在布尔列表中获取True值的索引 [英] Getting indices of True values in a boolean list

问题描述

我有一段代码应该在其中创建总机.我想返回所有打开的开关的列表.此处"on"等于True，"off"等于False.所以现在我只想返回所有True值及其位置的列表.这就是我所拥有的，但它只返回第一次出现的True的位置(这只是我的代码的一部分):

I have a piece of my code where I'm supposed to create a switchboard. I want to return a list of all the switches that are on. Here "on" will equal True and "off" equal False. So now I just want to return a list of all the True values and their position. This is all I have but it only return the position of the first occurrence of True (this is just a portion of my code):

self.states = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]

def which_switch(self):
    x = [self.states.index(i) for i in self.states if i == True]

这只会返回"4"

推荐答案

使用enumerate，list.index返回找到的第一个匹配项的索引.

Use enumerate, list.index returns the index of first match found.

>>> t = [False, False, False, False, True, True, False, True, False, False, False, False, False, False, False, False]
>>> [i for i, x in enumerate(t) if x]
[4, 5, 7]

对于庞大的列表，最好使用itertools.compress:

For huge lists, it'd be better to use itertools.compress:

>>> from itertools import compress
>>> list(compress(xrange(len(t)), t))
[4, 5, 7]
>>> t = t*1000
>>> %timeit [i for i, x in enumerate(t) if x]
100 loops, best of 3: 2.55 ms per loop
>>> %timeit list(compress(xrange(len(t)), t))
1000 loops, best of 3: 696 µs per loop

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