如何提取嵌套列表？

Question

重复:

• 在Python中展平浅层列表
• 理解扁平序列序列？

让我们说我有一个嵌套列表的列表:

[["a","b","c"], ["d","e","f"], ["g","h","i","j"]...]

将它转换为单个列表的最佳方法是什么？

["a", "b", "c", "d", "e"....]

Answer 1

用途itertools.chain:

from itertools import chain

list(chain.from_iterable(list_of_lists))

Answer 2

在itertools文档中有一个直接的例子(参见http://docs.python.org/library/itertools.html#recipes寻找flatten()),但它很简单:

>>> from itertools import chain
>>> list(chain(*x))
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

或者,它可以在单个列表理解中非常容易地完成:

>>> x=[["a","b","c"], ["d","e","f"], ["g","h","i","j"]]
>>> [j for i in x for j in i]
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

或者通过reduce():

>>> from operator import add
>>> reduce(add, x)
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']