Jack Twain Jack Twain - 1 month ago 4
Python Question

Finding which rows have all elements as zeros in a matrix with numpy

I have a large

. Some of the rows of the matrix have all of their elements as zero and I need to get the indices of those rows. The naive approach I'm considering is to loop through each row in the matrix and then check each elements. However I think there's a better and a faster approach to accomplish this using
. I hope you can help!


Here's one way. I assume numpy has been imported using import numpy as np.

In [20]: a
array([[0, 1, 0],
       [1, 0, 1],
       [0, 0, 0],
       [1, 1, 0],
       [0, 0, 0]])

In [21]: np.where(~a.any(axis=1))[0]
Out[21]: array([2, 4])

It's a slight variation of this answer: How to check that a matrix contains a zero column?

Here's what's going on:

The any method returns True if any value in the array is "truthy". Nonzero numbers are considered True, and 0 is considered False. By using the argument axis=1, the method is applied to each row. For the example a, we have:

In [32]: a.any(axis=1)
Out[32]: array([ True,  True, False,  True, False], dtype=bool)

So each value indicates whether the corresponding row contains a nonzero value. The ~ operator is the binary "not" or complement:

In [33]: ~a.any(axis=1)
Out[33]: array([False, False,  True, False,  True], dtype=bool)

(An alternative expression that gives the same result is (a == 0).all(axis=1).)

To get the row indices, we use the where function. It returns the indices where its argument is True:

In [34]: np.where(~a.any(axis=1))
Out[34]: (array([2, 4]),)

Note that where returned a tuple containing a single array. where works for n-dimensional arrays, so it always returns a tuple. We want the single array in that tuple.

In [35]: np.where(~a.any(axis=1))[0]
Out[35]: array([2, 4])