matplotlib.cbook.flatten() 将为粘贴列表工作,即使它们比示例更深地粘贴。

import matplotlib
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
print(list(matplotlib.cbook.flatten(l)))
l2 = [[1, 2, 3], [4, 5, 6], [7], [8, [9, 10, [11, 12, [13]]]]]
print(list(matplotlib.cbook.flatten(l2)))

结果:

[1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

这比 underscore 快 18 倍。

Average time over 1000 trials of matplotlib.cbook.flatten: 2.55e-05 sec
Average time over 1000 trials of underscore._.flatten: 4.63e-04 sec
(time for underscore._)/(time for matplotlib.cbook) = 18.1233394636

不是一个单行,但看到所有的答案在这里,我猜这个漫长的列表错过了一些模式匹配,所以在这里它是:)

这两种方法可能不是有效的,但无论如何,它很容易阅读(至少对我来说,也许我被功能编程所困扰):

def flat(x):
    match x:
        case []:
            return []
        case [[*sublist], *r]:
            return [*sublist, *flat(r)]

第二版考虑了列表列表的列表......不管什么:

def flat(x):
    match x:
        case []:
            return []
        case [[*sublist], *r]:
            return [*flat(sublist), *flat(r)]
        case [h, *r]:
            return [h, *flat(r)]

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]

这对我来说似乎是最简单的:

>>> import numpy as np
>>> l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
>>> print(np.concatenate(l))
[1 2 3 4 5 6 7 8 9]

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。

另一个不寻常的方法,适用于异常和均匀的整体列表:

from typing import List


def flatten(l: list) -> List[int]:
    """Flatten an arbitrary deep nested list of lists of integers.

    Examples:
        >>> flatten([1, 2, [1, [10]]])
        [1, 2, 1, 10]

    Args:
        l: Union[l, Union[int, List[int]]

    Returns:
        Flatted list of integer
    """
    return [int(i.strip('[ ]')) for i in str(l).split(',')]