根据布尔值列表过滤列表 [英] Filtering a list based on a list of booleans
问题描述
我有一个值列表,在给定布尔值列表中的值的情况下,我需要过滤这些值:
I have a list of values which I need to filter given the values in a list of booleans:
list_a = [1, 2, 4, 6]
filter = [True, False, True, False]
我用以下行生成一个新的过滤列表:
I generate a new filtered list with the following line:
filtered_list = [i for indx,i in enumerate(list_a) if filter[indx] == True]
其结果是:
print filtered_list
[1,4]
这条线有效,但是(对我而言)看起来有点过大,我想知道是否有更简单的方法来实现这一目标.
The line works but looks (to me) a bit overkill and I was wondering if there was a simpler way to achieve the same.
下面的答案中给出了两个好的建议的总结:
Summary of two good advices given in the answers below:
1-不要像我那样命名列表filter
,因为它是内置函数.
1- Don't name a list filter
like I did because it is a built-in function.
2-不要像我对if filter[idx]==True..
所做的那样将其与True
进行比较,因为这是不必要的.只需使用if filter[idx]
就足够了.
2- Don't compare things to True
like I did with if filter[idx]==True..
since it's unnecessary. Just using if filter[idx]
is enough.
推荐答案
您正在寻找 itertools.compress
:
You're looking for itertools.compress
:
>>> from itertools import compress
>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> list(compress(list_a, fil))
[1, 4]
时间比较(py3.x):
>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> %timeit list(compress(list_a, fil))
100000 loops, best of 3: 2.58 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v] #winner
100000 loops, best of 3: 1.98 us per loop
>>> list_a = [1, 2, 4, 6]*100
>>> fil = [True, False, True, False]*100
>>> %timeit list(compress(list_a, fil)) #winner
10000 loops, best of 3: 24.3 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]
10000 loops, best of 3: 82 us per loop
>>> list_a = [1, 2, 4, 6]*10000
>>> fil = [True, False, True, False]*10000
>>> %timeit list(compress(list_a, fil)) #winner
1000 loops, best of 3: 1.66 ms per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]
100 loops, best of 3: 7.65 ms per loop
不要使用filter
作为变量名,它是一个内置函数.
Don't use filter
as a variable name, it is a built-in function.
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