sort()返回None [英] sort() returns None

问题描述

代码:

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))

即使在两种情况下它应该返回一个空列表，代码也会输出None:

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

return []

和

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

详细信息: 该代码用于我为python练习制作的列表排序模块(我在python上相对较新). sort1是修改后的合并排序.

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.

推荐答案

@reut首先但是

return sorted(a2+b2)

不是

return (a2+b2).sort()

另外

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

应该是

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

您将两者都设置为[]都不表示a2为[1]且b2都不为您扔掉a2.我猜这是意料之外的.

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.

在代码中，在较低的sortClass中也有一个大写的S

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

另外 返回[] 永远不会返回，上述其他情况不允许它返回.

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

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