Как найти количество нулей в числе python

Eсть вот такой кусок кода
надо чтобы он при вводе числа в котором есть ноль, выдавал это кол-во нулей

def zero(value):
    x = 0
    if  value == '0':
        print(x)
        x = x + 1

F = input()

zero(F)

Не могу понять, что надо исправить

если надо работать только с числом:

num = -10203040506070809

count = 0
value = abs(num)

while value != 0:
    count += int(value % 10 == 0)
    value //= 10

print(count)

абсолютное значение берется, потому что -1 // 10 = -1 и цикл будет бесконечным для отрицательных чисел

если можно работать со строкой:

num = -10203040506070809

count = str(num).count('0')

print(count)

ответ дан 1 дек 2020 в 11:32

Если по честному работать с числом, то нужно делить:

def foo(n):
    if n < 0: 
        return foo(-n)
    i = 0
    while n:
        i += n % 10 == 0
        n //= 10
    return i

assert 3 == foo(10230240)

Для строки можно так:

def sfoo(sn):
    # if sn[0] == '-': return sfoo(sn[1:])
    # return sn.count('0')
    # len([*filterfalse(int, sn)])
    # return len([1 for d in sn if d == '0'])
    # return sum(d == '0' for d in sn)

    i = 0
    for d in sn:
        i += d == '0'

    return i

assert 3 == sfoo('10230240')

integer = int(raw_input("Enter a ten-digit number please: "))

zero_count = 0
even_count = 0
odd_count = 0

def digit_count(x):
    while x > 0:
        if x % 10 == 0:
            zero_count += 1
        elif x % 2 == 0:
            even_count += 1
        else:
            odd_count += 1
        x / 10

    print zero_count
    print even_count
    print odd_count

print digit_count(integer)

The user inputs a ten-digit integer and then the output should be the number of odd, even, and zero digits are in it. When I run this, currently it just has me input an integer and then does nothing else.

asked Jan 2, 2015 at 4:33

As others mentioned, it should be x = x / 10, but you also need to make your counter variables global.

Also,

print digit_count(integer)

Will display None. Since printing of the zero, odd and even counts is performed in digit_count(), you don’t need the additional print.

Another way to do this is to apply a Counter to mapped input:

from collections import Counter

s = raw_input("Enter a ten-digit number please: ")
c = Counter('zero' if d == '0' else ('odd' if int(d)%2 else 'even') for d in s)
print "zeros {}, odd {}, even {}".format(c['zero'], c['odd'], c['even'])

answered Jan 2, 2015 at 4:56

There are scoping issues in your code. The zero, even and odd counts are global variables and they’re modified within the function. And, you’re missing an assignment in the statement x = x / 10.

Instead of processing the big integer through division by 10, I’d like to highlight an alternative method. First convert your string to a list of single digits, and then process the list through the digit_count function:

def digit_count(digits):
    """ Takes a list of single digits and returns the 
        count of zeros, evens and odds
    """
    zero_count = 0
    even_count = 0
    odd_count = 0
    for i in digits:
        if i == 0:
            zero_count += 1
        elif i % 2 == 0:
            even_count += 1
        else:
            odd_count += 1
    return zero_count, even_count, odd_count

inp = raw_input("Enter a ten-digit number please: ")
digits = [int(i) for i in inp] # see comment at the bottom of the answer

zero_count, even_count, odd_count = digit_count(digits)

print('zero_count: %d' % zero_count)
print('even_count: %d' % even_count)
print('odd_count: %d' % odd_count)

The caveat with my code is that it will count leading zeros. While your method would not count leading zeros.
To get the exact behavior as your code, replace the line

digits = [int(i) for i in inp]

with

digits = [int(i) for i in  str(int(inp))]

answered Jan 2, 2015 at 5:05

By collections.Counter

1. Created Counter dictionary from the user input.
2. As we know 2,4,6,8 are even digit and 3,5,7,9 are odd digit. So add value of all even digit to get even count and all Odd digit to get odd count.

Code:

from collections import Counter
digit_counter = Counter(map(lambda x: int(x), raw_input("Enter a ten-digit number please: ")))

print "Zero count",  digit_counter[0]
print "Even count", digit_counter[2]+digit_counter[4]+digit_counter[6]+digit_counter[8]
print "Odd count", digit_counter[1]+digit_counter[3]+digit_counter[5]+digit_counter[7]+digit_counter[9]

Output:

python test.py 
Enter a ten-digit number please: 0112203344
Zero count 2
Even count 4
Odd count 4

answered Jan 2, 2015 at 7:11

With Python 3 you can do in a better way.

from collections import Counter
#s = '03246'
s = str(input ("Enter number --> "))
c = Counter('zero' if d == '0' else ('odd' if int(d)%2 else 'even') for d in s)
print("Zero --> " ,c['zero'], "Odd--> ", c['odd'],"Even-->", c['even'])

answered Apr 28, 2019 at 8:44

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

→ Ссылка

Автор решения: vp_arth

def foo(n):
    if n < 0: 
        return foo(-n)
    i = 0
    while n:
        i += n % 10 == 0
        n //= 10
    return i

assert 3 == foo(10230240)

def sfoo(sn):
    # if sn[0] == '-': return sfoo(sn[1:])
    # return sn.count('0')
    # len([*filterfalse(int, sn)])
    # return len([1 for d in sn if d == '0'])
    # return sum(d == '0' for d in sn)

    i = 0
    for d in sn:
        i += d == '0'

    return i

assert 3 == sfoo('10230240')

→ Ссылка

Автор решения: Zhihar

num = -10203040506070809

count = 0
value = abs(num)

while value != 0:
    count += int(value % 10 == 0)
    value //= 10

print(count)

num = -10203040506070809

count = str(num).count('0')

print(count)

→ Ссылка