python - Pythonic way to count the number of trailing zeros in base 2

Question

I'm looking for a Pythonic way to count the number of trailing zeros in the binary representation of a positive integer n (which will indicate the highest power of 2 which divides n without remainder).

A simple solution:

def CountZeros(n):
    c = 0
    while (n % 2) == 0:
        n /= 2
        c += 1
    return c

But in order to do it in a more Pythonic manner, I think that I can make use of:

• bin(n)[2:], which gives the binary representation of n
• bin(n)[:1:-1], which gives the reversed binary representation of n

So my question can be reduced to counting the number of trailing zeros in a string.

Is there any single-statement way to do this?

My ultimate goal is a Pythonic way for computing the highest power of 2 which divides n without remainder, so any ways to do this not by counting the trailing zeros in a string are also appreciated.

Answer 1

You could use str.rstrip:

def trailing(s):
    return len(s) - len(s.rstrip('0'))