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)]

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)]

我用 perfplot 测试了大多数建议的解决方案(我的宠物项目,基本上是时间周围的插槽),并发现

import functools
import operator
functools.reduce(operator.iconcat, a, [])

要成为最快的解决方案,无论是许多小列表还是很少的长列表都被混合(operator.iadd 同样快)。

更简单、更可接受的选择是

out = []
for sublist in a:
    out.extend(sublist)

如果字体列表的数量很大,这表现得比上面的建议略糟糕。

此分類上一篇

此分類上一篇

重复字符的代码:

import functools
import itertools
import operator

import numpy as np
import perfplot


def forfor(a):
    return [item for sublist in a for item in sublist]


def sum_brackets(a):
    return sum(a, [])


def functools_reduce(a):
    return functools.reduce(operator.concat, a)


def functools_reduce_iconcat(a):
    return functools.reduce(operator.iconcat, a, [])


def itertools_chain(a):
    return list(itertools.chain.from_iterable(a))


def numpy_flat(a):
    return list(np.array(a).flat)


def numpy_concatenate(a):
    return list(np.concatenate(a))


def extend(a):
    out = []
    for sublist in a:
        out.extend(sublist)
    return out


b = perfplot.bench(
    setup=lambda n: [list(range(10))] * n,
    # setup=lambda n: [list(range(n))] * 10,
    kernels=[
        forfor,
        sum_brackets,
        functools_reduce,
        functools_reduce_iconcat,
        itertools_chain,
        numpy_flat,
        numpy_concatenate,
        extend,
    ],
    n_range=[2 ** k for k in range(16)],
    xlabel="num lists (of length 10)",
    # xlabel="len lists (10 lists total)"
)
b.save("out.png")
b.show()

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(',')]

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”。

考虑到列表只有整体:

import re
l = [[1, 2, 3], [4, 5, 6], [7], [8, 9]]
list(map(int,re.sub('(\[|\])','',str(l)).split(',')))