计算 Python 列表中真布尔值的数量

Question

I have a list of Booleans:

[True, True, False, False, False, True]

and I am looking for a way to count the number of True in the list (so in the example above, I want the return to be 3 .) I have found examples of looking for the number of occurrences of specific elements, but is there a more efficient way to do it since I'm working with Booleans? I'm thinking of something analogous to all or any .

Answer 1

True is equal to 1 .

>>> sum([True, True, False, False, False, True])
3

Answer 2

list has a count method:

>>> [True,True,False].count(True)
2

This is actually more efficient than sum , as well as being more explicit about the intent, so there's no reason to use sum :

In [1]: import random

In [2]: x = [random.choice([True, False]) for i in range(100)]

In [3]: %timeit x.count(True)
970 ns ± 41.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [4]: %timeit sum(x)
1.72 µs ± 161 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 3

If you are only concerned with the constant True , a simple sum is fine. However, keep in mind that in Python other values evaluate as True as well. A more robust solution would be to use the bool builtin:

>>> l = [1, 2, True, False]
>>> sum(bool(x) for x in l)
3

UPDATE: Here's another similarly robust solution that has the advantage of being more transparent:

>>> sum(1 for x in l if x)
3

PS Python trivia: True could be true without being 1. Warning: do not try this at work!

>>> True = 2
>>> if True: print('true')
... 
true
>>> l = [True, True, False, True]
>>> sum(l)
6
>>> sum(bool(x) for x in l)
3
>>> sum(1 for x in l if x)
3

Much more evil:

True = False

Answer 4

You can use sum() :

>>> sum([True, True, False, False, False, True])
3

Answer 5

After reading all the answers and comments on this question, I thought to do a small experiment.

I generated 50,000 random booleans and called sum and count on them.

Here are my results:

>>> a = [bool(random.getrandbits(1)) for x in range(50000)]
>>> len(a)
50000
>>> a.count(False)
24884
>>> a.count(True)
25116
>>> def count_it(a):
...   curr = time.time()
...   counting = a.count(True)
...   print("Count it = " + str(time.time() - curr))
...   return counting
... 
>>> def sum_it(a):
...   curr = time.time()
...   counting = sum(a)
...   print("Sum it = " + str(time.time() - curr))
...   return counting
... 
>>> count_it(a)
Count it = 0.00121307373046875
25015
>>> sum_it(a)
Sum it = 0.004102230072021484
25015

Just to be sure, I repeated it several more times:

>>> count_it(a)
Count it = 0.0013530254364013672
25015
>>> count_it(a)
Count it = 0.0014507770538330078
25015
>>> count_it(a)
Count it = 0.0013344287872314453
25015
>>> sum_it(a)
Sum it = 0.003480195999145508
25015
>>> sum_it(a)
Sum it = 0.0035257339477539062
25015
>>> sum_it(a)
Sum it = 0.003350496292114258
25015
>>> sum_it(a)
Sum it = 0.003744363784790039
25015

And as you can see, count is 3 times faster than sum . So I would suggest to use count as I did in count_it .

Python version: 3.6.7
CPU cores: 4
RAM size: 16 GB
OS: Ubuntu 18.04.1 LTS

Answer 6

Just for completeness' sake ( sum is usually preferable), I wanted to mention that we can also use filter to get the truthy values. In the usual case, filter accepts a function as the first argument, but if you pass it None , it will filter for all "truthy" values. This feature is somewhat surprising, but is well documented and works in both Python 2 and 3.

The difference between the versions, is that in Python 2 filter returns a list, so we can use len :

>>> bool_list = [True, True, False, False, False, True]
>>> filter(None, bool_list)
[True, True, True]
>>> len(filter(None, bool_list))
3

But in Python 3, filter returns an iterator, so we can't use len , and if we want to avoid using sum (for any reason) we need to resort to converting the iterator to a list (which makes this much less pretty):

>>> bool_list = [True, True, False, False, False, True]
>>> filter(None, bool_list)
<builtins.filter at 0x7f64feba5710>
>>> list(filter(None, bool_list))
[True, True, True]
>>> len(list(filter(None, bool_list)))
3

Answer 7

It is safer to run through bool first. This is easily done:

>>> sum(map(bool,[True, True, False, False, False, True]))
3

Then you will catch everything that Python considers True or False into the appropriate bucket:

>>> allTrue=[True, not False, True+1,'0', ' ', 1, [0], {0:0}, set([0])]
>>> list(map(bool,allTrue))
[True, True, True, True, True, True, True, True, True]

If you prefer, you can use a comprehension:

>>> allFalse=['',[],{},False,0,set(),(), not True, True-1]
>>> [bool(i) for i in allFalse]
[False, False, False, False, False, False, False, False, False]

Answer 8

I prefer len([b for b in boollist if b is True]) (or the generator-expression equivalent), as it's quite self-explanatory. Less 'magical' than the answer proposed by Ignacio Vazquez-Abrams.

Alternatively, you can do this, which still assumes that bool is convertable to int, but makes no assumptions about the value of True: ntrue = sum(boollist) / int(True)