Get the the number of zeros and ones of a binary number in Python

I am trying to solve a binary puzzle, my strategy is to transform a grid in zeros and ones, and what I want to make sure is that every row has the same amount of 0 and 1.

Is there a any way to count how many 1s and 0s a number has without iterating through the number?

What I am currently doing is:

def binary(num, length=4):
    return format(num, '#0{}b'.format(length + 2)).replace('0b', '')

n = binary(112, 8)
// '01110000'
and then
n.count('0')
n.count('1')

Is there any more efficient computational (or maths way) of doing that?