Q:

Why is numpy.array so slow?

I am baffled by this
def main():
    for i in xrange(2560000):
        a = [0.0, 0.0, 0.0]

main()

$ time python test.py

real     0m0.793s

Let's now see with numpy:
import numpy

def main():
    for i in xrange(2560000):
        a = numpy.array([0.0, 0.0, 0.0])

main()

$ time python test.py

real    0m39.338s

Holy CPU cycles batman! 
Using numpy.zeros(3) improves, but still not enough IMHO
$ time python test.py

real    0m5.610s
user    0m5.449s
sys 0m0.070s

numpy.version.version = '1.5.1'
If you are wondering if the list creation is skipped for optimization in the first example, it is not:
  5          19 LOAD_CONST               2 (0.0)
             22 LOAD_CONST               2 (0.0)
             25 LOAD_CONST               2 (0.0)
             28 BUILD_LIST               3
             31 STORE_FAST               1 (a)

A:

Numpy is optimised for large amounts of data. Give it a tiny 3 length array and, unsurprisingly, it performs poorly.
Consider a separate test
import timeit

reps = 100

pythonTest = timeit.Timer('a = [0.] * 1000000')
numpyTest = timeit.Timer('a = numpy.zeros(1000000)', setup='import numpy')
uninitialised = timeit.Timer('a = numpy.empty(1000000)', setup='import numpy')
# empty simply allocates the memory. Thus the initial contents of the array 
# is random noise

print 'python list:', pythonTest.timeit(reps), 'seconds'
print 'numpy array:', numpyTest.timeit(reps), 'seconds'
print 'uninitialised array:', uninitialised.timeit(reps), 'seconds'

And the output is
python list: 1.22042918205 seconds
numpy array: 1.05412316322 seconds
uninitialised array: 0.0016028881073 seconds

It would seem that it is the zeroing of the array that is taking all the time for numpy. So unless you need the array to be initialised then try using empty.

A:

Holy CPU cycles batman!, indeed.  
But please rather consider something very fundamental related to numpy; sophisticated linear algebra based functionality (like random numbers or singular value decomposition). Now, consider these seamingly simple calculations:
In []: A= rand(2560000, 3)
In []: %timeit rand(2560000, 3)
1 loops, best of 3: 296 ms per loop
In []: %timeit u, s, v= svd(A, full_matrices= False)
1 loops, best of 3: 571 ms per loop

and please trust me that this kind of performance will not be beaten significantly by any package currently available.
So, please describe your real problem, and I'll try to figure out decent numpy based solution for it.
Update:
Here is some simply code for ray sphere intersection:
import numpy as np

def mag(X):
    # magnitude
    return (X** 2).sum(0)** .5

def closest(R, c):
    # closest point on ray to center and its distance
    P= np.dot(c.T, R)* R
    return P, mag(P- c)

def intersect(R, P, h, r):
    # intersection of rays and sphere
    return P- (h* (2* r- h))** .5* R

# set up
c, r= np.array([10, 10, 10])[:, None], 2. # center, radius
n= 5e5
R= np.random.rand(3, n) # some random rays in first octant
R= R/ mag(R) # normalized to unit length

# find rays which will intersect sphere
P, b= closest(R, c)
wi= b<= r

# and for those which will, find the intersection
X= intersect(R[:, wi], P[:, wi], r- b[wi], r)

Apparently we calculated correctly:
In []: allclose(mag(X- c), r)
Out[]: True

And some timings:
In []: % timeit P, b= closest(R, c)
10 loops, best of 3: 93.4 ms per loop
In []: n/ 0.0934
Out[]: 5353319 #=> more than 5 million detection's of possible intersections/ s
In []: %timeit X= intersect(R[:, wi], P[:, wi], r- b[wi])
10 loops, best of 3: 32.7 ms per loop
In []: X.shape[1]/ 0.0327
Out[]: 874037 #=> almost 1 million actual intersections/ s

These timings are done with very modest machine. With modern machine, a significant speed-up can be still expected.
Anyway, this is only a short demonstration how to code with numpy.

