Question

例如，我有这样的代码

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

我想将每个列表元素拆分为同一列表中的项目。所以我想得到这样的东西：

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

我该怎么办？谢谢你的回答。

Answer 1

仅使用字符串操作似乎最简单（当然这是主观的）和最快（与目前为止发布的其他解决方案相比，差距很大）。

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

证明/基准

按经过时间的升序排序：

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

Answer 2

您可以使用itertools.chain：

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

Answer 3

更具功能性的方法：

>>> 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 4

这是工作：

l = []
for item in ["a;b", "c;d", "e;f"]:
     l += item.split(";")

print l

它给出了：

['a', 'b', 'c', 'd', 'e', 'f']

Answer 5

a = ["a;b", "c;d","y;z"]
print [atom for pair in a for atom in pair.split(';')]

给出你想要的东西：

['a', 'b', 'c', 'd', 'y', 'z']

注意：我无法告诉你如何在阵列中间从'...'到'....'：）

Answer 6

l = []
for current in [c.split(';') for c in a]:
   l.extend(current)

您可能希望阅读列表推导http://docs.python.org/tutorial/datastructures.html#list-comprehensions

Answer 7

a = ["a;b", "c;d","e;f","y;z"]
b = []
for i in a:
    c = i.split(';')
    b = b + c

print b

Answer 8

比菲利克斯克林的回答有点长，但是这里有。首先将列表拆分为子列表

>>> a_split = [i.split(";", 1) for i in a]

这将产生一个表格列表：

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

您现在需要以某种方式“合并”内部和外部列表。内置reduce()函数非常适合：

>>> reduce(lambda x, y: x + y, a_split)

瞧：

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

Answer 9

字符串可用于此：

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

此解决方案是目前为止提出的最快的解决方案之一：

# Shawn Chin's solution (the fastest so far, by far):
python -mtimeit -s'a=["a;b","x;y","p;q"]*99' '";".join(a).split(";")'
10000 loops, best of 3: 27.4 usec per loop

# This solution:
python -mtimeit -s'a=["a;b","x;y","p;q"]*99' "list(''.join(a).replace(';', ''))"
10000 loops, best of 3: 33.5 usec per loop

士气是在这种情况下由字符串表示的列表非常有效，可能是因为更简单的内存处理（字符存储在连续的内存位置）。

Python列表及其拆分

9 个答案:

证明/基准