Checking input is a number in python

Question

I need help, my program is simulating the actions of a dice. I want to an error check to occur checking if the input string is a number and if it isn't I want to ask the question again again until he enters an integer

# This progam will simulate a dice with 4, 6 or 12 sides.

import random 

def RollTheDice():

    print("Roll The Dice")
    print()


    NumberOfSides = int(input("Please select a dice with 4, 6 or 12 sides: "))

    Repeat = True 

    while Repeat == True:


        if not NumberOfSides.isdigit() or NumberOfSides not in ValidNumbers:
            print("You have entered an incorrect value")
            NumberOfSides = int(input("Please select a dice with 4, 6 or 12 sides")

        print()
        UserScore = random.randint(1,NumberOfSides)
        print("{0} sided dice thrown, score {1}".format (NumberOfSides,UserScore))

        RollAgain = input("Do you want to roll the dice again? ")


        if RollAgain == "No" or RollAgain == "no":
            print("Have a nice day")
            Repeat = False

        else:
            NumberOfSides = int(input("Please select a dice with 4, 6 or 12 sides: "))

I understand this is one of your first python program, but please dont write in Python like it's C.... — Ant
– Ant, Commented May 3, 2013 at 10:41

Thomas Fenzl · Accepted Answer · 2013-05-07 18:48:56Z

2

As a commenter disliked my first answer with try: except ValueError and the OP asked about how to use isdigit, that's how you can do it:

valid_numbers = [4, 6, 12]
while repeat:
    number_of_sides = 0      
    while number_of_sides not in valid_numbers:
          number_of_sides_string = input("Please select a dice with 4, 6 or 12 sides: ")
          if (not number_of_sides_string.strip().isdigit() 
              or int(number_of_sides_string) not in valid_numbers):
              print ("please enter one of", valid_numbers)
          else:
              number_of_sides = int(number_of_sides_string)
    # do things with number_of_sides

the interesting line is not number_of_sides_string.strip().isdigit(). Whitespace at both ends of the input string is removed by strip, as a convenience. Then, isdigit() checks if the full string consists of numbers.

In your case, you could simply check

 if not number_of_sides_string not in ['4', '6', '12']:
     print('wrong')

but the other solution is more general if you want to accept any number.

As an aside, the Python coding style guidelines recommend lowercase underscore-separated variable names.

edited May 7, 2013 at 18:48

answered May 3, 2013 at 10:40

Thomas Fenzl

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7 Comments

Shane Over a year ago

I don't like the use of exception handling for flow control here. Much rather a call to something like text.isdigit()

Thomas Fenzl Over a year ago

it handles input with leading or trailing blanks, though. " 12".isdigit() is False

Thomas Fenzl Over a year ago

I know, I'm just wondering if text.strip().isdigit() really is more readable than the try-except.

Ahmed Abukar Over a year ago

okay if iff i use the 'isdigit()' function how do i put that into my code

Ahmed Abukar Over a year ago

i tried your code but it doesn't work can you just give me the whole code

|

John Zwinck · Accepted Answer · 2013-05-03 10:41:39Z

1

Capture the string in a variable, say text. Then do if text.isdigit().

answered May 3, 2013 at 10:41

John Zwinck

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1 Comment

Thomas Fenzl Over a year ago

If you want to be more forgiving to user input, you can strip() the input first.

Nux · Accepted Answer · 2013-05-03 10:47:14Z

0

Make a function out of:

while NumberOfSides != 4 and NumberOfSides != 6 and NumberOfSides != 12:
    print("You have selected the wrong sided dice")
    NumberOfSides = int(input("Please select a dice with 4, 6 or 12 sides: "))

And call it when you want to get input. You should also give an option to quit e.g. by pressing 0. Also you should try catch for invalid number. There is an exact example in Python doc. Note that input always try to parse as a number and will rise an exception of it's own.

answered May 3, 2013 at 10:47

Nux

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Comments

Sofia Velmer · Accepted Answer · 2013-05-03 12:18:45Z

0

You can use type method.

my_number = 4
if type(my_number) == int:
    # do something, my_number is int
else:
    # my_number isn't a int. may be str or dict or something else, but not int

Or more «pythonic» isinstance method:

my_number = 'Four'
if isinstance(my_number, int):
    # do something
raise Exception("Please, enter valid number: %d" % my_number)

answered May 3, 2013 at 12:18

Sofia Velmer

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