Merge sort implementation in python 3 I have implemented a Merge sort in Python 3, and it works well. If anything needs to be improved, I would appreciate the criticism.

def merge_sort(nums):

    if len(nums) == 1:                                  
        return

    middle_index = len(nums) // 2

    left_half = nums[:middle_index]
    right_half = nums[middle_index:]

    merge_sort(left_half)
    merge_sort(right_half)

    i = 0
    j = 0
    k = 0

    while i&lt;len(left_half) and j&lt;len(right_half):
        if left_half[i] &lt; right_half[j]:
            nums[k] = left_half[i]
            i = i + 1
        else:
            nums[k] = right_half[j]
            j = j + 1

        k = k + 1

    while i&lt;len(left_half):
        nums[k] = left_half[i]
        k = k + 1
        i = i + 1       

if __name__ == "__main__":

   nums = [-3,-2,-1,1,2,1,0,-1,-2,-3]
   merge_sort(nums)
   print(nums)
Everything I'm going to talk about is in the merge_sort function

General

i = 0
j = 0
k = 0


Can be defined as

i = j = k = 0



You should always leave spaces between operators as per PEP 8 rules

i&lt;len(left_half) should be i &lt; len(left_half)



Use x += y instead of x = x + y



In my opinion, I think using short and concise names such as mid or middle instead of middle_index would be better. If you don't wish to, you can leave it as it!



Use type hints



Add docstrings



Bug

Your function only takes into account the left_half of the array, and ignores what's left in the right_half

For example, if nums array was [3, 9, 0], The array would be [0, 3, 0]

This would happen as

merge_sort([3]) which won't change the left_half
merge_sort([9, 0]) which would make the right_half as [0, 9]

Then, 

left_half = [3]
right_half = [0, 9]

nums = [3, 9, 0]

i = 0
j = 0
k = 0

First, the else statement would be called as 3 &gt; 0.

i = 0
j = 1
k = 1

nums = [0, 9, 0]

Next, the if statement would be called as 3 &lt; 9

i = 1
j = 1
k = 2

nums = [0, 3, 0]

Now, the while loop will terminate as i = len(left_side)

Then, while i &lt; len(left_side) would immediately terminate as i = len(left_side)



Did you notice? right_side still has one element 9 waiting to be traversed, but it never will be.

To fix that, add the following to the end of the function

while j &lt; len(right_half):
    nums[k] = right_half[j]
    j += 1
    k += 1




Improvement

Now, instead of using a while loop at all, you can just use a[k:] = left_half[i:] + right_half[j:] to replace both the loops! This is true because one half must be empty and the other half must have the length of n - k.



Performance

If you are using this function in real time with an array of a really large size, this won't work efficiently.

len takes quite a bit of time. To make it even faster, use a parameter length which would be the length of the array



The final implementation of the function:

from typing import List, Any

def merge_sort(nums: List[Any], length: int) -&gt; None:
    """ Uses Merge Sort to sort an array """

    # Base case
    if length == 1:
        return

    mid = length // 2

    left, right = mid, length - mid

    left_half, right_half = nums[:mid], nums[mid:]

    merge_sort(left_half, left)
    merge_sort(right_half, right)

    i = j = k = 0

    while i &lt; left and j &lt; right:
        if left_half[i] &lt; right_half[j]:
            nums[k] = left_half[i]
            i += 1
        else:
            nums[k] = right_half[j]
            j += 1

        k += 1

    nums[k:] = left_half[i:] + right_half[j:]


Note: Any in typing means any datatype is allowed. The function can sort any datatype that is comparable with another element of the same datatype.