How can I limit the user input to only integers in Python

Question

I\'m trying to make a multiple choice survey that allows the user to pick from options 1-x. How can I make it so that if the user enters any characters besides numbers, retu

Answer 1

The best way would be to use a helper function which can accept a variable type along with the message to take input.

def _input(message,in_type=str):
    while True:
      try:
        return in_type (input(message))
    except:pass

if __name__ == '__main__':
    _input("Only accepting integer : ",int)
    _input("Only accepting float : ",float)
    _input("Accepting anything as string : ")

So when you want an integer , you can pass it that i only want integer, just in case you can accept floating number you pass the float as a parameter. It will make your code really slim so if you have to take input 10 times , you don't want to write try catch blocks ten times.

Answer 2

Your code would become:

def Survey():

    print('1) Blue')
    print('2) Red')
    print('3) Yellow')

    while True:
        try:
            question = int(input('Out of these options\(1,2,3), which is your favourite?'))
            break
        except:
            print("That's not a valid option!")

    if question == 1:
        print('Nice!')
    elif question == 2:
        print('Cool')
    elif question == 3:
        print('Awesome!')
    else:
        print('That\'s not an option!')

The way this works is it makes a loop that will loop infinitely until only numbers are put in. So say I put '1', it would break the loop. But if I put 'Fooey!' the error that WOULD have been raised gets caught by the except statement, and it loops as it hasn't been broken.

Answer 3

One solution amongst others : use the type function or isinstance function to check if you have an ̀int or a float or some other type

>>> type(1)
<type 'int'>

>>> type(1.5)
<type 'float'>

>>> isinstance(1.5, int)
False

>>> isinstance(1.5, (int, float))
True   

Answer 4

I would catch first the ValueError (not integer) exception and check if the answer is acceptable (within 1, 2, 3) or raise another ValueError exception

def survey():
    print('1) Blue')
    print('2) Red')
    print('3) Yellow')

    ans = 0
    while not ans:
        try:
            ans = int(input('Out of these options\(1, 2, 3), which is your favourite?'))
            if ans not in (1, 2, 3):
                raise ValueError
        except ValueError:
            ans = 0
            print("That's not an option!")

    if ans == 1:
        print('Nice!')
    elif ans == 2:
        print('Cool')
    elif ans == 3:
        print('Awesome!')
    return None