Python Multidimensional Arrays - most efficient way to count number of non-zero entries

Question

Hi there on a Saturday Fun Night,

I am getting around in python and I am quite enjoying it.

Assume I have a python array:

x = [1, 0, 0, 1, 3]

What is the fastest way to count all non zero elements in the list (ans: 3) ? Also I would like to do it without for loops if possible - the most succint and terse manner possibe, say something conceptually like

[counter += 1 for y in x if y > 0]

Now - my real problem is that I have a multi dimensional array and what I really want to avoid is doing the following:

for p in range(BINS):
    for q in range(BINS):
        for r in range(BINS):
            if (mat3D[p][q][r] > 0): some_feature_set_count += 1

From the little python I have seen, my gut feeling is that there is a really clean syntax (and efficient) way how to do this.

Ideas, anyone?

Answer 1

For the single-dimensional case:

sum(1 for i in x if i)

For the multi-dimensional case, you can either nest:

sum(sum(1 for i in row if i) for row in rows)

or do it all within the one construct:

sum(1 for row in rows
      for i in row if i)

Answer 2

If you are using numpy as suggested by the fact that you're using multi-dimensional arrays in Python, the following is similar to @Marcelo's answer, but a tad cleaner:

>>> a = numpy.array([[1,2,3,0],[0,4,2,0]])
>>> sum(1 for i in a.flat if i)
5
>>>