例如,我有以下代码:
我该如何做这样的事情?感谢您的答案。
a = ["a;b", "c;d",...,"y;z"]
我想将每个列表元素分成相同列表的两个项目。所以我想要得到这样的东西:
["a", "b", "c", "d", ...., "y", "z"]
我该如何做这样的事情?感谢您的答案。
仅使用字符串操作似乎是最简单的(当然这是主观的),并且速度最快(与迄今为止发布的其他解决方案相比,速度差距巨大)。
>>> a = ["a;b", "c;d", "y;z"]
>>> ";".join(a).split(";")
['a', 'b', 'c', 'd', 'y', 'z']
按照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
itertools.chain
:>>> a = ["a;b", "c;d","y;z"]
>>> list(itertools.chain(*(s.split(';') for s in a)))
['a', 'b', 'c', 'd', 'y', 'z']
更加功能化的方法:
>>> l = ["a;b", "c;d", "e;f", "y;z"]
>>> sum(map(lambda x: x.split(';'), l), [])
['a', 'b', 'c', 'd', 'e', 'f', 'y', 'z']
sum
来展开列表并不是邪恶的,但相当违反直觉。 - user395760[ y for z in [ x.split(';') for x in a ] for y in z ]
更易读,顺便说一下,后者是最高效的解决方案。 - Michal Chruszcz[] + ['a', 'b'] + ['c', 'd'] + ['e', 'f'] + ['y', 'z']
,因此它是那些列表的总和(这就是代码的作用)。 - ShreevatsaR这个可以工作:
l = []
for item in ["a;b", "c;d", "e;f"]:
l += item.split(";")
print l
它给出:
['a', 'b', 'c', 'd', 'e', 'f']
a = ["a;b", "c;d","y;z"]
print [atom for pair in a for atom in pair.split(';')]
提供您所需的内容:
['a', 'b', 'c', 'd', 'y', 'z']
注意:我无法告诉你如何在数组中从“...”到“....”
l = []
for current in [c.split(';') for c in a]:
l.extend(current)
你可能想要阅读一下关于列表推导式的内容http://docs.python.org/tutorial/datastructures.html#list-comprehensions
a = ["a;b", "c;d","e;f","y;z"]
b = []
for i in a:
c = i.split(';')
b = b + c
print b
>>> a_split = [i.split(";", 1) for i in a]
[[a,b], [c,d], ..., [y,z]]
现在你需要以某种方式“合并”内部和外部列表。内置的reduce()
函数非常适合这个任务:
>>> reduce(lambda x, y: x + y, a_split)
看这里:
['a', 'b', 'c', 'd', ... 'y', 'z']
可以使用字符串来实现这个功能:
>>> a = ["a;b", "c;d","y;z"]
>>> list(''.join(a).replace(';', ''))
['a', 'b', 'c', 'd', 'y', 'z']
这个解决方案是目前为止建议的最快速的之一:
# 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
结论是,在这种情况下,由字符串表示的列表可能非常高效,可能是因为内存处理更简单(字符存储在连续的内存位置中)。