I have implemented a radix sort algorithm in Python 3. It first finds the maximum number of digits in the list, then changes them to strings and adds 0's. For example, [7, 23, 107, 1, 53] to ['007', '023', '107', '001', '053']. I then make a new matrix, [[] * 10]. I append the numbers with last digit of 0 in lst[0], numbers with last digit of 1 in lst[1], etc. I then flatten the matrix into a list, and recurse with the lst, and with the position one less than the one before (in this case, the second to last digit). Can I get advice on how to improve it? Here is my code:

def radix_sort(lst):
    '''list -> list'''

    #finds maximum number of digits in list
    num_of_digits = max([len(str(x)) for x in lst])

    #Adds 0s to the front of the number if necessary and changes it to a string
    for x in range(len(lst)):
        lst[x] = '0' * (num_of_digits - len(str(lst[x]))) + str(lst[x])

    #helper function to sort based on the position
    def helper(lst, pos):
        '''list, int -> list'''
        #places numbers with 0 in the position in the ans[0], 1 in the position in ans[1], etc.
        ans = [[] for _ in range(10)]
        #adding numbers to the position in ans
        for x in lst:
        #flattened list, reusing lst to save memory
        lst = []
        for x in ans:
            for y in x:
        #we have sorted the whole list
        if pos == 0:
            return lst
        #recurse again with smaller position
        return helper(lst, pos - 1)

    #changing the strings to integers and returning
    return [int(x) for x in helper(lst, num_of_digits - 1)]
        #flattened list, reusing lst to save memory
        lst = []

That doesn't save memory. The caller of the current helper call still has a reference to the "old" list, so it'll remain in memory in addition to this new one. You'll have one list per call level. To actually save memory, you could do lst.clear() instead, so that all levels use the same single list.

To further save memory, do del ans, x, y after you rebuilt lst from them before you call helper recursively, otherwise they'll remain in memory as well, again once per call level.

For example for lst = ['1' * 800] * 1000, these improvements changed the peak memory usage from over 15 MB to less than 0.5 MB in this test of mine:

import tracemalloc

lst = ['1' * 800] * 1000
peak = tracemalloc.get_traced_memory()[1]

Or just do it iteratively instead of recursively, then you likely wouldn't have these issues in the first place and you wouldn't have to worry about recursion limit errors due to large numbers.

Here's an iterative one with a few other changes:

def radix_sorted(lst):
    '''list -> list'''
    lst = list(map(str, lst))
    width = max(map(len, lst))
    lst = [s.rjust(width, '0') for s in lst]
    for i in range(width)[::-1]:
        buckets = [[] for _ in range(10)]
        for s in lst:
        lst = [s for b in buckets for s in b]
    return list(map(int, lst))
  • \$\begingroup\$ Oh, thanks! I thought it was saving memory, I guess I was completely wrong. This will help my memory usage a lot. By the way, is there really a chance that I'll get a recursion limit error? The recursion depth is 1,000 which means it'll only pass recursion depth for 1000+ digit numbers, am I wrong? \$\endgroup\$ – Sola Sky Mar 23 at 23:04
  • \$\begingroup\$ @SolaSky Yes, roughly at 1000 digits you'd hit the limit. Whether there's a chance that you have such numbers, I don't know, only you do :-). Btw I added an iterative one now. \$\endgroup\$ – Kelly Bundy Mar 24 at 0:04

It isn't necessary to convert the numbers to strings and zero-pad them.

BASE = 10

def radix_sort(lst, base=BASE):
    biggest_number = max(lst)
    place = 1

    while place <= biggest_number:
        buckets = [[] for _ in range(BASE)]
        for number in lst:
            index = number // place % BASE

        lst = []
        for bucket in buckets

        place *= BASE

    return lst
  • \$\begingroup\$ You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why your solution is better than the original) so that the author and other readers can learn from your thought process. \$\endgroup\$ – Peilonrayz Mar 31 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.