Python拼合列表（但並非始終如此）

Question

我有一個清單清單清單...

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]

以下功能輸出[1, 3, 3, 5, 4, 4, 5, 3]

def flatten(a):
    b = []
    for c in a:
        if isinstance(c, list):
            b.extend(flatten(c))
        else:
            b.append(c)
    return b

但是，我想在最后一級停止展平，以便獲得[ [1,3], [3,5], [4,4], [5,3] ]

Answer 1

您可以在展平前測試包含的列表：

def flatten(a):
    b = []
    for c in a:
        if isinstance(c, list) and any(isinstance(i, list) for i in c):
            b.extend(flatten(c))
        else:
            b.append(c)
    return b

演示：

>>> def flatten(a):
...     b = []
...     for c in a:
...         if isinstance(c, list) and any(isinstance(i, list) for i in c):
...             b.extend(flatten(c))
...         else:
...             b.append(c)
...     return b
... 
>>> A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]
>>> flatten(A)
[[1, 3], [3, 5], [4, 4], [5, 3]]

這樣會盡可能地提高效率。 any()僅需要測試，直到找到列表為止，而不是所有元素。

Answer 2

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ]

print [child[0] if isinstance(child[0], list) else child for item in A for child in item]

輸出量

[[1, 3], [3, 5], [4, 4], [5, 3]]

注意：此解決方案僅用於此問題。 這不是通用的列表拼合解決方案。

同樣的想法，使用itertools.chain

from itertools import chain
print [item[0] if isinstance(child[0], list) else item for item in chain(*A)]

輸出量

[[1, 3], [3, 5], [4, 4], [5, 3]]