flat_list = []
map(flat_list.extend, list_of_lists)

最短!

import itertools
a = [['a','b'], ['c']]
print(list(itertools.chain.from_iterable(a)))

这给了

['a', 'b', 'c']

flat_list = []
map(flat_list.extend, list_of_lists)

最短!

来晚了，但是…

我是python的新手，有lisp的背景。这是我想出的(检查lulz的var名称):

def flatten(lst):
    if lst:
        car,*cdr=lst
        if isinstance(car,(list,tuple)):
            if cdr: return flatten(car) + flatten(cdr)
            return flatten(car)
        if cdr: return [car] + flatten(cdr)
        return [car]

似乎有用。测试:

flatten((1,2,3,(4,5,6,(7,8,(((1,2)))))))

返回:

[1, 2, 3, 4, 5, 6, 7, 8, 1, 2]

性能比较:

import itertools
import timeit
big_list = [[0]*1000 for i in range(1000)]
timeit.repeat(lambda: list(itertools.chain.from_iterable(big_list)), number=100)
timeit.repeat(lambda: list(itertools.chain(*big_list)), number=100)
timeit.repeat(lambda: (lambda b: map(b.extend, big_list))([]), number=100)
timeit.repeat(lambda: [el for list_ in big_list for el in list_], number=100)
[100*x for x in timeit.repeat(lambda: sum(big_list, []), number=1)]

生产:

>>> import itertools
>>> import timeit
>>> big_list = [[0]*1000 for i in range(1000)]
>>> timeit.repeat(lambda: list(itertools.chain.from_iterable(big_list)), number=100)
[3.016212113769325, 3.0148865239060227, 3.0126415732791028]
>>> timeit.repeat(lambda: list(itertools.chain(*big_list)), number=100)
[3.019953987082083, 3.528754223385439, 3.02181439266457]
>>> timeit.repeat(lambda: (lambda b: map(b.extend, big_list))([]), number=100)
[1.812084445152557, 1.7702404451095965, 1.7722977998725362]
>>> timeit.repeat(lambda: [el for list_ in big_list for el in list_], number=100)
[5.409658160700605, 5.477502077679354, 5.444318360412744]
>>> [100*x for x in timeit.repeat(lambda: sum(big_list, []), number=1)]
[399.27587954973444, 400.9240571138051, 403.7521153804846]

这是在Windows XP 32位的Python 2.7.1上，但上面评论中的@temoto得到from_iterable比map+extend更快，所以它相当依赖于平台和输入。

不要使用sum(big_list， [])

对于无限嵌套的元素，这是递归工作的:

def iterFlatten(root):
    if isinstance(root, (list, tuple)):
        for element in root:
            for e in iterFlatten(element):
                yield e
    else:
        yield root

结果:

>>> b = [["a", ("b", "c")], "d"]
>>> list(iterFlatten(b))
['a', 'b', 'c', 'd']