关于python：如何扁平列表/嵌套列表的列表？

How do I flatten a list of lists/nested lists?

本问题已经有最佳答案，请猛点这里访问。

我有这样的python代码：

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newlist =[[52, None, None], [129, None, None], [56, None, None], [111, None, None],
[22, None, None], [33, None, None], [28, None, None], [52, None, None],
[52, None, None], [52, None, None], [129, None, None], [56, None, None],
[111, None, None], [22, None, None], [33, None, None], [28, None, None]]

我要的是newlist式的：

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newlist =[52, None, None,129, None, None,56, None, None,111, None, None,22,
None, None,33, None, None,28, None, None,52, None, None,52, None,
None,52, None, None,129, None, None,56, None, None, 111, None,
None,22, None, None,33, None, None,28, None, None]

有什么办法可以解决吗？

你所要做的就是把名单拉平。根据Python的禅宗，你是在做正确的事情。从中引用

Flat is better than nested.

所以你可以用这样的列表理解

1	newlist = [item for items in newlist for item in items]

或者你可以这样使用EDOCX1的chain。

1 2	from itertools import chain newlist = list(chain(*newlist))

或者您可以使用chain.from_iterable，在不需要打开列表的情况下

1 2	from itertools import chain newlist = list(chain.from_iterable(newlist))

使用sum函数

1	newlist = sum(newlist, [])

使用reduce函数

1	newlist = reduce(lambda x,y: x+y, newlist)

使用operator.add。这将比使用lambda版本的reduce更快。

1 2	import operator newlist = reduce(operator.add, newlist)

编辑：为了完整起见，还包括在用python从列表中创建一个简单列表中找到的答案。

我试着用python 2.7来计时，就像这样

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from timeit import timeit
print(timeit("[item for items in newlist for item in items]","from __main__ import newlist"))
print(timeit("sum(newlist, [])","from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)","from __main__ import newlist"))
print(timeit("reduce(add, newlist)","from __main__ import newlist; from operator import add"))
print(timeit("list(chain(*newlist))","from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))","from __main__ import newlist; from itertools import chain"))

我机器上的输出

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2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

因此，最有效的方法是在python 2.7中使用list(chain.from_iterable(newlist))。在python 3.3中运行了相同的测试

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from timeit import timeit
print(timeit("[item for items in newlist for item in items]","from __main__ import newlist"))
print(timeit("sum(newlist, [])","from __main__ import newlist"))
print(timeit("reduce(lambda x,y: x+y, newlist)","from __main__ import newlist; from functools import reduce"))
print(timeit("reduce(add, newlist)","from __main__ import newlist; from operator import add; from functools import reduce"))
print(timeit("list(chain(*newlist))","from __main__ import newlist; from itertools import chain"))
print(timeit("list(chain.from_iterable(newlist))","from __main__ import newlist; from itertools import chain"))

我机器上的输出

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2.26074504852
2.45047688484
3.50180387497
2.56596302986
1.78825688362
1.61612296104

因此，无论是python 2.7还是3.3，都可以使用list(chain.from_iterable(newlist))来扁平嵌套列表。