Python：如何从大嵌套列表中生成“扁平化”列表

Question

假设我有一个如下列表：

[1, 2, [3, 4, [5, 6]], [7, 8], 9]

我想知道如何以下列方式展平这个列表：

[
    [7, 8],
    [5, 6],
    [3, 4, 5, 6],
    [1, 2, 3, 4, 5, 6, 7, 8, 9]
]

换句话说，我想知道如何从主列表的每个级别生成一个列表，生成的每个列表都是其所有子列表的扁平版本。

编辑：

如果该方法是左递归方法，则输出的列表很可能具有以下顺序（而不是上面的顺序）的列表：

[
    [5, 6],
    [3, 4, 5, 6],
    [7, 8],
    [1, 2, 3, 4, 5, 6, 7, 8, 9]
]

Answer 1

您可以使用递归生成器函数：

def yield_and_flatten(nested):
    """Yield sublists and flatten, recursively

    Produces a boolean and list on each yield; the boolean
    flags a merge; sublists are merged just once then
    passed down the recursion tree.

    """
    if not isinstance(nested, list):
        yield True, nested
        return
    res = []
    for elem in nested:
        for extend, sub in yield_and_flatten(elem):
            if isinstance(sub, list):
                if extend:
                    res.extend(sub)
                yield False, sub
            else:
                res.append(sub)
    yield True, res

这会在扩展当前级别之前传递子列表。

演示：

>>> sample = [1, 2, [3, 4, [5, 6]], [7, 8], 9]
>>> for _, res in yield_and_flatten(sample):
...     print res
... 
[5, 6]
[3, 4, 5, 6]
[7, 8]
[1, 2, 5, 6, 3, 4, 5, 6, 7, 8, 9]
>>> mlist = [1, 2, 3, [[4, [5, 6]], 7], 8, 9]
>>> for _, res in yield_and_flatten(mlist):
...     print res
... 
[5, 6]
[4, 5, 6]
[4, 5, 6, 7]
[1, 2, 3, 4, 5, 6, 7, 8, 9]