为什么'.sort()'在Python中导致列表为'None'? [英] Why does '.sort()' cause the list to be 'None' in Python?

问题描述

我正在尝试对int的Python列表进行排序，然后使用.pop()函数返回最高的列表.我尝试过用不同的方式编写方法:

I am attempting to sort a Python list of ints and then use the .pop() function to return the highest one. I have tried a writing the method in different ways:

def LongestPath(T):    
    paths = [Ancestors(T,x) for x in OrdLeaves(T)]
    #^ Creating a lists of lists of ints, this part works
    result =[len(y) for y in paths ]
    #^ Creating a list of ints where each int is a length of the a list in paths
    result = result.sort()
    #^meant to sort the result
    return result.pop()
    #^meant to return the largest int in the list (the last one)

我也尝试过

def LongestPath(T):
    return[len(y) for y in [Ancestors(T,x) for x in OrdLeaves(T)] ].sort().pop()

在两种情况下，.sort()都会导致列表为None(不具有.pop()函数并返回错误).当我删除.sort()时，它可以正常工作，但由于列表未排序，因此不会返回最大的int.

In both cases .sort() causes the list to be None (which has no .pop() function and returns an error). When I remove the .sort() it works fine but does not return the largest int since the list is not sorted.

推荐答案

只需从中删除作业

result = result.sort()

只留下

result.sort()

sort方法就地工作(它修改了现有列表)，因此不需要赋值，并且它返回None.将结果分配给列表名称时，就是在分配None.

The sort method works in-place (it modifies the existing list), so no assignment is necessary, and it returns None. When you assign its result to the name of the list, you're assigning None.

可以轻松地(并且更有效地)将其编写为单行代码:

It can easily (and more efficiently) be written as a one-liner:

max(len(Ancestors(T,x)) for x in OrdLeaves(T))

max在线性时间O(n)中运行，而排序为O(nlogn).您也不需要嵌套列表推导，只需一个生成器表达式就可以.

max operates in linear time, O(n), while sorting is O(nlogn). You also don't need nested list comprehensions, a single generator expression will do.

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