Python - 提取最内部的列表 [英] Python - Extracting inner most lists

问题描述

刚开始玩 Python，所以请耐心等待 :)

假设以下列表包含嵌套列表:

[[[[[[1, 3, 4, 5]], [1, 3, 8]], [[1, 7, 8]]], [[[6, 7, 8]]], [9]]

在不同的表示中:

<预><代码>[[[[[1, 3, 4, 5]],[1, 3, 8]],[[1, 7, 8]]],[[[6, 7, 8]]],[9]]

您将如何提取这些内部列表，以便返回具有以下形式的结果:

[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]

非常感谢！

编辑(感谢@falsetru):

空的内部列表或混合类型列表永远不会成为输入的一部分.

解决方案

这似乎可行，假设没有像 [1,2,[3]] 这样的混合"列表:

def get_inner(嵌套):if all(type(x) == list for x innested):对于嵌套中的 x:对于 get_inner(x) 中的 y:产量 y别的:产量嵌套

list(get_inner(nested_list)) 的输出:

[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]

或者更短，没有生成器，使用 sum组合结果列表:

def get_inner(嵌套):if all(type(x) == list for x innested):返回总和(地图(get_inner，嵌套)，[])返回 [嵌套]

Just started toying around with Python so please bear with me :)

Assume the following list which contains nested lists:

[[[[[1, 3, 4, 5]], [1, 3, 8]], [[1, 7, 8]]], [[[6, 7, 8]]], [9]]

In a different representation:

[
    [
        [
            [
                [1, 3, 4, 5]
            ], 
            [1, 3, 8]
        ], 
        [
            [1, 7, 8]
        ]
    ], 
    [
        [
            [6, 7, 8]
        ]
    ], 
    [9]
]

How would you go about extracting those inner lists so that a result with the following form would be returned:

[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]

Many thanks!

EDIT (Thanks @falsetru):

Empty inner-list or mixed type lists will never be part of the input.

解决方案

This seems to work, assuming no 'mixed' lists like [1,2,[3]]:

def get_inner(nested):
    if all(type(x) == list for x in nested):
        for x in nested:
            for y in get_inner(x):
                yield y
    else:
        yield nested

Output of list(get_inner(nested_list)):

[[1, 3, 4, 5], [1, 3, 8], [1, 7, 8], [6, 7, 8], [9]]

Or even shorter, without generators, using sum to combine the resulting lists:

def get_inner(nested):
    if all(type(x) == list for x in nested):
        return sum(map(get_inner, nested), [])
    return [nested]

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