How to quickly count number of 0(s) in a binary list?

Question

With an array like [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0], is there a quick way to return the number of 0(s), which is 5 in the example? Thanks!

There are plenty of duplicates here and online... – squiguy Mar 01 '13 at 07:52 — squiguy, Mar 01 '13 at 07:52

score 10 · Accepted Answer · answered Mar 01 '13 at 07:46

10

Use list.count:

your_list.count(0)

And the help:

>>> help(list.count)
Help on method_descriptor:

count(...)
    L.count(value) -> integer -- return number of occurrences of value

answered Mar 01 '13 at 07:46

Jon Clements

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I feel ashamed for asking this question. – Rock Mar 01 '13 at 07:52
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@Rock well - on the plus side - you won't forget it any time soon ;) – Jon Clements Mar 01 '13 at 07:59

score 2 · Answer 2 · answered Mar 01 '13 at 07:47

2

In [16]: l = [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0]

In [17]: l.count(0)
Out[17]: 5

answered Mar 01 '13 at 07:47

avasal

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score 2 · Answer 3 · answered Mar 01 '13 at 08:00

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li = [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0]
print len(li) - sum(li)

answered Mar 01 '13 at 08:00

eyquem

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JHolta · Answer 4 · 2013-03-01T08:07:38.620

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Your choice, whatever lets you sleep at night:

l = [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0]

print l.count(0)
# or maybe:
print len(filter(lambda a: a == 0, l))
# or maybe:
print len([0 for x in l if x==0])

edited Mar 01 '13 at 08:07

answered Mar 01 '13 at 08:01

JHolta

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score 1 · Answer 5 · answered Mar 01 '13 at 10:45

You can speed things up by a factor of 100 by using arrays (, which only becomes important for large lists)...

This should be 100 times faster than my_list.count(0):

(my_array==0).sum()

However it only helps, if your data is already arranged as a numpy array (or you can manage to put it into a numpy array when it is created). Otherwise the conversion my_array = np.array(my_list) eats the time.

How to quickly count number of 0(s) in a binary list?

5 Answers5