Как найти количество значащих нулей питон

задан 1 дек 2020 в 10:49

def count_zeroes(n):
    return str(n).count('0')

ответ дан 1 дек 2020 в 10:53

ZhiharZhihar

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ответ дан 1 дек 2020 в 11:03

def f(n):
    return len(list(filter(lambda x: x == "0", str(n)))) 

ответ дан 1 дек 2020 в 11:11

asked Dec 21, 2011 at 16:50

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

answered Dec 21, 2011 at 16:53

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'))

answered Dec 21, 2011 at 16:57

String-Methods a quite clearly answered – here is one with keeping the integer.
You can get it also with using modulo (%) with 10 in the loop, and then reduce your given number by dividing it by then in the next loop:

    def end_zeros(num: int):
        n = 0
        while num%10 == 0:
            n += 1
            num = num/10
            return n
        else:
            return 0

Without defining num as integer:

    def end_zeros(num):
        if num == 0:
            return 1
        elif num%10 == 0:
            n = 0
            while num%10 == 0:
                n += 1
                num = num/10
            return n
        else:
            return 0

answered May 16, 2021 at 19:45

Vlad BezdenVlad Bezden

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Kasem777Kasem777

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You can use bitwise operators:

>>> def trailing_zeros(x):
...     return (x & -x).bit_length() - 1
... 
>>> trailing_zeros(0b0110110000)
4
>>> trailing_zeros(0b0)
-1

answered Dec 19, 2021 at 17:15

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.

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

answered Nov 15, 2019 at 19:05

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 re.findall() function:

   from re import findall

   def end_zeros(num):
   return len(findall(“0*$”, str(num))[0])

answered Jul 4, 2020 at 20:37