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?

Psidom · Accepted Answer · 2017-05-11T02:59:44.123

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.

edited May 11 '17 at 02:59

answered May 11 '17 at 02:57

Psidom

is the `axis` correct? I still get the out of bounds error – pepe May 11 '17 at 02:58
If you extract values with `a[:,3]`, you get a 1d array, then you don't need the `axis` parameter. – Psidom May 11 '17 at 03:00

score 1 · Answer 2 · edited May 23 '17 at 12:02

1

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

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

edited May 23 '17 at 12:02

Community

answered May 11 '17 at 02:59

A.J. Uppal

2 Answers2