Split an integer into digits to compute an ISBN checksum

user_120456 0
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I'm writing a program which calculates the check digit of an ISBN number. I have to read the user's input (nine digits of an ISBN) into an integer variable, and then multiply the last digit by 2, the second last digit by 3 and so on. How can I "split" the integer into its constituent digits to do this? As this is a basic homework exercise I am not supposed to use a list.

decimal integer python
15 Answers

Just create a string out of it.

myinteger = 212345
number_string = str(myinteger)

That's enough. Now you can iterate over it:

for ch in number_string:
    print ch # will print each digit in order

Or you can slice it:

print number_string[:2] # first two digits
print number_string[-3:] # last three digits
print number_string[3] # forth digit

Or better, don't convert the user's input to an integer (the user types a string)

isbn = raw_input()
for pos, ch in enumerate(reversed(isbn)):
    print "%d * %d is %d" % pos + 2, int(ch), int(ch) * (pos + 2)

For more information read a tutorial.

user_171600 
while number:
    digit = number % 10

    # do whatever with digit

    # remove last digit from number (as integer)
    number //= 10

On each iteration of the loop, it removes the last digit from number, assigning it to digit. It's in reverse, starts from the last digit, finishes with the first

Alexandru_Nedelcu_32800 
list_of_ints = [int(i) for i in str(ISBN)]

Will give you a ordered list of ints. Of course, given duck typing, you might as well work with str(ISBN).

Edit: As mentioned in the comments, this list isn't sorted in the sense of being ascending or descending, but it does have a defined order (sets, dictionaries, etc in python in theory don't, although in practice the order tends to be fairly reliable). If you want to sort it:

list_of_ints.sort()

is your friend. Note that sort() sorts in place (as in, actually changes the order of the existing list) and doesn't return a new list.

user_684570
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