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(',')))

def flatten(alist):
    if alist == []:
        return []
    elif type(alist) is not list:
        return [alist]
    else:
        return flatten(alist[0]) + flatten(alist[1:])

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(',')]

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)]

你可以使用列表扩展方法. 它显示是最快的:

flat_list = []
for sublist in l:
    flat_list.extend(sublist)

表演:

import functools
import itertools
import numpy
import operator
import perfplot


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(numpy.array(a).flat)


def extend(a):
    n = []

    list(map(n.extend, a))

    return n


perfplot.show(
    setup = lambda n: [list(range(10))] * n,
    kernels = [
        functools_reduce_iconcat, extend, itertools_chain, numpy_flat
        ],
    n_range = [2**k for k in range(16)],
    xlabel = 'num lists',
    )

出口:

此分類上一篇

您也可以使用NumPy的公寓:

import numpy as np
list(np.array(l).flat)

它只有在超级列表具有相同的尺寸时才有效。