函数内部函数，python递归列表

Question

def minimizeMaximumPair(lst):
    lst.sort()

    def compute(aList):
        if len(aList) != 0:
            return [(aList[0], lst[len(aList) - 1])].extend(compute(aList[1:len(aList) - 1]))
        return []

    return compute(lst)

When I get the the last recursion step I get an

TypeError: 'NoneType' object is not iterable

I tried returning nothing, and a []

Answer 1

The issue is when you call .extend() .

Your compute function tries to return the value of .extend() , which is None . Similar to .sort() , .extend() modifies the object itself, rather than returning a modified copy of the object. This is known as mutability. Here's some working code:

def compute(aList):
    if len(aList) != 0:
        out = [(aList[0], lst[len(aList) - 1])]
        out.extend(compute(aList[1:len(aList) - 1]))
        return out
    return []

Answer 2

Instead of list.extend which returns None , you can use list.__iadd__

__iadd__ also extends the list inplace, but returns the list afterward

if you have an aversion to using special methods, you can use iadd from the operator module

from operator import iadd

...

def compute(aList):
    if len(aList) != 0:
        return iadd([(aList[0], aList[-1])], compute(aList[1: -1]))
    return []

Answer 3

That's because extend is a function that doesn't return anything, it simply changes the list in-place. A possible solution is to use + instead:

return [(aList[0], lst[len(aList) - 1])] + compute(aList[1:len(aList) - 1])