CODEWITHSUNDEEP
pythonpython-3.xquicksort
babrar
babrar

Reputation: 199

How to make the following quicksort more 'pythonic'

I just implemented Quicksort in python-3 and the following is a excerpt for the swapping mechanism involved in each sub array.

Now, I can't help but notice that it looks an awful lot like a C-based loop. Which makes sense since I'm coming over from a long time of C++ development. I was wondering if this loop can be made more pythonic.

I tried using for items in array but that's not helpful since the iterator should not increase for every round of the loop.

Any suggestions are appreciated! 

#  last elem is pivot at start
pivPos = end
#  iterate the list and bring pivot to correct position
count = 0

while (count != pivPos):

    #  swap places with elem just before pivot
    if arr[count] >= arr[pivPos]:
        temp = arr[count]
        arr[count] = arr[pivPos - 1]
        arr[pivPos - 1] = temp

        #  now swap the pivot
        arr[pivPos - 1] = arr[pivPos]
        arr[pivPos] = temp

        #  update pivPos
        pivPos -= 1

    else:
        count += 1
        continue

Upvotes: 1

Views: 72

Answers (1)

nosklo
nosklo

Reputation: 223122

This part:

temp = arr[count]
arr[count] = arr[pivPos - 1]
arr[pivPos - 1] = temp
arr[pivPos - 1] = arr[pivPos]
arr[pivPos] = temp

can be written without temp like this:

arr[count], arr[pivPos-1], arr[pivPos] = arr[pivPos-1], arr[pivPos], arr[count]

Also continue is redundant, you can remove it and the code will behave the same.

However the real pythonic way is:

arr.sort()

full edited code easy reference:

pivPos = end
count = 0
while count != pivPos:
    if arr[count] >= arr[pivPos]:
        arr[count], arr[pivPos-1], arr[pivPos] = arr[pivPos-1], arr[pivPos], arr[count]
        pivPos -= 1
    else:
        count += 1

Upvotes: 3

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