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”。

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

考虑到列表L的列表,

flat_list = [item for sublist in l for item in sublist]

意思是:

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

它比迄今为止发布的短篇文章更快(l 是表格的列表)。

下面是相应的功能:

def flatten(l):
    return [item for sublist in l for item in sublist]

作为证据,您可以在标准图书馆中使用时间模块:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

解释:基于 + 的缩写(包括在总中使用)是必然的 O(L**2)当有 L 列表时 - 因为中间结果列表保持长,每个步骤都会分配一个新的中间结果列表对象,前中间结果中的所有对象都必须复制(以及在结尾添加一些新的对象)。

列表理解只产生一个列表,一次,并复制每个项目(从其原始居住地到结果列表)也准确一次。

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)]

一个非回归功能,以便在任何深度的列表列表:

def flatten_list(list1):
    out = []
    inside = list1
    while inside:
        x = inside.pop(0)
        if isinstance(x, list):
            inside[0:0] = x
        else:
            out.append(x)
    return out

l = [[[1,2],3,[4,[[5,6],7],[8]]],[9,10,11]]
flatten_list(l)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

我会建议使用发电机与产量声明和产量从。

from collections.abc import Iterable

def flatten(items, ignore_types=(bytes, str)):
    """
       Flatten all of the nested lists to the one. Ignoring flatting of iterable types str and bytes by default.
    """
    for x in items:
        if isinstance(x, Iterable) and not isinstance(x, ignore_types):
            yield from flatten(x)
        else:
            yield x

values = [7, [4, 3, 5, [7, 3], (3, 4), ('A', {'B', 'C'})]]

for v in flatten(values):
    print(v)