How to check if all the digits of a number are odd in python?

Question

I was told to solve a problem in which I would have to find out the number of 4-digit numbers which are all composed of odd digits. I tried the following python code:

Answer 1

A 4 digit number that is composed only of odd digits can only use the digits 1, 3, 5, 7 and 9. That gives you 5 to the power 4 is 625 different numbers. That didn't require trying them all out.

You can still do that of course, using itertools.product() for example:

from itertools import product

print sum(1 for combo in product('13579', repeat=4))

because product('13579', repeat=4) will produce all possible combinations of 4 characters from the string of odd digits.

Your code needs to test if all digits are odd; break out early if any digit is not odd:

new_list =[] # A list which holds the numbers
for a in range(1111,10000):
    for b in str(a):
        if int(b) % 2 == 0:
            break
    else:
        # only executed if the loop wasn't broken out of
        new_list.append(a)

You could use the all() function with a generator expression for that test too:

new_list = []
for a in range(1111,10000):
    if all(int(b) % 2 == 1 for b in str(a)):
        new_list.append(a)

which then can be collapsed into a list comprehension:

new_list = [a for a in range(1111,10000) if all(int(b) % 2 == 1 for b in str(a))]