CODEWITHSUNDEEP
pythonstringfor-loop
Matthijs Abma
Matthijs Abma

Reputation: 1

Check if numbers in a comma-separated string are odd or even

I need to write a program which checks whether each number on a comma separated string is even or odd.

I have to do it with a for loop, the code I have is this:

getallen = '12, 3, 7, 25, 771, 45, 6, 98, 55, 546'

for number in getallen:
    if number %2 == 0:
        print(f'{int(number)} is even')
    else:
        print(f'{int(number)} is odd')

The output should be something like this:

12 is even
3 is odd
7 is odd etc etc.

Upvotes: 1

Views: 798

Answers (3)

MisterMiyagi
MisterMiyagi

Reputation: 52019

Your string is a tuple literal, so you can just use ast.literal_eval to safely convert it to a tuple of integers. This can be directly used in your for loop.

import ast
getallen = '12, 3, 7, 25, 771, 45, 6, 98, 55, 546'

for number in ast.literal_eval(getallen):
    if number %2 == 0:
        print(f'{int(number)} is even')
    else:
        print(f'{int(number)} is odd')

Upvotes: 0

Nick is tired
Nick is tired

Reputation: 7056

Matt is right that you're iterating over a string, and not an array, but after splitting you also need to convert each value to a number. There's a couple ways to do this:

You can use map to convert each item of the array to a number using int:

for number in map(int, getallen.split(',')):
    if number %2 == 0:
        print(f'{int(number)} is even')
    else:
        print(f'{int(number)} is odd')

Or you can convert each number as you test it:

for number in getallen.split(','):
    if int(number) %2 == 0:
        print(f'{int(number)} is even')
    else:
        print(f'{int(number)} is odd')

Both of these will yield:

12 is even
3 is odd
7 is odd
25 is odd
771 is odd
45 is odd
6 is even
98 is even
55 is odd
546 is even

Upvotes: 1

Matt
Matt

Reputation: 518

Your code is not working because getallen is a string, not an array. You need to convert it to an array first with getallen = getallen.split(',')

Upvotes: 1

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