python - Correct output for function that counts occurrences of each digit in a string

Question

I want the output of the code to be something like this if the user enters a string of numbers like let's say... 122033

Enter string of numbers: 122033
0 occurs 1 time
1 occurs 1 time
2 occurs 2 times
3 occurs 2 times


def count_digits(s):
    res = [0]*10
    for x in s:
        res[int(x)] += 1
    while 0 in res:
        res.remove(0)
    return res

def main():
    s=input("Enter string of numbers: ")

    print(count_digits(s))
main()

This is the program that I have so far. At it's current state, if a user enters something like 122033 the output is: [1,1,2,2]

Note: I cannot use collections for this.

Answer 1

You're pretty close to a working solution, but removing all the 0-count entries changes the indices of your list. You already need to write some custom pretty printing code, so just leave the 0s in and skip elements where the count is 0. Maybe something like this:

def count_digits(s):
    res = [0]*10
    for x in s:
        res[int(x)] += 1
    return res

def print_counts(counts):
    for (index, count) in enumerate(counts):
        if count == 1:
            print("%d occurs %d time" % (index, count))
        elif count > 1:
            print("%d occurs %d times" % (index, count))

def main():
    s=input("Enter string of numbers: ")

    print_counts(count_digits(s))