I got an idea to make a Roman/Arabic number converter, and I would like to hear your thoughts, where I could improve it, if there are Python specific things that I missed, so on and so forth.# Decodes roman numeralsdef get_arabic_numbers(t):    # Hold the numbers in the list    l = list()    # The format    format = 0    # Convert from roman numeral to arabic number    for i in range(0, len(t)):        if t[i] == 'I':            l.append(int(1))        elif t[i] == 'V':            l.append(int(5))        elif t[i] == 'X':            l.append(int(10))        elif t[i] == 'L':            l.append(int(50))        elif t[i] == 'C':            l.append(int(100))        elif t[i] == 'D':            l.append(int(500))        elif t[i] == 'M':            l.append(int(1000))    # Calculate the format as follows:    i = 0    while i < len(l) - 1:        # If the next numeral is greater than the current one, add the next one minus the current one = format = format + next_numeral - current_numeral        if l[i] < l[i+1]:            format = format + l[i+1] - l[i];            i = i + 2        # Else add it normally        else:            format = format + l[i]            i = i + 1    # Fix in case the last two numerals are equal    if i == len(l):        if l[i-2] == l[i-1]:            format = format + l[i-1]    else:        format = format + l[i]    # Return the format    return format# Encode an arabic digit to a roman numeraldef encode(l, c, one, five, nine):    # The format to be returned    format =     # Special case 5.1: 5 has its own numeral 'V'    # Special case 5.2: 50 has its own numeral 'A'    # Special case 5.3: 500 has its own numeral 'D'    if l[c] == 5:        format = format + five    elif l[c] < 5:        for i in range(0, l[c]):            format = format + one    elif l[c] > 5:    # Special case 9.1: 9 depends on 10, so it's 'IX'    # Special case 9.2: 90 depends on 1000, so it's 'XC'    # Special case 9.3: 900 depends on 1000, so it's 'CM'        if l[c] == 9:            format = format + nine        else:            format = format + five            for i in range(6, l[c] + 1):                format = format + one    return format# Encodes arabic numbersdef get_roman_numerals(t):    # The arabic number to be converted    t = int(t)    # The list contains the number's digits    l = list()    # int is not iterable, so I wrote this to get the digits    while t >= 1:        l.append(int(t % 10))        t = t / 10    # They are in the wrong order, so the list has to be reversed    l.reverse()    # The format    format =     # Current position in the number    c = 0    # If the number is at least 1000    if len(l) >= 4:        num = 0    # All the digits after 1000    i = int(len(l) - 1)    while i > 3:        num = num + 1        i = i - 1       # A list containing these digits    li = list()    for i in range(0, num + 1):        li.append(l[i])    # Converts the list to an int    n = map(str, li)    n = ''.join(n)    n = int(n)        # Add an 'M' for every 1000    for i in range(0, n):        format = format + 'M'    c = num + 1    # If the number is also least 100    if len(l) >= 3:        format = format + encode(l, c, 'C', 'D', CM)        c = c + 1    # If the number is at least 10    if len(l) >= 3:        format = format + encode(l, c, 'X', 'L', XC)        c = c + 1    # If the number is at least 100 but smaller than 1000    if len(l) > 2:        format = format + encode(l, c, 'I', 'V', IX)    # If the number is at least 10 but smaller than 100    if len(l) == 2:        format = format + encode(l, c, 'X', 'L', XC)        c = c + 1        format = format + encode(l, c, 'I', 'V', IX)    # If the number is at least 1 but smaller than 10       if len(l) == 1:        format = format + encode(l, c, 'I', 'V', IX)    return format
Converting from Roman numerals to Arabic numbers
python;beginner;roman numerals
ImpressionsThat's a very long solution.  The comments were generally helpful, but the code itself could have been easier to understand if it were shorter and more expressive.The variable names were generally not helpful: l, i, t, c, li were all cryptic.get_arabic_numbers(t) is not really named appropriately:get implies retrieving something that already exists.  This is more of a calculation.Why is numbers plural?Why is the parameter named t  what does it stand for?Strictly speaking, the result is an int.  It's not Arabic, nor is it even in base ten.  It's just an abstract integer, which is conventionally rendered for you as base-ten Arabic digits when str() is called on it (whether explicitly or implicitly).  (And by Arabic, we mean 0123456789, not .)Based on these concerns, I'd call it decode_roman_numeral(roman).  It happens to be what you wrote as the comment!  Similarly, I'd call the inverse function encode_roman_numeral(num).Arabic  Roman# The formatformat = 0That's not what I would call a format  I would call it a result.# Hold the numbers in the listl = list()# Convert from roman numeral to arabic numberfor i in range(0, len(t)):    if t[i] == 'I':        l.append(int(1))    elif t[i] == 'V':        l.append(int(5))    elif t[i] == 'X':        l.append(int(10))    elif t[i] == 'L':        l.append(int(50))    elif t[i] == 'C':        l.append(int(100))    elif t[i] == 'D':        l.append(int(500))    elif t[i] == 'M':        l.append(int(1000))As I mentioned before, l is a cryptic name.  There is no need to convert 1 to int(1).  Lookups are typically done in Python using a dictionary.  Also, whenever you populate a list by creating an empty list and appending to it repeatedly, that is a good candidate for a list comprehension:trans = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}values = [trans[r] for r in roman]# Calculate the format as follows:i = 0while i < len(l) - 1:    # If the next numeral is greater than the current one, add the next one minus the current one = format = format + next_numeral - current_numeral    if l[i] < l[i+1]:        format = format + l[i+1] - l[i];        i = i + 2    # Else add it normally    else:        format = format + l[i]        i = i + 1Do you really need a case that advances i by two?  How about this instead:result = 0for i in range(0, len(l) - 1):    if l[i] < l[i+1]:        result -= l[i]    else:        result += l[i]As I mentioned before, the code could be more expressive.  This loop is summing values in a list, so we should use the sum() built-in function to say what we mean.result = sum(    val if val >= next_val else -val    for val, next_val in zip(values[:-1], values[1:]))All together, then, I'd condense your function down to this:def decode_roman_numeral(roman):    Calculate the numeric value of a Roman numeral (in capital letters)    trans = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}    values = [trans[r] for r in roman]    return sum(        val if val >= next_val else -val        for val, next_val in zip(values[:-1], values[1:])    ) + values[-1]Arabic  RomanThe encode() function encodes a single Arabic digit, so encode_digit() would be a better name.  There is no point in passing l and c separately, since you only ever care about l[c].  Instead of looping to append characters to format, use the * operator to construct a repeated string.I would write this helper function using a single expression with four cases:def encode_digit(digit, one, five, nine):    return (        nine                     if digit == 9 else        five + one * (digit - 5) if digit >= 5 else        one + five               if digit == 4 else        one * digit                  )The get_roman_numerals() function is rather convoluted.  I had a hard time following the code, so I went by your comments instead.  In the end, I figured out that it was essentially doing this:def encode_roman_numeral(num):    num = int(num)    return (        'M' * (num // 1000) +        encode_digit((num // 100) % 10, 'C', 'D', 'CM') +        encode_digit((num //  10) % 10, 'X', 'L', 'XC') +        encode_digit( num         % 10, 'I', 'V', 'IX')     )