在Python中计算10位数字中的零，几率和偶数的数量？

Question

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.

Answer 1

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'])

Answer 2

there are some flaw in your code:

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

zero_count = 0
even_count = 0
odd_count = 0


def digit_count(x):
    global zero_count,even_count,odd_count
    while x > 0:
        num = x%10  # store the last digit of x
        if  num % 10 == 0:
            zero_count += 1
        elif num % 2 == 0:
            even_count += 1
        else:
           odd_count += 1
        x /= 10

digit_count(integer)       # you need to call the function 1st
print zero_count
print even_count
print odd_count

Answer 3

That's because you are not re-assigning the x value after each iteration of 'while'.

Simply change x/10 , which does nothing, to x = x/10

ALSO

Your declarations:

zero_count = 0
even_count = 0
odd_count = 0

are outside of your function. Therefore, you will get a 'referenced before assignment' error unless you:

1. declare them as nonlocal inside your digit_count function. by

nonlocal zero_count
nonlocal even_count
nonlocal odd_count

If you are using Python 2, will you need to use 'global' instead of nonlocal.

Or

2. initialize them inside your function.

For your specific purpose, I'd say the second option of declaring and initializing them inside your function makes more sense. You only need to use them inside the function.

Answer 4

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))]

Answer 5

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

Answer 6

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'])