Python 列表及其分割

Question

比如，我有这样的代码

a = ["a;b", "c;d",...,"y;z"]

我想把每个列表里的元素分成两个相同的项。所以我想得到像这样的结果：

["a", "b", "c", "d", ...., "y", "z"]

我该怎么做呢？谢谢大家的回答。

Answer 1

这里有一种更实用的方法：

>>> l = ["a;b", "c;d", "e;f", "y;z"]
>>> sum(map(lambda x: x.split(';'), l), [])
['a', 'b', 'c', 'd', 'e', 'f', 'y', 'z']

Answer 2

你可以使用 itertools.chain 这个工具：

>>> a = ["a;b", "c;d","y;z"]
>>> list(itertools.chain(*(s.split(';') for s in a)))
['a', 'b', 'c', 'd', 'y', 'z']

Answer 3

仅仅使用字符串操作似乎是最简单的（当然这只是个人看法），而且是最快的（与目前发布的其他解决方案相比，速度差距很大）。

>>> a = ["a;b", "c;d", "y;z"]
>>> ";".join(a).split(";")
['a', 'b', 'c', 'd', 'y', 'z']

证明 / 基准测试

按照耗时从少到多排序：

python -mtimeit -s'a=["a;b","x;y","p;q"]*99' '";".join(a).split(";")'
10000 loops, best of 3: 48.2 usec per loop

python -mtimeit -s'a=["a;b","x;y","p;q"]*99' '[single for pair in a for single in pair.split(";")]'
1000 loops, best of 3: 347 usec per loop

python -mtimeit -s'from itertools import chain; a=["a;b","x;y","p;q"]*99' 'list(chain(*(s.split(";") for s in a)))'
1000 loops, best of 3: 350 usec per loop

python -mtimeit -s'a=["a;b","x;y","p;q"]*99' 'sum([x.split(";") for x in a],[])'
1000 loops, best of 3: 1.13 msec per loop

python -mtimeit -s'a=["a;b","x;y","p;q"]*99' 'sum(map(lambda x: x.split(";"), a), [])'
1000 loops, best of 3: 1.22 msec per loop

python -mtimeit -s'a=["a;b","x;y","p;q"]*99' 'reduce(lambda x,y:x+y, [pair.split(";") for pair in a])'
1000 loops, best of 3: 1.24 msec per loop