How to implement a Quick Sort in Python

Question

I have tried to do it but I can't figure it out. I have this Quick Sort that is searching at index 1 but how to I change the string at that index to ints?

def quick_sort(list):
    if not list:
        return list
    pivot = list[0]
    lesser = quick_sort([x for x in list[1:] if x[1] < pivot[1]], 1)
    greater = quick_sort([x for x in list[1:] if x[1] >= pivot[1]], 1)
    return lesser + [pivot] + greater

Any help out there?

Is it a list of strings? Or a list of integers?

Simeon Visser
– Simeon Visser

2014-01-15 23:37:08 +00:00
Commented Jan 15, 2014 at 23:37 — Simeon Visser
– Simeon Visser, Commented Jan 15, 2014 at 23:37

jonrsharpe · Accepted Answer · 2014-01-15 23:43:40Z

2

Here try this out.

def quick_sort(list):
    if not list: 
        return list 
    pivot = list[0] 
    lesser = quick_sort([x for x in list[1:] if float(x[1]) < float(pivot[1])], 1) 
    greater = quick_sort([x for x in list[1:] if float(x[1]) >= float(pivot[1])], 1) 
    return lesser + [pivot] + greater

edited Jan 15, 2014 at 23:43

jonrsharpe

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answered Jan 15, 2014 at 23:37

Pythonist

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1 Comment

Pythonist Over a year ago

Thanks for the edit @jonrsharpe, I forgot that hanging return statment!

Simeon Visser · Accepted Answer · 2014-01-15 23:39:16Z

0

You don't need the index variable:

def quick_sort(list):
    if not list:
        return list
    pivot = list[0]
    lesser = quick_sort([x for x in list[1:] if x < pivot])
    greater = quick_sort([x for x in list[1:] if x >= pivot])
    return lesser + [pivot] + greater

answered Jan 15, 2014 at 23:39

Simeon Visser

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