sort() returns None | 易学教程

问题

Code:

import math
import time
import random
class SortClass(object):
    def sort1(self, l):
        if len(l)==1:
            return l
        elif len(l)==2:
            if l[0]<l[1]:
                return l
            else:
                return l[::-1]
        else:
            pivot=math.floor(len(l)/2)
            a=l[pivot:]
            b=l[:pivot]
            a2=self.sort1(a)
            b2=self.sort1(b)
            if a2==None or b2==None:
                a2=[]
                b2=[]
            return (a2+b2).sort()
        return []
Sort=SortClass()
x=[20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1]
print(Sort.sort1(x))

The code outputs None even though it should return an empty list in two cases:

return []

and

a2=self.mergeSort(a)
b2=self.mergeSort(b)
if a2==None or b2==None:
    a2=[]
    b2=[]
return (a2+b2).sort()

Details: The code is for a list sorting module I am making for python practice (I am relatively new at python). sort1 is a modified mergesort.

回答1:

@reut got to it first but

return sorted(a2+b2)

not

return (a2+b2).sort()

Additionally

if a2 == None or b2 == None:
    a2 = []
    b2 = []

Should be

if a2 == None:
    a2 = []
if b2 == None:
    b2 = []

Your setting both to [] if either is none meaning if a2 is [1] and b2 is none your throwing away a2. I'm guessing this is unintended.

Also in your code you have an uppercase S in the lower sortClass

in addition return[] will never return, the above else does not allow it to.

回答2:

list.sort returns None (sorts in-place):

>>> [].sort() is None
True

You're using it here:

return (a2+b2).sort()

Use sorted to sort to a new list and not in-place:

return sorted(a2+b2)

来源：https://stackoverflow.com/questions/34710129/sort-returns-none

标签

python

sorting

python-3.x

recursion

return