Mergesort with Python

I couldn't find any working Python 3.3 mergesort algorithm codes, so I made one myself. Is there any way to speed it up? It sorts 20,000 numbers in about 0.3-0.5 seconds

def msort(x):
    result = []
    if len(x) < 2:
        return x
    mid = int(len(x)/2)
    y = msort(x[:mid])
    z = msort(x[mid:])
    while (len(y) > 0) or (len(z) > 0):
        if len(y) > 0 and len(z) > 0:
            if y[0] > z[0]:
                result.append(z[0])
                z.pop(0)
            else:
                result.append(y[0])
                y.pop(0)
        elif len(z) > 0:
            for i in z:
                result.append(i)
                z.pop(0)
        else:
            for i in y:
                result.append(i)
                y.pop(0)
    return result

Solution 1:

The first improvement would be to simplify the three cases in the main loop: Rather than iterating while some of the sequence has elements, iterate while both sequences have elements. When leaving the loop, one of them will be empty, we don't know which, but we don't care: We append them at the end of the result.

def msort2(x):
    if len(x) < 2:
        return x
    result = []          # moved!
    mid = int(len(x) / 2)
    y = msort2(x[:mid])
    z = msort2(x[mid:])
    while (len(y) > 0) and (len(z) > 0):
        if y[0] > z[0]:
            result.append(z[0])
            z.pop(0)
        else:
            result.append(y[0])
            y.pop(0)
    result += y
    result += z
    return result

The second optimization is to avoid popping the elements. Rather, have two indices:

def msort3(x):
    if len(x) < 2:
        return x
    result = []
    mid = int(len(x) / 2)
    y = msort3(x[:mid])
    z = msort3(x[mid:])
    i = 0
    j = 0
    while i < len(y) and j < len(z):
        if y[i] > z[j]:
            result.append(z[j])
            j += 1
        else:
            result.append(y[i])
            i += 1
    result += y[i:]
    result += z[j:]
    return result

A final improvement consists in using a non recursive algorithm to sort short sequences. In this case I use the built-in sorted function and use it when the size of the input is less than 20:

def msort4(x):
    if len(x) < 20:
        return sorted(x)
    result = []
    mid = int(len(x) / 2)
    y = msort4(x[:mid])
    z = msort4(x[mid:])
    i = 0
    j = 0
    while i < len(y) and j < len(z):
        if y[i] > z[j]:
            result.append(z[j])
            j += 1
        else:
            result.append(y[i])
            i += 1
    result += y[i:]
    result += z[j:]
    return result

My measurements to sort a random list of 100000 integers are 2.46 seconds for the original version, 2.33 for msort2, 0.60 for msort3 and 0.40 for msort4. For reference, sorting all the list with sorted takes 0.03 seconds.

Solution 2:

Code from MIT course. (with generic cooperator )

import operator


def merge(left, right, compare):
    result = []
    i, j = 0, 0
    while i < len(left) and j < len(right):
        if compare(left[i], right[j]):
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            j += 1
    while i < len(left):
        result.append(left[i])
        i += 1
    while j < len(right):
        result.append(right[j])
        j += 1
    return result


def mergeSort(L, compare=operator.lt):
    if len(L) < 2:
        return L[:]
    else:
        middle = int(len(L) / 2)
        left = mergeSort(L[:middle], compare)
        right = mergeSort(L[middle:], compare)
        return merge(left, right, compare)

Solution 3:

def merge_sort(x):

    if len(x) < 2:return x

    result,mid = [],int(len(x)/2)

    y = merge_sort(x[:mid])
    z = merge_sort(x[mid:])

    while (len(y) > 0) and (len(z) > 0):
            if y[0] > z[0]:result.append(z.pop(0))   
            else:result.append(y.pop(0))

    result.extend(y+z)
    return result