Checking how many times a digit appears in a dictionary in python?

Question

I'm trying to check how many times a digit appears in a dictionary.

This code only works if I input a single digit.

numbers = input("Enter numbers ")

d = {}
d['0'] = 0
d['1'] = 0
d['2'] = 0
d['3'] = 0
d['4'] = 0
d['5'] = 0
d['6'] = 0
d['7'] = 0
d['8'] = 0
d['9'] = 0

for i in numbers:
    d[numbers] += 1

print(d)

For example if I input 8 the output will be

{'8': 1, '9': 0, '4': 0, '5': 0, '6': 0, '7': 0, '0': 0, '1': 0, '2': 0, '3': 0}

But if I input 887655 then it gives me a builtins.KeyError: '887655'

If I input 887655 the output should actually be

{'8': 2, '9': 0, '4': 0, '5': 2, '6': 1, '7': 1, '0': 0, '1': 0, '2': 0, '3': 0}

Answer 1

You should probably change

d[numbers] += 1

=>

d[i] += 1

Answer 2

Use collections.Counter for this instead - no need to reinvent the wheel.

>>> import collections
>>> collections.Counter("887655")
Counter({'8': 2, '5': 2, '6': 1, '7': 1})

Answer 3

I think what you want is actually this:

for number in numbers:
    for digit in str(number):
        d[digit] += 1

Answer 4

You should use collections.Counter

from collections import Counter
numbers = input("Enter numbers: ")
count = Counter(numbers)
for c in count:
    print c, "occured", count[c], "times"

Answer 5

I would recommend collections.Counter but here's an improved version of your code:

numbers = input("Enter numbers ")

d = {} # no need to initialize each key

for i in numbers:
    d[i] = d.get(i, 0) + 1 # we can use dict.get for that, default val of 0

print(d)