Get count of digits of a number that divide the number

So, here's a program in python to get the count of digits that divide the number. If you give the program a list like [24,10245], you get an output of:

2
2

This is the code:

def count_div_digits(l):
    for number in l:
        count=0
        digits=str(number)
        for digit in digits:
            if digit=='0':
                continue
            else:
                if number%(int(digit))==0:
                    count+=1
        print count

The above code obviously works. How can I further optimize this, make it short or do it differently?

Do one thing

Let's start with your top level:

def count_div_digits(l):
    for number in l:

Firstly, l is a terrible variable name. It looks like 1. Avoid it at all costs. Secondly, this function is less useful than it could be simply because it requires a list of numbers. It'd be better if the function took a single number, then you could call it on a list separately if you want:

def count_div_digits(number):
    ...

for number in lst:
    count_div_digits(number)

Return, don't print

Right now, your count_div_digits simply prints its result. That's all well and good, but it'd be better if you just returned the number. If you want to print it, you can always just do:

print count_div_digits(number)

And this way, you can store the results somewhere else for further analysis later.

Avoid continue:

The less weird hops your loops make, the easier it is to follow along the logic. Right now you have:

if digit=='0':
    continue
else:    
    # stuff

But it'd be clearer to just do:

if digit != '0':
    # same stuff

Proposed Solution

def count_div_digits(number):
    count = 0
    for digit in str(number):
        if digit != '0' and number % int(digit) == 0:
            count += 1
    return count

Generator Expressions

You could also write the above with the help of a generator expression by factoring out a length:

def sizeof(iterable):
    return sum(1 for _ in iterable)

def count_div_digits(number):
    return sizeof(digit for digit in str(number)
                  if digit != '0' and number % int(digit) == 0)

Whether you consider that an improvement or not is a matter of opinion.

StackExchange Code Review Q#105274, answer score: 7