ISBN final digit finder

Question

I am working in python 3 and I am making a program that will take in a 10 digit ISBN Number and applying a method to it to find the 11th number.

Here is my current c

Answer 1

I think this should do what you want.

def get_isbn_number(isbn):
    digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]
    digit_11 = 11 - (sum(digits) % 11)
    if digit_11 == 10:
        digit_11 = 'X'    
    digits.append(digit_11)
    isbn_number = "".join(map(str, digits))
    return isbn_number

EXAMPLE

>>> print(get_isbn_number('2345432681'))
22303640281810242428
>>> print(get_isbn_number('2345432680'))
2230364028181024240X

Explanation of second line:

digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]

Could be written out like:

isbn_letters = list(isbn) # turn a string into a list of characters
isbn_numbers = map(int, isbn_letters) # run the function int() on each of the items in the list
digits = [] # empty list to hold the digits
for i, num in enumerate(isbn_numbers): # loop over the numbers - i is a 0 based counter you get for free when using enumerate
    digits.append((11 - i) * num) # If you notice the pattern, if you subtract the counter value (starting at 0) from 11 then you get your desired multiplier

Terms you should look up to understand the one line version of the code:
map,
enumerate,
list conprehension