How to get a new numpy array based on conditions of two other numpy arrays using only numpy operations?

Question

Let's say I have np.array of a = [0, 1, 1, 0, 0, 1] and b = [1, 1, 0, 0, 0, 1]

I want a new matrix c such that if a[i] = 0 and b[i] = 0 then c[i] = True

I've tried using c = np.where(a == 0 & b==0) but seems this won't work on two arrays.

Ultimately I want a count of '0's that are true in both cases and '1's that are true in both cases without using any loops and external libraries. Please advise.

Answer 1

Here is my solution:

Your logic expression is nothing but c = NOT (a OR b). This means, you want c[i] = True , only if both a[i] and b[i] are zero. The OR can be achieved by adding the two arrays. We then convert the array into a boolean type and invert it.

import numpy as np

a = np.array([0, 1, 1, 0, 0, 1])
b = np.array([1, 1, 0, 0, 0, 1])

c = np.invert((a+b).astype('bool'))

If you then want to count the number of zeros you can simply perform

n_zeros = np.sum(c)

More general solution:

If you want your array c[i] = True if a[i] == a0 and b[i] == b0 , you can do:

c = (a == a0) & (b == b0)

The conditions a == a0 and b == b0 return each a boolean array with the truth value of the condition for each individual array element. The bitwise operator & performs an element-wise logical AND. For more bitwise operators see https://wiki.python.org/moin/BitwiseOperators .