How to count the number of zeros in a decimal?

Question

What is the most efficient way to calculate the number of 0s in a decimal before the first non 0 number?

Desired outcome:

0.0000456 ---> 4
0.00406 ---> 2
0.000456 ---> 3

So far I have tried mathematically this (which is slow):

from math import floor, log

def num_zeros(decimal):
    return 10 - floor(log(decimal *10e10,10))

and this (which is not right as 0 after a first number will be counted):

def count_0(number):
    n = format(number, 'f').count('0')
    return n - 1

Answer 1

Why not simply:

from math import floor, log10, inf

def num_zeros(decimal):
    return inf if decimal == 0 else -floor(log10(abs(decimal))) - 1

This handles 0.001 correctly (2 zeros), whereas some answers would print 3. It also now works for negative numbers and zero.

I'd be shocked if string methods were faster.

---

To handle non-finite numbers correctly as well, simply swap out the definition of floor as follows:

def floor(decimal):
    return math.floor(decimal) if math.isfinite(decimal) else decimal

Or equivalently:

from math import floor, log10, inf, isinf, nan, isnan

def num_zeros(decimal):
    if isnan(decimal):
        return nan
    if isinf(decimal):
        return -inf
    return inf if decimal == 0 else -floor(log10(abs(decimal))) - 1

Answer 2

Here's a way to approach this using a regular expression:

# Update, simplified sol thanks to @Aran-Fey's comments
import re
s = '0.0000456'
len(re.search('\d+\.(0*)', s).group(1))
#4