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这不是最舒适的数据结构。我建议重构代码并选择3倍嵌套列表之外的其他内容来保存数据,但如果当前不可能,我建议使用以下方法:

import itertools


def flatten(iterable):
    yield from itertools.chain.from_iterable(iterable)


result = []
for elements in zip(*map(flatten, samples)):
    result.append(list(flatten(elements)))

举例来说,它给出了:

[['A', 'E', '1'], 
 ['B', 'F', '2'], 
 ['C', 'G', '3'], 
 ['D', 'H', '4']]

测试2个以上的样品:

samples = [[[['A', 'E'], ['B', 'F']],
            [['C', 'G'], ['D', 'H']]],
           [[['1'], ['2']], 
            [['3'], ['4']]], 
           [[['5'], ['6']],
            [['7'], ['8']]]]

给予:

[['A', 'E', '1', '5'],
 ['B', 'F', '2', '6'],
 ['C', 'G', '3', '7'],
 ['D', 'H', '4', '8']]

说明:

这个 flatten generator function 只需展平嵌套iterable的1个级别。它基于 itertools.chain.from_iterable 功能。在 map(flatten, samples) 我们将此函数应用于 samples :

>>> map(flatten, samples)
<map at 0x3c6685fef0>  # <-- map object returned, to see result wrap it in `list`:

>>> list(map(flatten, samples))
[<generator object flatten at 0x0000003C67A2F9A8>,  # <-- will flatten the 1st sample
 <generator object flatten at 0x0000003C67A2FA98>,  # <-- ... the 2nd
 <generator object flatten at 0x0000003C67A2FB10>]  # <-- ... the 3rd and so on if there are more

# We can see what each generator will give by applying `list` on each one of them
>>> list(map(list, map(flatten, samples)))
[[['A', 'E'], ['B', 'F'], ['C', 'G'], ['D', 'H']],
 [['1'], ['2'], ['3'], ['4']],
 [['5'], ['6'], ['7'], ['8']]]

接下来,我们可以使用 zip 在平坦的样本上迭代。请注意,我们不能将其应用于 map 直接对象:

>>> list(zip(map(flatten, samples)))
[(<generator object flatten at 0x0000003C66944138>,),
 (<generator object flatten at 0x0000003C669441B0>,),
 (<generator object flatten at 0x0000003C66944228>,)]

我们应该 unpack 首先:

>>> list(zip(*map(flatten, samples)))
[(['A', 'E'], ['1'], ['5']),
 (['B', 'F'], ['2'], ['6']),
 (['C', 'G'], ['3'], ['7']),
 (['D', 'H'], ['4'], ['8'])]

# or in a for loop:
>>> for elements in zip(*map(flatten, samples)):
...     print(elements)
(['A', 'E'], ['1'], ['5'])
(['B', 'F'], ['2'], ['6'])
(['C', 'G'], ['3'], ['7'])
(['D', 'H'], ['4'], ['8'])

最后,我们只需要加入每个 elements 把元组放在一起。我们可以用同样的 扁平化 功能:

>>> for elements in zip(*map(flatten, samples)):
...     print(list(flatten(elements)))
['A', 'E', '1', '5']
['B', 'F', '2', '6']
['C', 'G', '3', '7']
['D', 'H', '4', '8']

您只需将其全部放回一个列表中,如第一个代码示例所示。