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!
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There are plenty of duplicates here and online...squiguy– squiguy2013-03-01 07:52:13 +00:00Commented Mar 1, 2013 at 7:52
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5 Answers
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
2 Comments
Rock
I feel ashamed for asking this question.
Jon Clements
@Rock well - on the plus side - you won't forget it any time soon ;)
li = [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0]
print len(li) - sum(li)
Comments
In [16]: l = [0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0]
In [17]: l.count(0)
Out[17]: 5
Comments
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])
Comments
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.
