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, return something like "That's an invalid answer"

def Survey():
    print('1) Blue')
    print('2) Red')
    print('3) Yellow')
    question = int(input('Out of these options\(1,2,3), which is your favourite?'))
    if question == 1:
        print('Nice!')
    elif question == 2:
        print('Cool')
    elif question == 3:
        print('Awesome!')
    else:
        print('That\'s not an option!')

Solution 1

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.

Solution 2

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, input_type=str):
    while True:
      try:
        return input_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.

Solution 3

def func():
    choice = "Wrong"
    
    while choice.isdigit()==False :
        choice = input("Enter a number: ")
        
        if choice.isdigit()==False:
            print("Wrongly entered: ")
        else:
            return int(choice)

Solution 4

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   

Solution 5

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

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Comments

[Ahsan Roy]

It does not answer the question specifically. He wants to convert entered str type into int.

[Ahsan Roy]

Can be made lot cleaner with a helper function

[Cyrille]

@AhsanRoy You are right. I'm not sure why I answered that 5 years ago ! I think I answered to this specific part: "How can I make it so that if the user enters any characters besides numbers, return ..."

[kodahScripts]

This is my favorite and I think should be the accepted answer, thank you @ashan-roy appreciate it.