How do I count the trailing zeros in integer?

Question

I am trying to write a function that returns the number of trailing 0s in a string or integer. Here is what I am trying and it is not returning the correct values.

Answer 1

May be you can try doing this. This may be easier than counting each trailing '0's

def trailing_zeros(longint):
    manipulandum = str(longint)
    return len(manipulandum)-len(manipulandum.rstrip('0'))

Answer 2

You could just:

1. Take the length of the string value of what you're checking
2. Trim off trailing zeros from a copy of the string
3. Take the length again, of the trimmed string
4. Subtract the new length from the old length to get the number of zeroes trailing.

Answer 3

Here are two examples of doing it:

1. Using rstrip and walrus := operator. Please notice that it only works in Python 3.8 and above.

def end_zeros(num):
    return len(s := str(num)) - len(s.rstrip("0"))

2. Using findall() regular expression (re) function:

from re import findall
    
def end_zeros(num):
    return len(findall("0*$", str(num))[0])

Answer 4

The question asks to count the trailing zeros in a string or integer. For a string, len(s) - len(s.rstrip('0')) is fine. But for an integer, presumably you don't want to convert it to a string first. In that case, use recursion:

def trailing_zeros(longint):
    assert(longint)
    return ~longint % 2 and trailing_zeros(longint/2) + 1

Answer 5

For strings, it is probably the easiest to use rstrip():

In [2]: s = '23989800000'

In [3]: len(s) - len(s.rstrip('0'))
Out[3]: 5

Answer 6

I found two ways to achieve this, one is already mentioned above and the other is almost similar:

manipulandum.count('0') - manipulandum.rstrip('0').count('0')

But still, I'm looking for some better answer.