How do I count the number of nonzero values in a given array column?

Question

I have an array that is as follows:

[[0, 0, 0, 1],
 [1, 0, 0, 0],
 [0, 1, 0, 0],
 [0, 0, 1, 0],
 [0, 0, 0, 1],
 etc

I'd like to determine the sum of index [3] in each line. For example, here I'd like to get 2 as a result.

I'm trying

np.sum(np.count_nonzero(array[:][3], axis=1))

but I get an out of bounds error. Any ideas?

akuiper · Accepted Answer · 2017-05-11 02:59:44Z

4

You need to index the array as a[:, 3] (the third column of all rows), then you can do:

# if the array contains only 0 and 1
a[:,3].sum()
# 2

# if the array can have other values besides 0 and 1
np.count_nonzero(a[:,3])
# 2

Here is more info about numpy indexing.

answered May 11, 2017 at 2:57

akuiper

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2 Comments

is the axis correct? I still get the out of bounds error

If you extract values with a[:,3], you get a 1d array, then you don't need the axis parameter.

Community · Accepted Answer · 2017-05-23 12:02:38Z

1

Simply get the list as n[:,3] then sum normally since summing zero does nothing:

>>> numpy.sum(n[:,3])
2

CommunityBot

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answered May 11, 2017 at 2:59

A.J. Uppal

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