我想要一個解決方案,可以處理多種<unk>(<unk>,<unk>,<unk>,<unk>),<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>,<unk>。

这就是我所带来的:

def _flatten(l) -> Iterator[Any]:
    stack = l.copy()
    while stack:
        item = stack.pop()
        if isinstance(item, list):
            stack.extend(item)
        else:
            yield item


def flatten(l) -> Iterator[Any]:
    return reversed(list(_flatten(l)))

和测试:

@pytest.mark.parametrize('input_list, expected_output', [
    ([1, 2, 3], [1, 2, 3]),
    ([[1], 2, 3], [1, 2, 3]),
    ([[1], [2], 3], [1, 2, 3]),
    ([[1], [2], [3]], [1, 2, 3]),
    ([[1], [[2]], [3]], [1, 2, 3]),
    ([[1], [[[2]], [3]]], [1, 2, 3]),
])
def test_flatten(input_list, expected_output):
    assert list(flatten(input_list)) == expected_output

你的功能不起作用的原因是因为延伸延伸一个序列在现场,并且不会返回它。

reduce(lambda x,y: x.extend(y) or x, l)

注意:扩展比 + 列表更有效。

注意: 下面适用于 Python 3.3+ 因为它使用 yield_from. six 也是第三方包,尽管它是稳定的。

在obj = [1, 2,], [3, 4], [5, 6]的情况下,这里的所有解决方案都很好,包括列表理解和 itertools.chain.from_iterable。

但是,考虑这个稍微复杂的案例:

>>> obj = [[1, 2, 3], [4, 5], 6, 'abc', [7], [8, [9, 10]]]

这里有几个问题:

您可以以以下方式解决此问题:

>>> from collections import Iterable
>>> from six import string_types

>>> def flatten(obj):
...     for i in obj:
...         if isinstance(i, Iterable) and not isinstance(i, string_types):
...             yield from flatten(i)
...         else:
...             yield i


>>> list(flatten(obj))
[1, 2, 3, 4, 5, 6, 'abc', 7, 8, 9, 10]

在这里,您可以检查(一)的子元素(一)与(一)的Iterable(一)无效,从(一)的ABC,但也希望确保(二)的元素(一)不是“类似于(一)的”。

对于包含多个列表的列表,这里是一个重复的解决方案,为我工作,我希望它是正确的:

# Question 4
def flatten(input_ls=[]) -> []:
    res_ls = []
    res_ls = flatten_recursive(input_ls, res_ls)

    print("Final flatten list solution is: \n", res_ls)

    return res_ls


def flatten_recursive(input_ls=[], res_ls=[]) -> []:
    tmp_ls = []

    for i in input_ls:
        if isinstance(i, int):
            res_ls.append(i)
        else:
            tmp_ls = i
            tmp_ls.append(flatten_recursive(i, res_ls))

    print(res_ls)
    return res_ls


flatten([0, 1, [2, 3], 4, [5, 6]])  # test
flatten([0, [[[1]]], [[2, 3], [4, [[5, 6]]]]])

出口:

[0, 1, 2, 3]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6]
Final flatten list solution is: 
 [0, 1, 2, 3, 4, 5, 6]
[0, 1]
[0, 1]
[0, 1]
[0, 1, 2, 3]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6]
Final flatten list solution is: 
 [0, 1, 2, 3, 4, 5, 6]

我创建了一点功能,基本上可以平滑任何东西. 你可以用管道:管道安装平滑一切

from flatten_everything import flatten_everything
withoutprotection=list(
    flatten_everything(
        [
            1,
            1,
            2,
            [3, 4, 5, [6, 3, [2, 5, ["sfs", "sdfsfdsf",]]]],
            1,
            3,
            34,
            [
                55,
                {"brand": "Ford", "model": "Mustang", "year": 1964, "yearxx": 2020},
                pd.DataFrame({"col1": [1, 2], "col2": [3, 4]}),
                {"col1": [1, 2], "col2": [3, 4]},
                55,
                {"k32", 34},
                np.array([[[1, 2], [3, 4]], [[5, 6], [7, 8]]]),
                (np.arange(22), np.eye(2, 2), 33),
            ],
        ]
    )
)
print(withoutprotection)
output:
[1, 1, 2, 3, 4, 5, 6, 3, 2, 5, 'sfs', 'sdfsfdsf', 1, 3, 34, 55, 'Ford', 'Mustang', 1964, 2020, 1, 2, 3, 4, 1, 2, 3, 4, 55, 34, 'k32', 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 1.0, 0.0, 0.0, 1.0, 33]

你甚至可以保护物体免受闪烁:

from flatten_everything import ProtectedDict,ProtectedList,ProtectedTuple
withprotection=list(
    flatten_everything(
        [
            1,
            1,
            2,
            [3, 4, 5, [6, 3, [2, 5, ProtectedList(["sfs", "sdfsfdsf",])]]],
            1,
            3,
            34,
            [
                55,
                ProtectedDict({"brand": "Ford", "model": "Mustang", "year": 1964, "yearxx": 2020}),
                pd.DataFrame({"col1": [1, 2], "col2": [3, 4]}),
                {"col1": [1, 2], "col2": [3, 4]},
                55,
                {"k32", 34},
                np.array([[[1, 2], [3, 4]], [[5, 6], [7, 8]]]),
                ProtectedTuple((np.arange(22), np.eye(2, 2), 33)),
            ],
        ]
    )
)
print(withprotection)
output:
[1, 1, 2, 3, 4, 5, 6, 3, 2, 5, ['sfs', 'sdfsfdsf'], 1, 3, 34, 55, {'brand': 'Ford', 'model': 'Mustang', 'year': 1964, 'yearxx': 2020}, 1, 2, 3, 4, 1, 2, 3, 4, 55, 34, 'k32', 1, 2, 3, 4, 5, 6, 7, 8, (array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,17, 18, 19, 20, 21]), array([[1., 0.], [0., 1.]]), 33)]

根據您的列表(1, 2, 3), [4, 5, 6], [7], [8, 9] 是 1 列表水平,我們可以簡單地使用數量(列表),而不使用任何圖書館。

sum([[1, 2, 3], [4, 5, 6], [7], [8, 9]],[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]

延伸此方法的优势,当内部存在一个<unk>或数字时,简单地将每个元素的地图函数添加到列表中

#For only tuple
sum(list(map(list,[[1, 2, 3], (4, 5, 6), (7,), [8, 9]])),[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]

#In general

def convert(x):
    if type(x) is int or type(x) is float:
           return [x]
    else:
           return list(x)

sum(list(map(convert,[[1, 2, 3], (4, 5, 6), 7, [8, 9]])),[])
# [1, 2, 3, 4, 5, 6, 7, 8, 9]

在这里,有一个明确的解释的缺点在记忆的这个方法。 简而言之,它重复创建列表对象,应该避免( )