu{_{03}} - c{_4} \mu{_{11}} - 3 c{_4} \mu{_{13}} - c{_5} \mu{_{10}} - 3 c{_5} \mu{_{11}} - 3 c{_5} \mu{_{12}} \end{align*} We are solving for moments up to `m_order = 3`, and in the equations encounter moments up to `exp_order = 5`. Use the Bernoulli variable properties to eliminate redundant equations to see how they simplify: ```julia bernoulli_eqs = bernoulli_moment_eqs(raw_eqs, binary_vars) latexify(bernoulli_eqs) ``` \begin{align*} \frac{d\mu{_{10}}}{dt} =& c{_5} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{01}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} - c{_2} \mu{_{10}} - c{_3} \mu{_{01}} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{11}}}{dt} =& c{_5} + c{_1} \mu{_{10}} + c{_5} \mu{_{01}} - c{_3} \mu{_{11}} - c{_4} \mu{_{12}} - c{_5} \mu{_{10}} - c{_5} \mu{_{11}} \\ \frac{d\mu{_{02}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + c{_3} \mu{_{01}} + c{_4} \mu{_{11}} + 2 c{_1} \mu{_{11}} + 2 c{_2} \mu{_{01}} + 2 c{_5} \mu{_{01}} - c{_2} \mu{_{10}} - 2 c{_2} \mu{_{11}} - 2 c{_3} \mu{_{02}} - 2 c{_4} \mu{_{12}} - c{_5} \mu{_{10}} - 2 c{_5} \mu{_{11}} \\ \frac{d\mu{_{12}}}{dt} =& c{_5} + c{_1} \mu{_{10}} + c{_3} \mu{_{11}} + c{_5} \mu{_{02}} + 2 c{_1} \mu{_{11}} + 2 c{_5} \mu{_{01}} - 2 c{_3} \mu{_{12}} - c{_4} \mu{_{13}} - c{_5} \mu{_{10}} - 2 c{_5} \mu{_{11}} - c{_5} \mu{_{12}} \\ \frac{d\mu{_{03}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + 3 c{_1} \mu{_{11}} + 3 c{_1} \mu{_{12}} + 3 c{_2} \mu{_{01}} + 3 c{_2} \mu{_{02}} + 3 c{_3} \mu{_{02}} + 3 c{_4} \mu{_{12}} + 3 c{_5} \mu{_{01}} + 3 c{_5} \mu{_{02}} - c{_2} \mu{_{10}} - 3 c{_2} \mu{_{11}} - 3 c{_2} \mu{_{12}} - c{_3} \mu{_{01}} - 3 c{_3} \mu{_{03}} - c{_4} \mu{_{11}} - 3 c{_4} \mu{_{13}} - c{_5} \mu{_{10}} - 3 c{_5} \mu{_{11}} - 3 c{_5} \mu{_{12}} \end{align*} ### Closing the moment equations Finally, we can apply the selected moment closure method on the system of raw moment equations: ```julia closed_raw_eqs = moment_closure(raw_eqs, "conditional derivative matching", binary_vars) latexify(closed_raw_eqs) ``` \begin{align*} \frac{d\mu{_{10}}}{dt} =& c{_5} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{01}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} - c{_2} \mu{_{10}} - c{_3} \mu{_{01}} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} \\ \frac{d\mu{_{11}}}{dt} =& c{_5} + c{_1} \mu{_{10}} + c{_5} \mu{_{01}} - c{_3} \mu{_{11}} - c{_4} \mu{_{12}} - c{_5} \mu{_{10}} - c{_5} \mu{_{11}} \\ \frac{d\mu{_{02}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + c{_3} \mu{_{01}} + c{_4} \mu{_{11}} + 2 c{_1} \mu{_{11}} + 2 c{_2} \mu{_{01}} + 2 c{_5} \mu{_{01}} - c{_2} \mu{_{10}} - 2 c{_2} \mu{_{11}} - 2 c{_3} \mu{_{02}} - 2 c{_4} \mu{_{12}} - c{_5} \mu{_{10}} - 2 c{_5} \mu{_{11}} \\ \frac{d\mu{_{12}}}{dt} =& c{_5} + c{_1} \mu{_{10}} + c{_3} \mu{_{11}} + c{_5} \mu{_{02}} + 2 c{_1} \mu{_{11}} + 2 c{_5} \mu{_{01}} - 2 c{_3} \mu{_{12}} - c{_5} \mu{_{10}} - 2 c{_5} \mu{_{11}} - c{_5} \mu{_{12}} - c{_4} \mu{_{10}} \mu{_{11}}^{-3} \mu{_{12}}^{3} \\ \frac{d\mu{_{03}}}{dt} =& c{_2} + c{_5} + c{_1} \mu{_{10}} + 3 c{_1} \mu{_{11}} + 3 c{_1} \mu{_{12}} + 3 c{_2} \mu{_{01}} + 3 c{_2} \mu{_{02}} + 3 c{_3} \mu{_{02}} + 3 c{_4} \mu{_{12}} + 3 c{_5} \mu{_{01}} + 3 c{_5} \mu{_{02}} - c{_2} \mu{_{10}} - 3 c{_2} \mu{_{11}} - 3 c{_2} \mu{_{12}} - c{_3} \mu{_{01}} - 3 c{_3} \mu{_{03}} - c{_4} \mu{_{11}} - c{_5} \mu{_{10}} - 3 c{_5} \mu{_{11}} - 3 c{_5} \mu{_{12}} - 3 c{_4} \mu{_{10}} \mu{_{11}}^{-3} \mu{_{12}}^{3} \end{align*} We can also print out the closure functions for each higher order moment: ```julia latexify(closed_raw_eqs, :closure) ``` \begin{align*} \mu{_{13}} =& \mu{_{10}} \mu{_{11}}^{-3} \mu{_{12}}^{3} \\ \mu{_{04}} =& \mu{_{01}}^{4} \mu{_{02}}^{-6} \mu{_{03}}^{4} \end{align*} ### Numerical solution The closed moment equations can be solved using DifferentialEquations.jl (or just OrdinaryDiffEq.jl which is more lightweight and sufficient for this particular case. ```julia # PARAMETER INITIALISATION pmap = [c₁ => 0.01, c₂ => 40, c₃ => 1, c₄ => 1, c₅ => 1] # DETERMINISTIC INITIAL CONDITIONS μ₀ = [1., 0.001] u₀map = deterministic_IC(μ₀, closed_raw_eqs) # time interval to solve on tspan = (0., 1000.0) dt = 1 @time oprob = ODEProblem(closed_raw_eqs, u₀map, tspan, pmap); @time sol_CDM = solve(oprob, Tsit5(), saveat=dt); ``` 12.183509 seconds (17.83 M allocations: 933.692 MiB, 2.85% gc time) 5.671233 seconds (12.59 M allocations: 608.756 MiB, 11.10% gc time) ```julia using Plots plot(sol_CDM.t, sol_CDM[1,:], label = "CDM", legend = true, xlabel = "Time [s]", ylabel = "Mean gene number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ``` ```julia plot(sol_CDM.t, sol_CDM[2,:], label = "CDM", legend = :bottomright, xlabel = "Time [s]", ylabel = "Mean protein number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ``` ```julia std_CDM = sqrt.(sol_CDM[4,2:end] .- sol_CDM[2,2:end].^2) plot(sol_CDM.t[2:end], std_CDM, label = "CDM", legend = true, xlabel = "Time [s]", ylabel = "standard deviation of the protein number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ``` ### SSA ```julia using DiffEqJump # parameters [c₁, c₂, c₃, c₄, c₅] p = [param[2] for param in pmap] # initial conditions [g, p] # NOTE: define as FLOATS as otherwise will encounter problems due to parsing from Int to Float u₀ = [1., 0.] # time interval to solve on tspan = (0., 1000.) # create a discrete problem to encode that our species are integer valued dprob = DiscreteProblem(rn, u₀, tspan, p) # create a JumpProblem and specify Gillespie's Direct Method as the solver: jprob = JumpProblem(rn, dprob, Direct(), save_positions=(false, false)) # SET save_positions to (false, false) as otherwise time of each reaction occurence is saved dt = 1 # time resolution at which numerical solution is saved # solve and plot ensembleprob = EnsembleProblem(jprob) @time sol_SSA = solve(ensembleprob, SSAStepper(), saveat=dt, trajectories=10000); ``` 205.463844 seconds (646.80 M allocations: 15.332 GiB, 56.95% gc time) ```julia using DiffEqJump # This is not equal to μ₀ as the initial values should be Int (not 0.001) # However, these still must be defined as Floats, otherwise will encounter # problems due to parsing from Int to Float using EnsembleAnalysis u₀ = [1., 0.] # create a discrete problem to encode that our species are integer valued dprob = DiscreteProblem(rn, u₀, tspan, pmap) # create a JumpProblem and specify Gillespie's Direct Method as the solver: jprob = JumpProblem(rn, dprob, Direct(), save_positions=(false, false)) # SET save_positions to (false, false) as otherwise time of each reaction occurence is saved dt = 1 # time resolution at which numerical solution is saved # solve and plot ensembleprob = EnsembleProblem(jprob) @time sol_SSA = solve(ensembleprob, SSAStepper(), saveat=dt, trajectories=10000); ``` 245.071448 seconds (645.00 M allocations: 15.269 GiB, 65.19% gc time) Can compute all sample moments up to chosen order ```julia @time SSA_μ = sample_raw_moments(sol_SSA, 2); @time SSA_M = sample_central_moments(sol_SSA, 2); ``` 6.137514 seconds (2.82 M allocations: 1.325 GiB, 21.46% gc time) 2.501856 seconds (2.79 M allocations: 744.488 MiB) ```julia plot(sol_CDM.t, [sol_CDM[1,:], SSA_μ[1,0]], label = ["CDM" "SSA"], legend = true, xlabel = "Time [s]", ylabel = "Mean gene number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ``` ```julia plot(sol_CDM.t, [sol_CDM[2,:], SSA_μ[0,1]], label = ["CDM" "SSA"], legend = :bottomright, xlabel = "Time [s]", ylabel = "Mean protein number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ``` ```julia std_CDM = sqrt.(sol_CDM[4,2:end] .- sol_CDM[2,2:end].^2) std_p_SSA = sqrt.(SSA_M[0,2][2:end]) plot(sol_CDM.t[2:end], [std_CDM, std_p_SSA], label = ["CDM" "SSA"], legend = true, xlabel = "Time [s]", ylabel = "standard deviation of the protein number", lw=2, legendfontsize=8, xtickfontsize=10, ytickfontsize=10, dpi=100) ```

fb7923c865a0af7e118573c7723fd305ec0182f1

ipynb

Jupyter Notebook

examples/conditional_TEST_J.ipynb

palmtree2013/MomentClosure.jl

171d18818657bf1e93240e4b24d393fac3eea72d

2021-02-21T00:44:05.000Z

2022-03-25T23:48:52.000Z

examples/conditional_TEST_J.ipynb

palmtree2013/MomentClosure.jl

171d18818657bf1e93240e4b24d393fac3eea72d

2021-02-26T15:44:04.000Z

2022-03-16T12:48:27.000Z

examples/conditional_TEST_J.ipynb

palmtree2013/MomentClosure.jl

171d18818657bf1e93240e4b24d393fac3eea72d

2021-02-21T01:20:10.000Z

2022-03-24T13:18:07.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

```python import sys if not '..' in sys.path: sys.path.insert(0, '..') import control import sympy import numpy as np import matplotlib.pyplot as plt import ulog_tools as ut import ulog_tools.control_opt as opt %matplotlib inline %load_ext autoreload %autoreload 2 ``` The autoreload extension is already loaded. To reload it, use: %reload_ext autoreload # System Identification ```python log_file = ut.ulog.download_log('http://review.px4.io/download?log=35b27fdb-6a93-427a-b634-72ab45b9609e', '/tmp') data = ut.sysid.prepare_data(log_file) res = ut.sysid.attitude_sysid(data) res ``` {'pitch': {'model': {'delay': 0.051503721021908873, 'f_s': 232.99287433805779, 'fit': 0.57135828183122261, 'gain': 29.538707721830491, 'sample_delay': 12}, 't_end': 80, 't_start': 75}, 'roll': {'model': {'delay': 0.072963604781037569, 'f_s': 232.99287433805779, 'fit': 0.80970246292599435, 'gain': 45.686710321167887, 'sample_delay': 17}, 't_end': 105, 't_start': 100}, 'yaw': {'model': {'delay': 0.11159139554746923, 'f_s': 232.99287433805779, 'fit': 0.87415539224859207, 'gain': 41.274521147805387, 'sample_delay': 26}, 't_end': 5, 't_start': 0}} # Continuous Time Optimization ```python opt.attitude_loop_design(res['roll']['model'], 'ROLL', d) ``` ```python attitude_loop_design(res['pitch']['model'], 'PITCH') ``` {'MC_PITCHRATE_D': 0.015662896675004697, 'MC_PITCHRATE_I': 0.48847645640076243, 'MC_PITCHRATE_P': 0.51104029619426683, 'MC_PITCH_P': 5.8666514695501988} ```python attitude_loop_design(res['yaw']['model'], 'YAW') ``` {'MC_YAWRATE_D': 0.017251069591687748, 'MC_YAWRATE_I': 0.19498248018478978, 'MC_YAWRATE_P': 0.18924319337905329, 'MC_YAW_P': 3.598452484267229}

aa968dd62ad677f36cf9425507b2e8b35e29dbb9

ipynb

Jupyter Notebook

notebooks/old/System ID.ipynb

dronecrew/ulog_tools

65aabdf8c729d8307b6bab2e0897ecf2a222d6bb

2017-07-07T15:26:45.000Z

2021-01-29T11:00:37.000Z

notebooks/old/System ID.ipynb

dronecrew/ulog_tools

65aabdf8c729d8307b6bab2e0897ecf2a222d6bb

2017-09-05T18:56:57.000Z

2021-09-12T09:35:19.000Z

notebooks/old/System ID.ipynb

dronecrew/ulog_tools

65aabdf8c729d8307b6bab2e0897ecf2a222d6bb

2017-07-03T19:48:30.000Z

2021-07-06T14:26:27.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__yue_Hant

# Jupyter notebook example ## Simple plots Loading the necessary modules (maybe _numpy_ is superceeded by _scipy_) ```python import numpy as npy import scipy as scy # import sympy as spy # import timeit ``` The sine function, $\sin(t)$, and its Fourier-transformation. $$f_{max}=\frac1{2\Delta t},\quad \Delta f=\frac1{t_{max}}$$ ```python # tmax, t = 200/4, scy.linspace(0,tmax,201) t = [ i/4 for i in range(201) ] y = [ npy.sin(t[i]/10*2*npy.pi) for i in range(201) ] Y = scy.fft(y) Y = abs(Y[:101])*2/200 f = scy.linspace(0,1/t[1]/2,101) ``` ### Plotting the graphs ```python %matplotlib inline import matplotlib.pyplot as plt ``` ```python plt.figure(1,figsize=(10,4)) plt.subplot(121) plt.plot(t,y,'g.') plt.title('$\sin(t)$') plt.xlabel('$t$[s]') plt.grid('on') plt.subplot(122) plt.plot(f,Y,'b') plt.title('FFT') plt.xlabel(r'$\nu$[Hz]') plt.grid('on') plt.show() ``` ```html ``` ## Fázissík módszer ### Elsőrendű differenciálegyenletek $$\begin{array}{rcl} \dot x &=& (a^2x^2+y^2-1)y\\ \dot y &=& (x^2+a^2y^2-1)x \end{array} $$ ### Nullvonalak és fázisportré ```python # %matplotlib inline from pylab import * ``` ```python a = 2 x, y = meshgrid(arange(-1.5, 1.5, 0.1), arange(-1.5, 1.5, 0.1)) fxy = -((a*x)**2+y**2-1)*y gxy = (x**2+(a*y)**2-1)*x streamplot(x, y, fxy, gxy) # f(x,y)=0 görbék: ellipszis+egyenes contour(x,y,fxy,1,colors="red") # g(x,y)=0 görbék: ellipszis+egyenes contour(x,y,gxy,1,colors="green") grid(); show() ``` ## Benchmarking ```python import time import timeit ``` ```python # benchmark from https://devblogs.nvidia.com/drop-in-acceleration-gnu-octave/ N = 8192 # def f(N): A = scy.single(scy.rand(N,N)) B = scy.single(scy.rand(N,N)) # return A * B #tic = time.clock() #print(tic) #C = A * B etime = timeit.timeit('A*B',globals=globals(),number=1) #toc = time.clock() #print(toc) elapsedTime = etime #toc-tic # elapsedTime = etime(clock(), start); print("Elapsed time: {0:.2g} sec (".format(elapsedTime)) gFlops = 2*N*N*N/(elapsedTime * 1e+9) # disp(gFlops); print("{0:.2f} GFlops)\n".format(gFlops)) ``` Elapsed time: 0.64 sec ( 1713.92 GFlops) ```python ```

07c24d3ae27da03a61f5c9414a3344b13009e817

ipynb

Jupyter Notebook

python_ex1.ipynb

szazs89/jupyter_ex

366079f54a8ac8f3d6e65d45ec79d4b2318bed40

python_ex1.ipynb

szazs89/jupyter_ex

366079f54a8ac8f3d6e65d45ec79d4b2318bed40

python_ex1.ipynb

szazs89/jupyter_ex

366079f54a8ac8f3d6e65d45ec79d4b2318bed40

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

<a id='heavy-tails'></a> <div id="qe-notebook-header" align="right" style="text-align:right;"> <a href="https://quantecon.org/" title="quantecon.org"> </a> </div> # Heavy-Tailed Distributions <a id='index-0'></a> ## Contents - [Heavy-Tailed Distributions](#Heavy-Tailed-Distributions) - [Overview](#Overview) - [Visual Comparisons](#Visual-Comparisons) - [Failure of the LLN](#Failure-of-the-LLN) - [Classifying Tail Properties](#Classifying-Tail-Properties) - [Exercises](#Exercises) - [Solutions](#Solutions) In addition to what’s in Anaconda, this lecture will need the following libraries: ```python !pip install --upgrade quantecon --user !pip install --upgrade yfinance --user ``` Requirement already up-to-date: quantecon in c:\users\harve\appdata\roaming\python\python36\site-packages (0.4.6) Requirement already satisfied, skipping upgrade: numba>=0.38 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from quantecon) (0.38.0) Requirement already satisfied, skipping upgrade: numpy in c:\users\harve\appdata\roaming\python\python36\site-packages (from quantecon) (1.17.4) Requirement already satisfied, skipping upgrade: requests in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from quantecon) (2.22.0) Requirement already satisfied, skipping upgrade: scipy>=1.0.0 in c:\users\harve\appdata\roaming\python\python36\site-packages (from quantecon) (1.2.0) Requirement already satisfied, skipping upgrade: sympy in c:\users\harve\appdata\roaming\python\python36\site-packages (from quantecon) (1.3) Requirement already satisfied, skipping upgrade: llvmlite>=0.23.0dev0 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from numba>=0.38->quantecon) (0.23.2+1.g67a89c7) Requirement already satisfied, skipping upgrade: certifi>=2017.4.17 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests->quantecon) (2019.9.11) Requirement already satisfied, skipping upgrade: idna<2.9,>=2.5 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests->quantecon) (2.8) Requirement already satisfied, skipping upgrade: urllib3!=1.25.0,!=1.25.1,<1.26,>=1.21.1 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests->quantecon) (1.24.2) Requirement already satisfied, skipping upgrade: chardet<3.1.0,>=3.0.2 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests->quantecon) (3.0.4) Requirement already satisfied, skipping upgrade: mpmath>=0.19 in c:\users\harve\appdata\roaming\python\python36\site-packages (from sympy->quantecon) (1.1.0) Requirement already up-to-date: yfinance in c:\users\harve\appdata\roaming\python\python36\site-packages (0.1.54) Requirement already satisfied, skipping upgrade: numpy>=1.15 in c:\users\harve\appdata\roaming\python\python36\site-packages (from yfinance) (1.17.4) Requirement already satisfied, skipping upgrade: multitasking>=0.0.7 in c:\users\harve\appdata\roaming\python\python36\site-packages (from yfinance) (0.0.9) Requirement already satisfied, skipping upgrade: pandas>=0.24 in c:\users\harve\appdata\roaming\python\python36\site-packages (from yfinance) (0.25.3) Requirement already satisfied, skipping upgrade: requests>=2.20 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from yfinance) (2.22.0) Requirement already satisfied, skipping upgrade: pytz>=2017.2 in c:\users\harve\appdata\roaming\python\python36\site-packages (from pandas>=0.24->yfinance) (2018.7) Requirement already satisfied, skipping upgrade: python-dateutil>=2.6.1 in c:\users\harve\appdata\roaming\python\python36\site-packages (from pandas>=0.24->yfinance) (2.7.5) Requirement already satisfied, skipping upgrade: urllib3!=1.25.0,!=1.25.1,<1.26,>=1.21.1 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests>=2.20->yfinance) (1.24.2) Requirement already satisfied, skipping upgrade: chardet<3.1.0,>=3.0.2 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests>=2.20->yfinance) (3.0.4) Requirement already satisfied, skipping upgrade: certifi>=2017.4.17 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests>=2.20->yfinance) (2019.9.11) Requirement already satisfied, skipping upgrade: idna<2.9,>=2.5 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from requests>=2.20->yfinance) (2.8) Requirement already satisfied, skipping upgrade: six>=1.5 in c:\program files (x86)\microsoft visual studio\shared\anaconda3_64\lib\site-packages (from python-dateutil>=2.6.1->pandas>=0.24->yfinance) (1.13.0) ```python ``` ## Overview Most commonly used probability distributions in classical statistics and the natural sciences have either bounded support or light tails. When a distribution is light-tailed, extreme observations are rare and draws tend not to deviate too much from the mean. Having internalized these kinds of distributions, many researchers and practitioners use rules of thumb such as “outcomes more than four or five standard deviations from the mean can safely be ignored.” However, some distributions encountered in economics have far more probability mass in the tails than distributions like the normal distribution. With such **heavy-tailed** distributions, what would be regarded as extreme outcomes for someone accustomed to thin tailed distributions occur relatively frequently. Examples of heavy-tailed distributions observed in economic and financial settings include - the income distributions and the wealth distribution (see, e.g., [[Vil96]](zreferences.ipynb#pareto1896cours), [[BB18]](zreferences.ipynb#benhabib2018skewed)), - the firm size distribution ([[Axt01]](zreferences.ipynb#axtell2001zipf), [[Gab16]](zreferences.ipynb#gabaix2016power)}), - the distribution of returns on holding assets over short time horizons ([[Man63]](zreferences.ipynb#mandelbrot1963variation), [[Rac03]](zreferences.ipynb#rachev2003handbook)), and - the distribution of city sizes ([[RRGM11]](zreferences.ipynb#rozenfeld2011area), [[Gab16]](zreferences.ipynb#gabaix2016power)). These heavy tails turn out to be important for our understanding of economic outcomes. As one example, the heaviness of the tail in the wealth distribution is one natural measure of inequality. It matters for taxation and redistribution policies, as well as for flow-on effects for productivity growth, business cycles, and political economy - see, e.g., [[AR02]](zreferences.ipynb#acemoglu2002political), [[GSS03]](zreferences.ipynb#glaeser2003injustice), [[BEGS18]](zreferences.ipynb#bhandari2018inequality) or [[AKM+18]](zreferences.ipynb#ahn2018inequality). This lecture formalizes some of the concepts introduced above and reviews the key ideas. Let’s start with some imports: ```python import numpy as np import matplotlib.pyplot as plt %matplotlib inline ``` The following two lines can be added to avoid an annoying FutureWarning, and prevent a specific compatibility issue between pandas and matplotlib from causing problems down the line: ```python from pandas.plotting import register_matplotlib_converters register_matplotlib_converters() ``` ## Visual Comparisons One way to build intuition on the difference between light and heavy tails is to plot independent draws and compare them side-by-side. ### A Simulation The figure below shows a simulation. (You will be asked to replicate it in the exercises.) The top two subfigures each show 120 independent draws from the normal distribution, which is light-tailed. The bottom subfigure shows 120 independent draws from [the Cauchy distribution](https://en.wikipedia.org/wiki/Cauchy_distribution), which is heavy-tailed. <a id='light-heavy-fig1'></a> In the top subfigure, the standard deviation of the normal distribution is 2, and the draws are clustered around the mean. In the middle subfigure, the standard deviation is increased to 12 and, as expected, the amount of dispersion rises. The bottom subfigure, with the Cauchy draws, shows a different pattern: tight clustering around the mean for the great majority of observations, combined with a few sudden large deviations from the mean. This is typical of a heavy-tailed distribution. ### Heavy Tails in Asset Returns Next let’s look at some financial data. Our aim is to plot the daily change in the price of Amazon (AMZN) stock for the period from 1st January 2015 to 1st November 2019. This equates to daily returns if we set dividends aside. The code below produces the desired plot using Yahoo financial data via the `yfinance` library. ```python import yfinance as yf import pandas as pd s = yf.download('AMZN', '2015-1-1', '2019-11-1')['Adj Close'] r = s.pct_change() fig, ax = plt.subplots() ax.plot(r, linestyle='', marker='o', alpha=0.5, ms=4) ax.vlines(r.index, 0, r.values, lw=0.2) ax.set_ylabel('returns', fontsize=12) ax.set_xlabel('date', fontsize=12) plt.show() type(r) ``` Five of the 1217 observations are more than 5 standard deviations from the mean. Overall, the figure is suggestive of heavy tails, although not to the same degree as the Cauchy distribution the figure above. If, however, one takes tick-by-tick data rather daily data, the heavy-tailedness of the distribution increases further. ## Failure of the LLN One impact of heavy tails is that sample averages can be poor estimators of the underlying mean of the distribution. To understand this point better, recall [our earlier discussion](lln_clt.ipynb) of the Law of Large Numbers, which considered IID $ X_1, \ldots, X_n $ with common distribution $ F $ If $ \mathbb E |X_i| $ is finite, then the sample mean $ \bar X_n := \frac{1}{n} \sum_{i=1}^n X_i $ satisfies <a id='equation-lln-as2'></a> $$ \mathbb P \left\{ \bar X_n \to \mu \text{ as } n \to \infty \right\} = 1 \tag{1} $$ where $ \mu := \mathbb E X_i = \int x F(x) $ is the common mean of the sample. The condition $ \mathbb E | X_i | = \int |x| F(x) < \infty $ holds in most cases but can fail if the distribution $ F $ is very heavy tailed. For example, it fails for the Cauchy distribution Let’s have a look at the behavior of the sample mean in this case, and see whether or not the LLN is still valid. ```python from scipy.stats import cauchy np.random.seed(1234) N = 1_000 distribution = cauchy() fig, ax = plt.subplots() data = distribution.rvs(N) # Compute sample mean at each n sample_mean = np.empty(N) for n in range(1, N): sample_mean[n] = np.mean(data[:n]) # Plot ax.plot(range(N), sample_mean, alpha=0.6, label='$\\bar X_n$') ax.plot(range(N), np.zeros(N), 'k--', lw=0.5) ax.legend() plt.show() ``` The sequence shows no sign of converging. Will convergence occur if we take $ n $ even larger? The answer is no. To see this, recall that the [characteristic function](https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29) of the Cauchy distribution is <a id='equation-lln-cch'></a> $$ \phi(t) = \mathbb E e^{itX} = \int e^{i t x} f(x) dx = e^{-|t|} \tag{2} $$ Using independence, the characteristic function of the sample mean becomes $$ \begin{aligned} \mathbb E e^{i t \bar X_n } & = \mathbb E \exp \left\{ i \frac{t}{n} \sum_{j=1}^n X_j \right\} \\ & = \mathbb E \prod_{j=1}^n \exp \left\{ i \frac{t}{n} X_j \right\} \\ & = \prod_{j=1}^n \mathbb E \exp \left\{ i \frac{t}{n} X_j \right\} = [\phi(t/n)]^n \end{aligned} $$ In view of [(2)](#equation-lln-cch), this is just $ e^{-|t|} $. Thus, in the case of the Cauchy distribution, the sample mean itself has the very same Cauchy distribution, regardless of $ n $! In particular, the sequence $ \bar X_n $ does not converge to any point. <a id='cltail'></a> ## Classifying Tail Properties To keep our discussion precise, we need some definitions concerning tail properties. We will focus our attention on the right hand tails of nonnegative random variables and their distributions. The definitions for left hand tails are very similar and we omit them to simplify the exposition. ### Light and Heavy Tails A distribution $ F $ on $ \mathbb R_+ $ is called **heavy-tailed** if <a id='equation-defht'></a> $$ \int_0^\infty \exp(tx) F(x) = \infty \; \text{ for all } t > 0. \tag{3} $$ We say that a nonnegative random variable $ X $ is **heavy-tailed** if its distribution $ F(x) := \mathbb P\{X \leq x\} $ is heavy-tailed. This is equivalent to stating that its **moment generating function** $ m(t) := \mathbb E \exp(t X) $ is infinite for all $ t > 0 $. - For example, the lognormal distribution is heavy-tailed because its moment generating function is infinite everywhere on $ (0, \infty) $. A distribution $ F $ on $ \mathbb R_+ $ is called **light-tailed** if it is not heavy-tailed. A nonnegative random variable $ X $ is **light-tailed** if its distribution $ F $ is light-tailed. - Example: Every random variable with bounded support is light-tailed. (Why?) - Example: If $ X $ has the exponential distribution, with cdf $ F(x) = 1 - \exp(-\lambda x) $ for some $ \lambda > 0 $, then its moment generating function is finite whenever $ t < \lambda $. Hence $ X $ is light-tailed. One can show that if $ X $ is light-tailed, then all of its moments are finite. The contrapositive is that if some moment is infinite, then $ X $ is heavy-tailed. The latter condition is not necessary, however. - Example: the lognormal distribution is heavy-tailed but every moment is finite. ### Pareto Tails One specific class of heavy-tailed distributions has been found repeatedly in economic and social phenomena: the class of so-called power laws. Specifically, given $ \alpha > 0 $, a nonnegative random variable $ X $ is said to have a **Pareto tail** with **tail index** $ \alpha $ if <a id='equation-plrt'></a> $$ \lim_{x \to \infty} x^\alpha \, \mathbb P\{X > x\} = c. \tag{4} $$ Evidently [(4)](#equation-plrt) implies the existence of positive constants $ b $ and $ \bar x $ such that $ \mathbb P\{X > x\} \geq b x^{- \alpha} $ whenever $ x \geq \bar x $. The implication is that $ \mathbb P\{X > x\} $ converges to zero no faster than $ x^{-\alpha} $. In some sources, a random variable obeying [(4)](#equation-plrt) is said to have a **power law tail**. The primary example is the **Pareto distribution**, which has distribution <a id='equation-pareto'></a> $$ F(x) = \begin{cases} 1 - \left( \bar x/x \right)^{\alpha} & \text{ if } x \geq \bar x \\ 0 & \text{ if } x < \bar x \end{cases} \tag{5} $$ for some positive constants $ \bar x $ and $ \alpha $. It is easy to see that if $ X \sim F $, then $ \mathbb P\{X > x\} $ satisfies [(4)](#equation-plrt). Thus, in line with the terminology, Pareto distributed random variables have a Pareto tail. ### Rank-Size Plots One graphical technique for investigating Pareto tails and power laws is the so-called **rank-size plot**. This kind of figure plots log size against log rank of the population (i.e., location in the population when sorted from smallest to largest). Often just the largest 5 or 10% of observations are plotted. For a sufficiently large number of draws from a Pareto distribution, the plot generates a straight line. For distributions with thinner tails, the data points are concave. A discussion of why this occurs can be found in [[NOM04]](zreferences.ipynb#nishiyama2004estimation). The figure below provides one example, using simulated data. The rank-size plots shows draws from three different distributions: folded normal, chi-squared with 1 degree of freedom and Pareto. In each case, the largest 5% of 1,000 draws are shown. The Pareto sample produces a straight line, while the lines produced by the other samples are concave. <a id='rank-size-fig1'></a> ## Exercises ### Exercise 1 Replicate [the figure presented above](#light-heavy-fig1) that compares normal and Cauchy draws. Use `np.random.seed(11)` to set the seed. ```python from scipy.stats import cauchy, uniform from random import uniform import pandas as pd np.random.seed(11) N=100 var1=2 var2=12 distribution0 = np.random.randn(N) * var1 distribution1=np.random.randn(N) * var2 distribution2=cauchy() datas = np.array([ distribution0, distribution1, distribution2.rvs(N)]) for data in datas: fig, ax = plt.subplots() ax.plot(data, linestyle='', marker='o', alpha=1, ms=2) ax.vlines(list(range(N)), 0, data, lw=0.2) plt.show() ``` ### Exercise 2 Prove: If $ X $ has a Pareto tail with tail index $ \alpha $, then $ \mathbb E[X^r] = \infty $ for all $ r \geq \alpha $. ### Exercise 3 Repeat exercise 1, but replace the three distributions (two normal, one Cauchy) with three Pareto distributions using different choices of $ \alpha $. For $ \alpha $, try 1.15, 1.5 and 1.75. Use `np.random.seed(11)` to set the seed. ```python from scipy.stats import pareto import matplotlib.pyplot as plt np.random.seed(11) def plot_pareto(α,N): distribution=pareto(α) data =distribution.rvs(N) fig, ax = plt.subplots() ax.plot(data, linestyle='', marker='o', alpha=1, ms=2, label=f"α={α}") ax.legend() ax.vlines(list(range(N)), 0, data, lw=0.2) plt.show() N=100 αs=np.array([1.15,1.5,1.75]) for α in αs: plot_pareto(α,N) ``` ### Exercise 4 Replicate the rank-size plot figure [presented above](#rank-size-fig1). Use `np.random.seed(13)` to set the seed. ```python from scipy.stats import pareto, norm import matplotlib.pyplot as plt import numpy as np np.random.seed(13) def plot(data): #print(data) log_size=np.sort(np.log(data)) log_size_rank=np.log(log_size.argsort()[::-1]+1) fig, ax = plt.subplots() ax.scatter(y=log_size,x=log_size_rank, marker='o', alpha=0.5) ax.set_xlabel("log rank") ax.set_ylabel("log size") plt.show() α=1 N=1000 datas=[abs(norm().rvs(N)),np.exp(norm().rvs(N)),pareto(α).rvs(N)] for data in datas: plot(data) ``` ### Exercise 5 There is an ongoing argument about whether the firm size distribution should be modeled as a Pareto distribution or a lognormal distribution (see, e.g., [[FDGA+04]](zreferences.ipynb#fujiwara2004pareto), [[KLS18]](zreferences.ipynb#kondo2018us) or [[ST19]](zreferences.ipynb#schluter2019size)). This sounds esoteric but has real implications for a variety of economic phenomena. To illustrate this fact in a simple way, let us consider an economy with 100,000 firms, an interest rate of `r = 0.05` and a corporate tax rate of 15%. Your task is to estimate the present discounted value of projected corporate tax revenue over the next 10 years. Because we are forecasting, we need a model. We will suppose that 1. the number of firms and the firm size distribution (measured in profits) remain fixed and 1. the firm size distribution is either lognormal or Pareto. Present discounted value of tax revenue will be estimated by 1. generating 100,000 draws of firm profit from the firm size distribution, 1. multiplying by the tax rate, and 1. summing the results with discounting to obtain present value. The Pareto distribution is assumed to take the form [(5)](#equation-pareto) with $ \bar x = 1 $ and $ \alpha = 1.05 $. (The value the tail index $ \alpha $ is plausible given the data [[Gab16]](zreferences.ipynb#gabaix2016power).) To make the lognormal option as similar as possible to the Pareto option, choose its parameters such that the mean and median of both distributions are the same. Note that, for each distribution, your estimate of tax revenue will be random because it is based on a finite number of draws. To take this into account, generate 100 replications (evaluations of tax revenue) for each of the two distributions and compare the two samples by - producing a [violin plot](https://en.wikipedia.org/wiki/Violin_plot) visualizing the two samples side-by-side and - printing the mean and standard deviation of both samples. For the seed use `np.random.seed(1234)`. What differences do you observe? (Note: a better approach to this problem would be to model firm dynamics and try to track individual firms given the current distribution. We will discuss firm dynamics in later lectures.) ```python ```

5c0a12343a1d2353c65f8c6a0b927aad8f590457

ipynb

Jupyter Notebook

homework/HarveyT47/heavy_tails.ipynb

QuantEcon/summer_course_2019

8715ce171dbe371aac44bb7cda00c44aea8a8690

2019-11-01T06:33:00.000Z

2020-03-20T10:28:26.000Z

homework/HarveyT47/heavy_tails.ipynb

QuantEcon/summer_course_2019

8715ce171dbe371aac44bb7cda00c44aea8a8690

2019-12-14T07:26:59.000Z

2019-12-20T06:03:28.000Z

homework/HarveyT47/heavy_tails.ipynb

QuantEcon/summer_course_2019

8715ce171dbe371aac44bb7cda00c44aea8a8690

2019-12-14T07:08:04.000Z

2021-11-17T13:48:56.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Task Three: Quantum Gates and Circuits ```python from qiskit import * from qiskit.visualization import plot_bloch_multivector ``` ## Pauli Matrices \begin{align} I = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}, \quad X = \begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}, \quad Y = \begin{pmatrix} 0&i \\ -i&0 \end{pmatrix}, \quad Z = \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix} \quad \end{align} ## X-gate The X-gate is represented by the Pauli-X matrix: $$ X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = |0\rangle\langle1| + |1\rangle\langle0| $$ Effect a gate has on a qubit: $$ X|0\rangle = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = |1\rangle$$ ```python # Let's do an X-gate on a |0> qubit qc=QuantumCircuit(1) qc.x(0) qc.draw('mpl')#mpl stands for the matplotlib argument ``` ```python # Let's see the result backend = Aer.get_backend('statevector_simulator') out = execute(qc, backend).result().get_statevector() print(out) ``` [0.+0.j 1.+0.j] ## Z & Y-Gate $$ Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \quad\quad\quad\quad Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ $$ Y = -i|0\rangle\langle1| + i|1\rangle\langle0| \quad\quad Z = |0\rangle\langle0| - |1\rangle\langle1| $$ ```python # Do Y-gate on qubit 0 qc.y(0) # Do Z-gate on qubit 0 qc.z(0) qc.draw('mpl') ``` ## Hadamard Gate $$ H = \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ We can see that this performs the transformations below: $$ H|0\rangle = |+\rangle $$ $$ H|1\rangle = |-\rangle $$ ```python #create circuit with three qubit qc = QuantumCircuit(3) # Apply H-gate to each qubit: for qubit in range(3): qc.h(qubit) # See the circuit: qc.draw('mpl') ``` ## Identity Gate $$ I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $$ $$ I = XX $$ ```python qc.i(0) qc.draw('mpl') ``` **Other Gates: S-gate , T-gate, U-gate** # That's all for Task 3 ## Thank You! Where can you find me? LinkedIn : https://www.linkedin.com/in/arya--shah/ Twitter : https://twitter.com/aryashah2k Github : https://github.com/aryashah2k If you Like My Work, Follow me/ Connect with me on these platforms. Show some Love ❤️ Sponsor me on Github! ```python ```

f21e707097ed88f3b52fd3bce07b1d3b12d40d34

ipynb

Jupyter Notebook

Guided Project - Programming a Quantum Computer with Qiskit - IBM SDK/Task 3/Task 3 - Quantum Gates and Circuits.ipynb

bobsub218/exercise-qubit

32e1b851f65b98dcdf90ceaca1bd52ac6553e63a

2020-11-13T07:11:20.000Z

2022-03-06T02:27:45.000Z

Guided Project - Programming a Quantum Computer with Qiskit - IBM SDK/Task 3/Task 3 - Quantum Gates and Circuits.ipynb

bobsub218/Exercise-qubit

32e1b851f65b98dcdf90ceaca1bd52ac6553e63a

2020-12-25T17:25:14.000Z

2021-04-26T07:56:06.000Z

Guided Project - Programming a Quantum Computer with Qiskit - IBM SDK/Task 3/Task 3 - Quantum Gates and Circuits.ipynb

trial1user/Quantum-Computing-Collection-Of-Resources

6953854769adef05e705fa315017c459e5e581ce

2020-11-13T08:55:28.000Z

2022-03-14T21:16:07.000Z

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# Robust Data-Driven Portfolio Diversification ### Francisco A. Ibanez 1. RPCA on the sample 2. Singular Value Hard Thresholding (SVHT) 3. Truncated SVD 4. Maximize portfolio effective bets - regualization, s.t.: - Positivity constraint - Leverage 1x The combination of (1), (2), and (3) should limit the possible permutations of the J vector when doing the spectral risk parity. ## Methodology The goal of the overall methodology is to arrive to a portfolio weights vector which provides a well-balanced portfolio exposure to each one of the spectral risk factors present in an given investable universe. We start with the data set $X_{T \times N}$ which containst the historical excess returns for each one of the assets that span the investable universe of the portfolio. Before performing the eigendecomposition on $X$, we need to clean the set from noisy trading observations and outliers. We apply Robust Principal Components (RPCA) on $X$ to achieve this, which seeks to decompose $X$ into a structured low-rank matrix $R$ and a sparse matrix $C$ containing outliers and corrupt data: \begin{aligned} X=R_0+C_0 \end{aligned} The principal components of $R$ are robust to outliers and corrupt data in $C$. Mathematically, the goal is to find $R$ and $C$ that satisfy the following: \begin{aligned} \min_{R,C} ||R||_{*} + \lambda ||C||_{1} \\ \text{subject to} \\ R + C = X \end{aligned} ```python import pandas as pd import numpy as np from rpca import RobustPCA import matplotlib.pyplot as plt from scipy.linalg import svd from optht import optht raw = pd.read_pickle('etf_er.pkl') sample = raw.dropna() # Working with even panel for now # Outlier detection & cleaning X = (sample - sample.mean()).div(sample.std()).values # Normalization t, n = X.shape lmb = 4 / np.sqrt(max(t, n)) # Hyper-parameter rob = RobustPCA(lmb=lmb, max_iter=int(1E6)) R, C = rob.fit(X) # Robust, Corrupted # Low-rank representation (compression) through hard thresholding Truncated-SVD U, S, Vh = svd(R, full_matrices=False, compute_uv=True, lapack_driver='gesdd') S = np.diag(S) k = optht(X, sv=np.diag(S), sigma=None) V = Vh.T Vhat = V.copy() Vhat[:, k:] = 0 Shat = S.copy() Shat[k:, k:] = 0 cum_energy = np.cumsum(np.diag(S)) / np.sum(np.diag(S)) print(f'SVHT: {k}, {round(cum_energy[k] * 100, 2)}% of energy explained') ``` SVHT: 8, 58.43% of energy explained ```python D = np.diag(sample.std().values) t, n = X.shape w = np.array([1 / n] * n).reshape(-1, 1) eigen_wts = V.T @ D @ w p = np.divide(np.diag(eigen_wts.flatten()) @ S.T @ S @ eigen_wts, w.T @ D @ R.T @ R @ D @ w) p = p.flatten() eta_p = np.exp(-np.sum(np.multiply(p, np.log(p)))) eta_p ``` 1.4535327279694732 ```python def effective_bets(weights, singular_values_matrix, eigen_vector_matrix, volatilities, k=None): w = weights.reshape(-1, 1) eigen_wts = eigen_vector_matrix.T @ np.diag(volatilities) @ w p = (np.diag(eigen_wts.flatten()) @ singular_values_matrix.T @ singular_values_matrix @ eigen_wts).flatten() if k != None: p = p[:k] p_norm = np.divide(p, p.sum()) eta = np.exp(-np.sum(np.multiply(p_norm, np.log(p_norm)))) return eta effective_bets(np.array([1 / n] * n), S, V, sample.std().values) ``` 1.4535327279694745 ```python def objfunc(weights, singular_values_matrix, eigen_vector_matrix, volatilities, k=None): return -effective_bets(weights, singular_values_matrix, eigen_vector_matrix, volatilities, k) # Testing if minimizing p.T @ p yields the same results as maximizing the effective numbers of bets def objfunc2(weights, singular_values_matrix, eigen_vector_matrix, volatilities, k=None): w = weights.reshape(-1, 1) eigen_wts = eigen_vector_matrix.T @ np.diag(volatilities) @ w p = np.diag(eigen_wts.flatten()) @ singular_values_matrix.T @ singular_values_matrix @ eigen_wts if k != None: p = p[:k] n = p.shape[0] p_norm = np.divide(p, p.sum()) c = np.divide(np.ones((n, 1)), n) return ((p_norm - c).T @ (p_norm - c)).item() ``` ```python # POSITIVE ONLY from scipy.optimize import minimize cons = ( {'type': 'ineq', 'fun': lambda x: x}, {'type': 'ineq', 'fun': lambda x: np.sum(x) - 1} ) opti = minimize( fun=objfunc, x0=np.array([1 / n] * n), args=(S, V, sample.std().values), constraints=cons, method='SLSQP', tol=1E-12, options={'maxiter': 1E9} ) w_star = opti.x w_star /= w_star.sum() pd.Series(w_star, index=sample.columns).plot.bar() print(-opti.fun) ``` ```python # UNCONSTRAINED from scipy.optimize import minimize cons = ( {'type': 'ineq', 'fun': lambda x: x}, {'type': 'ineq', 'fun': lambda x: np.sum(x) - 1} ) opti = minimize( fun=objfunc, x0=np.array([1 / n] * n), args=(S, V, sample.std().values), # constraints=cons, method='SLSQP', tol=1E-12, options={'maxiter': 1E9} ) w_star = opti.x w_star /= w_star.sum() pd.Series(w_star, index=sample.columns).plot.bar() print(-opti.fun) ``` ```python eigen_wts = V.T @ np.diag(sample.std().values) @ w_star.reshape(-1, 1) p = (np.diag(eigen_wts.flatten()) @ S.T @ S @ eigen_wts).flatten() p = np.divide(p, p.sum()) pd.Series(p).plot.bar() ``` ```python # RC.T @ RC... different? cons = ( {'type': 'ineq', 'fun': lambda x: x}, {'type': 'ineq', 'fun': lambda x: np.sum(x) - 1} ) opti = minimize( fun=objfunc2, x0=np.array([1 / n] * n), args=(S, V, sample.std().values), constraints=cons, method='SLSQP', tol=1E-12, options={'maxiter': 1E9} ) w_star = opti.x w_star /= w_star.sum() #pd.Series(w_star, index=sample.columns).plot.bar() print(effective_bets(w_star, S, V, sample.std().values)) S, V, sample.std().values eigen_wts = V.T @ np.diag(sample.std().values) @ w_star.reshape(-1, 1) p = (np.diag(eigen_wts.flatten()) @ S.T @ S @ eigen_wts).flatten() p = np.divide(p, p.sum()) pd.Series(p).plot.bar() ``` ```python np.array([1, 2, 3]).shape ``` (3,) ```python ``` ```python D = np.diag(sample.std().values) n = sample.shape[0] Sigma = 1 / (n - 1) * D @ X.T @ X @ D Sigma_b = 1 / (n - 1) * D @ (R + C).T @ (R + C) @ D pd.DataFrame(R.T @ C) pd.DataFrame(R.T @ C) + pd.DataFrame(C.T @ R) pd.DataFrame(R.T @ R) + pd.DataFrame(C.T @ C) ``` \begin{aligned} X &= R + C \\ R &= USV^{T} \end{aligned} using the Singular Value Hard Thresholding (SVHT) obtained above we can approximate $R$: \begin{aligned} R &\approx \tilde{U}\tilde{S}\tilde{V}^{T} \end{aligned} Check the algebra so everything add up and the first matrix $X$ can be recovered from this point. \begin{aligned} \Sigma &= \frac{1}{(n - 1)}DX^{T}XD \\ \Sigma &= \frac{1}{(n - 1)}D(R + C)^{T}(R + C))D \end{aligned} then, portfolio risk will be given by: \begin{aligned} w^{T}\Sigma w &= \frac{1}{(n - 1)}w^{T}D(R + C)^{T}(R + C))D w \\ w^{T}\Sigma w &= \frac{1}{(n - 1)}w^{T}D(R^{T}R + R^{T}C + C^{T}R + C^{T}C ) D w \\ \end{aligned} \begin{aligned} w^{T}\Sigma w &= \frac{1}{(n - 1)} \lbrack w^{T}D(R^{T}R)Dw + w^{T} D(R^{T}C + C^{T}R + C^{T}C ) D w \rbrack \end{aligned} Taking the Singular Value Decomposition of R \begin{aligned} R &= USV^{T} \\ \end{aligned} we can express R in terms of its singular values and eigenvectors: \begin{aligned} w^{T}\Sigma w &= (n - 1)^{-1} \lbrack w^{T}D(VSU^{T}USV^{T})Dw + w^{T} D(R^{T}C + C^{T}R + C^{T}C) D w \rbrack \\ w^{T}\Sigma w &= (n - 1)^{-1} \lbrack w^{T}D(V S^{2} V^{T})Dw + w^{T} D(R^{T}C + C^{T}R + C^{T}C) D w \rbrack \end{aligned} where $S^{2}$ contains the eigenvalues of $R$ in its diagonal entries \begin{aligned} w^{T}\Sigma w &= (n - 1)^{-1} \lbrack \underbrace{w^{T}DV S^{2} V^{T}Dw}_\text{Robust Component} + \underbrace{w^{T} D(R^{T}C + C^{T}R + C^{T}C) D w}_\text{Noisy Component} \rbrack \end{aligned} TO DO: There has to be a way of reducing the noisy component, or at least interpret/explain it. The portfolio risk contribution is then given by \begin{aligned} diag(w)\Sigma w &= (n - 1)^{-1} \lbrack \underbrace{\theta}_\text{Robust Component} + \underbrace{\gamma}_\text{Noisy Component} \rbrack \\ \theta &= diag(V^{T}Dw)S^{2} V^{T}Dw \end{aligned} \begin{align} \eta (w) & \equiv \exp \left( -\sum^{N}_{n=1} p_{n} \ln{(p_{n})} \right) \end{align} Now we look for: \begin{align} \arg \max_{w} \eta(w) \end{align}

78043cf83883c1f7b5ef0baa28b812b0c49cbcd2

ipynb

Jupyter Notebook

notebooks/spectral_diversification.ipynb

fcoibanez/eigenportfolio

6e0f6c0239448a191aecf9137d545abf12cb344e

notebooks/spectral_diversification.ipynb

fcoibanez/eigenportfolio

6e0f6c0239448a191aecf9137d545abf12cb344e

notebooks/spectral_diversification.ipynb

fcoibanez/eigenportfolio

6e0f6c0239448a191aecf9137d545abf12cb344e

Qwen/Qwen-72B

1. YES 2. YES

__label__eng_Latn

# The Laplace Transform *This Jupyter notebook is part of a [collection of notebooks](../index.ipynb) in the bachelors module Signals and Systems, Communications Engineering, Universität Rostock. Please direct questions and suggestions to [[email protected]](mailto:[email protected]).* ## Theorems The theorems of the Laplace transformation relate basic time-domain operations to their equivalents in the Laplace domain. They are of use for the computation of Laplace transforms of signals composed from modified [standard signals](../continuous_signals/standard_signals.ipynb) and for the computation of the response of systems to an input signal. The theorems allow further to predict the consequences of modifying a signal or system by certain operations. ### Temporal Scaling Theorem A signal $x(t)$ is given for which the Laplace transform $X(s) = \mathcal{L} \{ x(t) \}$ exists. The Laplace transform of the [temporally scaled signal](../continuous_signals/operations.ipynb#Temporal-Scaling) $x(a t)$ with $a \in \mathbb{R} \setminus \{0\}$ reads \begin{equation} \mathcal{L} \{ x(a t) \} = \frac{1}{|a|} \cdot X \left( \frac{s}{a} \right) \end{equation} The Laplace transformation of a temporally scaled signal is given by weighting the inversely scaled Laplace transform of the unscaled signal with $\frac{1}{|a|}$. The scaling of the Laplace transform can be interpreted as a scaling of the complex $s$-plane. The region of convergence (ROC) of the temporally scaled signal $x(a t)$ is consequently the inversely scaled ROC of the unscaled signal $x(t)$ \begin{equation} \text{ROC} \{ x(a t) \} = \left\{ s: \frac{s}{a} \in \text{ROC} \{ x(t) \} \right\} \end{equation} Above relation is known as scaling theorem of the Laplace transform. The scaling theorem can be proven by introducing the scaled signal $x(a t)$ into the definition of the Laplace transformation \begin{equation} \mathcal{L} \{ x(a t) \} = \int_{-\infty}^{\infty} x(a t) \, e^{- s t} \; dt = \frac{1}{|a|} \int_{-\infty}^{\infty} x(t') \, e^{-\frac{s}{a} t'} \; dt' = \frac{1}{|a|} \cdot X \left( \frac{s}{a} \right) \end{equation} where the substitution $t' = a t$ was used. Note that a negative value of $a$ would result in a reversal of the integration limits. In this case a second reversal of the integration limits together with the sign of the integration element $d t'= a \, dt$ was consolidated into the absolute value of $a$. ### Convolution Theorem The convolution theorem states that the Laplace transform of the convolution of two signals $x(t)$ and $y(t)$ is equal to the scalar multiplication of their Laplace transforms $X(s)$ and $Y(s)$ \begin{equation} \mathcal{L} \{ x(t) * y(t) \} = X(s) \cdot Y(s) \end{equation} under the assumption that both Laplace transforms $X(s) = \mathcal{L} \{ x(t) \}$ and $Y(s) = \mathcal{L} \{ y(t) \}$ exist, respectively. The ROC of the convolution $x(t) * y(t)$ includes at least the intersection of the ROCs of $x(t)$ and $y(t)$ \begin{equation} \text{ROC} \{ x(t) * y(t) \} \supseteq \text{ROC} \{ x(t) \} \cap \text{ROC} \{ y(t) \} \end{equation} The theorem can be proven by introducing the [definition of the convolution](../systems_time_domain/convolution.ipynb) into the [definition of the Laplace transform](definition.ipynb) and changing the order of integration \begin{align} \mathcal{L} \{ x(t) * y(t) \} &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} x(\tau) \cdot y(t-\tau) \; d \tau \right) e^{-s t} \; dt \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} y(t-\tau) \, e^{-s t} \; dt \right) x(\tau) \; d\tau \\ &= Y(s) \cdot \int_{-\infty}^{\infty} x(\tau) \, e^{-s \tau} \; d \tau \\ &= Y(s) \cdot X(s) \end{align} The convolution theorem is very useful in the context of linear time-invariant (LTI) systems. The output signal $y(t)$ of an LTI system is given as the convolution of the input signal $x(t)$ with the impulse response $h(t)$. The signals can be represented either in the time or Laplace domain. This leads to the following equivalent representations of an LTI system in the time and Laplace domain, respectively Calculation of the system response by transforming the problem into the Laplace domain can be beneficial since this replaces the evaluation of the convolution integral by a scalar multiplication. In many cases this procedure simplifies the calculation of the system response significantly. A prominent example is the [analysis of a passive electrical network](network_analysis.ipynb). The convolution theorem can also be useful to derive an unknown Laplace transform. The key is here to express the signal as convolution of two other signals for which the Laplace transforms are known. This is illustrated by the following example. **Example** The Laplace transform of the convolution of a causal cosine signal $\epsilon(t) \cdot \cos(\omega_0 t)$ with a causal sine signal $\epsilon(t) \cdot \sin(\omega_0 t)$ is derived by the convolution theorem \begin{equation} \mathcal{L} \{ \epsilon(t) \cdot ( \cos(\omega_0 t) * \sin(\omega_0 t) \} = \frac{s}{s^2 + \omega_0^2} \cdot \frac{\omega_0}{s^2 + \omega_0^2} = \frac{\omega_0 s}{(s^2 + \omega_0^2)^2} \end{equation} where the [Laplace transforms of the causal cosine and sine signals](properties.ipynb#Transformation-of-the-cosine-and-sine-signal) were used. The ROC of the causal cosine and sine signal is $\Re \{ s \} > 0$. The ROC for their convolution is also $\Re \{ s \} > 0$, since no poles and zeros cancel out. Above Laplace transform has one zero $s_{00} = 0$, and two poles of second degree $s_{\infty 0} = s_{\infty 1} = j \omega_0$ and $s_{\infty 2} = s_{\infty 3} = - j \omega_0$. This example is evaluated numerically in the following. First the convolution of the causal cosine and sine signal is computed ```python %matplotlib inline import sympy as sym sym.init_printing() t, tau = sym.symbols('t tau', real=True) s = sym.symbols('s', complex=True) w0 = sym.symbols('omega0', positive=True) x = sym.integrate(sym.cos(w0*tau) * sym.sin(w0*(t-tau)), (tau, 0, t)) x = x.doit() x ``` For the sake of illustration let's plot the signal for $\omega_0 = 1$ ```python sym.plot(x.subs(w0, 1), (t, 0, 50), xlabel=r'$t$', ylabel=r'$x(t)$'); ``` The Laplace transform is computed ```python X, a, cond = sym.laplace_transform(x, t, s) X, a ``` which exists for $\Re \{ s \} > 0$. Its zeros are given as ```python sym.roots(sym.numer(X), s) ``` and its poles as ```python sym.roots(sym.denom(X), s) ``` ### Temporal Shift Theorem The [temporal shift of a signal](../continuous_signals/operations.ipynb#Temporal-Shift) $x(t - \tau)$ for $\tau \in \mathbb{R}$ can be expressed by the convolution of the signal $x(t)$ with a shifted Dirac impulse \begin{equation} x(t - \tau) = x(t) * \delta(t - \tau) \end{equation} This follows from the sifting property of the Dirac impulse. Applying a two-sided Laplace transform to the left- and right-hand side and exploiting the convolution theorem yields \begin{equation} \mathcal{L} \{ x(t - \tau) \} = X(s) \cdot e^{- s \tau} \end{equation} where $X(s) = \mathcal{L} \{ x(t) \}$ is assumed to exist. Note that $\mathcal{L} \{ \delta(t - \tau) \} = e^{- s \tau}$ can be derived from the definition of the two-sided Laplace transform together with the sifting property of the Dirac impulse. The Laplace transform of a shifted signal is given by multiplying the Laplace transform of the original signal with $e^{- s \tau}$. The ROC does not change \begin{equation} \text{ROC} \{ x(t-\tau) \} = \text{ROC} \{ x(t) \} \end{equation} This result is known as shift theorem of the Laplace transform. For a causal signal $x(t)$ and $\tau > 0$ the shift theorem of the one-sided Laplace transform is equal to the shift theorem of the two-sided transform. #### Transformation of the rectangular signal The Laplace transform of the [rectangular signal](../continuous_signals/standard_signals.ipynb#Rectangular-Signal) $x(t) = \text{rect}(t)$ is derived by expressing it by the Heaviside signal \begin{equation} \text{rect}(t) = \epsilon \left(t + \frac{1}{2} \right) - \epsilon \left(t - \frac{1}{2} \right) \end{equation} Applying the shift theorem to the [transform of the Heaviside signal](definition.ipynb#Transformation-of-the-Heaviside-Signal) and the linearity of the Laplace transform yields \begin{equation} \mathcal{L} \{ \text{rect}(t) \} = \frac{1}{s} e^{s \frac{1}{2}} - \frac{1}{s} e^{- s \frac{1}{2}} = \frac{\sinh \left( \frac{s}{2} \right) }{\frac{s}{2}} \end{equation} where $\sinh(\cdot)$ denotes the [hyperbolic sine function](https://en.wikipedia.org/wiki/Hyperbolic_function#Sinh). The ROC of the Heaviside signal is given as $\Re \{ s \} > 0$. Applying [l'Hopitals rule](https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule) the pole at $s=0$ can be disregarded leading to \begin{equation} \text{ROC} \{ \text{rect}(t) \} = \mathbb{C} \end{equation} For illustration, the magnitude of the Laplace transform $|X(s)|$ is plotted in the $s$-plane, as well as $X(\sigma)$ and $X(j \omega)$ for the real and imaginary part of the complex frequency $s = \sigma + j \omega$. ```python sigma, omega = sym.symbols('sigma omega') X = sym.sinh(s/2)*2/s sym.plotting.plot3d(abs(X.subs(s, sigma+sym.I*omega)), (sigma, -5, 5), (omega, -20, 20), xlabel=r'$\Re\{s\}$', ylabel=r'$\Im\{s\}$', title=r'$|X(s)|$') sym.plot(X.subs(s, sigma) , (sigma, -5, 5), xlabel=r'$\Re\{s\}$', ylabel=r'$X(s)$', ylim=(0, 3)) sym.plot(X.subs(s, sym.I*omega) , (omega, -20, 20), xlabel=r'$\Im\{s\}$', ylabel=r'$X(s)$'); ``` **Exercise** * Derive the Laplace transform $X(s) = \mathcal{L} \{ x(t) \}$ of the causal rectangular signal $x(t) = \text{rect} (a t - \frac{1}{2 a})$ * Derive the Laplace transform of the [triangular signal](../fourier_transform/theorems.ipynb#Transformation-of-the-triangular-signal) $x(t) = \Lambda(a t)$ with $a \in \mathbb{R} \setminus \{0\}$ ### Differentiation Theorem Derivatives of signals are the fundamental operations of differential equations. Ordinary differential equations (ODEs) with constant coefficients play an important role in the theory of time-invariant (LTI) systems. Consequently, the representation of the derivative of a signal in the Laplace domain is of special interest. #### Two-sided transform A differentiable signal $x(t)$ whose temporal derivative $\frac{d x(t)}{dt}$ exists is given. Using the [derivation property of the Dirac impulse](../continuous_signals/standard_signals.ipynb#Dirac-Impulse), the derivative of the signal can be expressed by the convolution \begin{equation} \frac{d x(t)}{dt} = \frac{d \delta(t)}{dt} * x(t) \end{equation} Applying a two-sided Laplace transformation to the left- and right-hand side together with the [convolution theorem](#Convolution-Theorem) yields the Laplace transform of the derivative of $x(t)$ \begin{equation} \mathcal{L} \left\{ \frac{d x(t)}{dt} \right\} = s \cdot X(s) \end{equation} where $X(s) = \mathcal{L} \{ x(t) \}$. The two-sided Laplace transform $\mathcal{L} \{ \frac{d \delta(t)}{dt} \} = s$ can be derived by applying the definition of the Laplace transform together with the derivation property of the Dirac impulse. The ROC is given as a superset of the ROC for $x(t)$ \begin{equation} \text{ROC} \left\{ \frac{d x(t)}{dt} \right\} \supseteq \text{ROC} \{ x(t) \} \end{equation} due to the zero at $s=0$ which may cancel out a pole. Above result is known as differentiation theorem of the two-sided Laplace transform. It states that the differentiation of a signal in the time domain is equivalent to a multiplication of its spectrum by $s$. #### One-sided transform Many practical signals and systems are causal, hence $x(t) = 0$ for $t < 0$. A causal signal is potentially discontinuous for $t=0$. The direct application of above result for the two-sided Laplace transform is not possible since it assumes that the signal is differentiable for every time $t$. The potential discontinuity at $t=0$ has to be considered explicitly for the derivation of the differentiation theorem for the one-sided Laplace transform [[Girod et al.](index.ipynb#Literature)] \begin{equation} \mathcal{L} \left\{ \frac{d x(t)}{dt} \right\} = s \cdot X(s) - x(0+) \end{equation} where $x(0+) := \lim_{\epsilon \to 0} x(0+\epsilon)$ denotes the right sided limit value of $x(t)$ for $t=0$. The ROC is given as a superset of the ROC of $x(t)$ \begin{equation} \text{ROC} \left\{ \frac{d x(t)}{dt} \right\} \supseteq \text{ROC} \{ x(t) \} \end{equation} due to the zero at $s=0$ which may cancel out a pole. The one-sided Laplace transform of a causal signal is equal to its two-sided transform. Above result holds therefore also for the two-sided transform of a causal signal. The main application of the differentiation theorem is the transformation and solution of differential equations under consideration of initial values. Another application area is the derivation of transforms of signals which can be expressed as derivatives of other signals. ### Integration Theorem An integrable signal $x(t)$ for which the integral $\int_{-\infty}^{t} x(\tau) \; d\tau$ exists is given. The integration can be represented as convolution with the rectangular signal $\epsilon(t)$ \begin{equation} \int_{-\infty}^{t} x(\tau) \; d\tau = \int_{-\infty}^{\infty} x(\tau) \cdot \epsilon(t - \tau) \; d\tau = \epsilon(t) * x(t) \end{equation} as illustrated below Two-sided Laplace transformation of the left- and right-hand side of above equation, application of the convolution theorem and using the Laplace transform of the Heaviside signal $\epsilon(t)$ yields \begin{equation} \mathcal{L} \left\{ \int_{-\infty}^{t} x(\tau) \; d\tau \right\} = \frac{1}{s} \cdot X(s) \end{equation} The ROC is given as a superset of the intersection of the ROC of $x(t)$ and the right $s$-half-plane \begin{equation} \text{ROC} \left\{ \int_{-\infty}^{t} x(\tau) \; d\tau \right\} \supseteq \text{ROC} \{ x(t) \} \cap \{s : \Re \{ s \} > 0\} \end{equation} due to the pole at $s=0$. This integration theorem holds also for the one-sided Laplace transform. #### Transformation of the ramp signal The Laplace transform of the causal [ramp signal](https://en.wikipedia.org/wiki/Ramp_function) $t \cdot \epsilon(t)$ is derived by the integration theorem. The ramp signal can be expressed as integration over the Heaviside signal \begin{equation} t \cdot \epsilon(t) = \int_{-\infty}^{t} \tau \cdot \epsilon(\tau) \; d \tau \end{equation} Laplace transformation of the left- and right-hand side and application of the integration theorem together with the Laplace transform of the Heaviside signal yields \begin{equation} \mathcal{L} \{ t \cdot \epsilon(t) \} = \frac{1}{s^2} \end{equation} with \begin{equation} \text{ROC} \{ t \cdot \epsilon(t) \} = \{s : \Re \{ s \} > 0\} \end{equation} **Exercise** * Derive the Laplace transform $X(s) = \mathcal{L} \{ x(t) \}$ of the signal $x(t) = t^n \cdot \epsilon(t)$ with $n \geq 0$ by repeated application of the integration theorem. * Compare your result to the numerical result below. Note that $\Gamma(n+1) = n!$ for $n \in \mathbb{N}$. ```python n = sym.symbols('n', integer=True) X, a, cond = sym.laplace_transform(t**n, t, s) X, a, cond ``` ### Modulation Theorem The complex modulation of a signal $x(t)$ is defined as $e^{s_0 t} \cdot x(t)$ with $s_0 \in \mathbb{C}$. The Laplace transform of a modulated signal is derived by introducing it into the definition of the two-sided Laplace transform \begin{align} \mathcal{L} \left\{ e^{s_0 t} \cdot x(t) \right\} &= \int_{-\infty}^{\infty} e^{s_0 t} x(t) \, e^{-s t} \; dt = \int_{-\infty}^{\infty} x(t) \, e^{- (s - s_0) t} \; dt \\ &= X(s-s_0) \end{align} where $X(s) = \mathcal{L} \{ x(t) \}$. Modulation of the signal $x(t)$ leads to a translation of the $s$-plane into the direction given by the complex value $s_0$. Consequently, the ROC is also shifted \begin{equation} \text{ROC} \{ e^{s_0 t} \cdot x(t) \} = \{s: s - \Re \{ s_0 \} \in \text{ROC} \{ x(t) \} \} \end{equation} This relation is known as modulation theorem. **Example** The Laplace transform of the signal $t^n \cdot \epsilon(t)$ \begin{equation} \mathcal{L} \{ t^n \cdot \epsilon(t) \} = \frac{n!}{s^{n+1}} \end{equation} for $\Re \{ s \} > 0$ was derived in the previous example. This result can be extended to the class of signals $t^n e^{-s_0 t} \epsilon(t)$ with $s_0 \in \mathbb{C}$ using the modulation theorem \begin{equation} \mathcal{L} \{ t^n e^{-s_0 t} \epsilon(t) \} = \frac{n!}{(s + s_0)^{n+1}} \qquad \text{for } \Re \{ s \} > \Re \{ - s_0 \}. \end{equation} **Copyright** The notebooks are provided as [Open Educational Resource](https://de.wikipedia.org/wiki/Open_Educational_Resources). Feel free to use the notebooks for your own educational purposes. The text is licensed under [Creative Commons Attribution 4.0](https://creativecommons.org/licenses/by/4.0/), the code of the IPython examples under the [MIT license](https://opensource.org/licenses/MIT). Please attribute the work as follows: *Lecture Notes on Signals and Systems* by Sascha Spors.

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Jupyter Notebook

laplace_transform/theorems.ipynb

swchao/signalsAndSystemsLecture

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2019-01-27T12:39:27.000Z

2022-03-15T10:26:12.000Z

laplace_transform/theorems.ipynb

swchao/signalsAndSystemsLecture

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laplace_transform/theorems.ipynb

swchao/signalsAndSystemsLecture

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2020-09-18T06:26:48.000Z

2021-12-10T06:11:45.000Z

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# Calculating Times of Rise, Set, and Culmination Suppose we want to calculate when a given celestial object rises above the horizon, sets below the horizon, or reaches the highest point above the horizon (*culminates*), as seen by an observer at a given location on the surface of the Earth. ### Azimuth and altitude Let us consider how this observer can describe the apparent location of the celestial object, say a star, at a specific moment in time using a pair of angles. Let the *azimuth* $A$ be the angle measured westward from due south along the horizon needed to face the star. Let the *altitude* $h$ be the angle of the star above the horizon. (If the star is below the horizon, $h < 0$.) We define a pair of co-rotating Cartesian coordinate systems for such an observer, called *horizontal* coordinates and *local equatorial* coordinates. In both systems we treat the entire sky as a sphere of unit radius centered on the observer. ### Horizontal coordinates In the horizontal coordinate system we have three directions $x$, $y$, and $z$ such that - $x$ points due south toward the horizon, - $y$ points due west toward the horizon, and - $z$ points straight up toward the zenith. Thus it is possible to relate the angles $A$ and $h$ with horizontal Cartesian coordinates $x$, $y$, and $z$ by \begin{align} x & = \cos{h} \cos{A} \\ y & = \cos{h} \sin{A} \tag{1} \\ z & = \sin{h} \end{align} To understand these equations a little more intuitively, consider a star that is exactly on the horizon and due west. Then $h=0$, which means $\cos{h}=1$ and $\sin{h}=0$. Also, $A=90^{\circ}$, so $\cos{A}=0$ and $\sin{A}=1$. This results in $x=0$, $y=1$, $z=0$, which is consistent with the above descriptions of the three coordinates. Another example is where $h=90^{\circ}$ and $A$ has any value, resulting in the vertical vector $x=0$, $y=0$, $z=1$. ### Declination and hour angle Usually in astronomy when you see the phrase "equatorial coordinates," it refers to right ascension $\alpha$ and declination $\delta$. We need those two quantities for this discussion, but they are not the same as local equatorial coordinates. Here I will refer to right ascension and declination as "celestial equatorial coordinates" to avoid ambiguity. Like their celestial counterparts, local equatorial coordinates are defined with respect to the plane of the Earth's equator. But being local means they rotate along with the observer, while celestial equatorial coordinates are fixed to the starry background of the celestial sphere. In other words, an earthbound observer experiences stationary local equatorial coordinates, but sees stars rotating around him. Meanwhile, an observer far out in space sees fixed celestial equatorial coordinates, but sees a rotating Earth. *Declination* $\delta$ is the angle of a star north ($\delta>0$) or south ($\delta<0$) of the celestial equator. For example, the north celestial pole has $\delta=90^{\circ}$. I will describe right ascension later. *Hour angle* $\tau$ is the angle from the local meridian westward toward a given star. It is measured around a celestial circle having same declination as the star. It represents the amount of time that has passed since the star last crossed the meridian, or equivalently, since the star culminated (reached its highest point above the horizon). Hour angle can be measured in degrees (0 to 360) or in sidereal hours (0 to 24). You can divide degrees by 15 to get sidereal hours. ### Local equatorial coordinates Specifically, local equatorial coordinates are Cartesian coordinates $\hat{x}$, $\hat{y}$, $\hat{z}$ such that - $\hat{x}$ points to where the celestial equator intersects the meridian, - $\hat{y}$ points due west toward the horizon, and - $\hat{z}$ points to the north celestial pole. Using $\delta$ as the star's declination and $\tau$ as the star's hour angle at a given time, we can write the local equatorial coordinates of the star as \begin{align} \hat{x} & = \cos{\delta} \cos{\tau} \\ \hat{y} & = \cos{\delta} \sin{\tau} \tag{2} \\ \hat{z} & = \sin{\delta} \nonumber \end{align} Note that $\tau$ changes with the Earth's rotation but $\delta$ does not. Therefore, for a fixed star, $\hat{x}$ and $\hat{y}$ change with time but $\hat{z}$ remains constant. ### Relating the two local systems We can write a system of equations that relates these two local rotating coordinate systems. First note that the $y$-axis is the same as the $\hat{y}$-axis; they are both aimed at the point on the horizon due west from the observer. The $\hat{z}$-axis, pointing toward the north celestial pole, has an angle $\phi$ above the due-north point on the horizon, where $\phi$ is the geographic latitude of the observer. (If the observer is in the southern hemisphere, then $\phi$ is negative and the north celestial pole is below the horizon.) That northernmost point on the horizon is in the $-x$ direction, because $x$ points due south. Similarly, the $\hat{x}$-axis is rotated by $\phi$ from the vertical $z$-axis. In general, $x$ and $z$ each depend on both $\hat{x}$ and $\hat{z}$ through a rotation of $\phi$. We can express horizontal coordinates in terms of local equatorial coordinates as \begin{align} x & = \hat{x} \sin{\phi} - \hat{z} \cos{\phi} \\ y & = \hat{y} \tag{3} \\ z & = \hat{x} \cos{\phi} + \hat{z} \sin{\phi} \end{align} ### Relating (azimuth, altitude) to (hour angle, declination) Substituting equations (1) and (2) into (3), we can write a new system of equations that relate azimuth, altitude, hour angle, and declination: \begin{align} \cos{h} \cos{A} &= \cos{\delta} \cos{\tau} \sin{\phi} - \sin{\delta} \cos{\phi} \\ \cos{h} \sin{A} &= \cos{\delta} \sin{\tau} \tag{4} \\ \sin{h} &= \cos{\delta} \cos{\tau} \cos{\phi} + \sin{\delta} \sin{\phi} \end{align} The third equation in (4) is most helpful for finding rise and set times because it eliminates the azimuth as a factor. ### Calculating hour angle for a given star at a given time To calculate hour angle $\tau$ we need to use \begin{equation} \tau = \theta + \frac{\lambda}{15} - \alpha \tag{5} \end{equation} where - $\theta$ is the current Greenwich Apparent Sidereal Time (GAST) at the moment of observation, - $\lambda$ is the geographic longitude of the observer east of Greenwich in degrees, and - $\alpha$ is the right ascension of the star. Note that $\tau$, $\theta$, and $\alpha$ are all expressed in sidereal hours. We divide $\lambda$ by 15 to convert degrees to hours. ### Determining when the object is highest or lowest in the sky The object reaches maximum altitude angle $h_1$ when its hour angle $\tau=0$ and its minimum value of $h_2$ when $\tau = 12$ hours. Let's look at the general case where we are trying to find the *event time* when the object reaches an arbitrary hour angle $\tau$: \begin{align} \theta + \frac{\lambda}{15} - \alpha & = \tau \\ \theta = \tau + \alpha - \frac{\lambda}{15} \tag{6} \end{align} A star's right ascension $\alpha$ is almost constant over time, and we assume the observer's longitude $\lambda$ is fixed. In practice, $\alpha$ for the Sun, the Moon, or a planet is not constant. The Moon especially will move in right ascension significantly over the course of a few hours. So we will end up needing to numerically iterate to find the exact event time. The approach I use in the Astronomy library is to determine the GAST at a start search time. This start search time represents a time after which the next event is to be found. This initial guess $\theta_{x}$ of the GAST value will be incorrect by the approximate amount \begin{align} \Delta \theta = \left( \tau + \alpha - \frac{\lambda}{15} \right) - \theta_x \tag{7} \end{align} On the first iteration, if $\Delta\theta$ is negative, it is corrected by adding 24 sidereal hours to guarantee finding a culmination time after the start search time. On subsequent iterations, the correction may be positive or negative to home in on the correct time. At this point, it is important to note that sidereal hours are not exactly the same as clock hours. The Earth rotates with respect to a fixed star once every 23 hours, 56 minutes, and 4.091 seconds, or 86164.091 seconds. This is compared to a mean solar day of 24 hours = 86400 seconds. This gives a ratio of $\rho = 0.99726957$ mean solar days per sidereal day. After each iteration, we correct by adding $\rho \Delta \theta$ days of terrestrial time and calculating the right ascension $\alpha$ and declination $\delta$ of the object, and GAST, again at the new time. We plug the updated values of $\theta_x$ and $\alpha$ back into (7) to obtain a more accurate estimate. We keep iterating until $\Delta \theta$ is tolerably small (less than 0.1 seconds). This typically takes 4 or 5 iterations. ### Calculating rise and set times Now that we can determine when an object reaches its highest and lowest altitudes in the sky for a given observer, we know that *if* the object rises or sets, it will be bounded by those two events. At locations close to one of the Earth's poles, an object may stay in the sky for weeks at a time, never setting, or it may stay below the horizon for weeks at a time, never rising. Specifically, if the object is at its lowest at time $t_L$ and at its highest at $t_H \approx t_L + 12$ hours, then we can calculate the object's altitudes $h_L$ and $h_H$ at those respective times. If $h_L \lt 0$ and $h_H \gt 0$, we know the object must have risen at some time $t_R$ such that $t_L \lt t_R \lt t_H$.

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ipynb

Jupyter Notebook

theory/rise_set_culm.ipynb

matheo/astronomy

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theory/rise_set_culm.ipynb

matheo/astronomy

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