Only allowing Integer input in python 3.3.2

Question

Hi i am trying to make the program only accept the numbers 0, 4, 6, and 12, and not allow anything else to be inputted. So far i have been successful in only allowing certain integers to be entered, however i am having trouble with not allowing any letters to be entered. When a letter is entered, the entire program crashes. Please could you help me only allow integers to be entered? Thank you.

My code is below:

from random import randint 
def simul():
    dice = int(input("What sided dice would you like to roll? 4, 6 or 12? 0 to not roll:"))
    if dice != 4 and dice!=6 and dice!=12 and dice!=0:
        print('You must either enter 4, 6, or 12')
        simul()
    elif dice==0:
        exit()
    else:
        while dice !=0 and dice==4 or dice==6 or dice ==12 :
            print (randint(1,dice))
            dice = int(input("What sided dice would you like to roll? 4, 6 or 12? press 0 to stop."))
simul()

It's because you're casting the input to an int and then checking the values. Check the values before you call int() — reggaeguitar, Dec 30 '14 at 17:38
So should i convert the input into an int after checking the values? — pythoner888, Dec 30 '14 at 17:39

score 1 · Answer 1 · edited Oct 15 '22 at 17:25

1

put it in a try catch block like so:

try:
    choice = int(raw_input("Enter choice 1, 2 or 3:"))
    if not (1 <= choice <= 3):
        raise ValueError()
except ValueError:
    print "Invalid Option, you needed to type a 1, 2 or 3...."
else:
    print "Your choice is", choice

copied from: limit input to integer only (text crashes PYTHON program)

edited Oct 15 '22 at 17:25

miken32

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answered Dec 30 '14 at 17:40

Ganesh Kamath - 'Code Frenzy'

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score 1 · Accepted Answer · edited Dec 30 '14 at 20:13

1

A couple of things you could look for in your code:

using try/catch is the recommended way to test input for many reasons including knowing the exact cause of the error
you can reduce some of your ifs and elses by thinking a little more about how they are nested
having the function call itself and using a while loop isn't the best way, use one or the other
in your case, you don't really need to allow only integer input, what you're looking for is to only allow a 0, 4, 6, or 12, which you do with the if statement

from random import randint
def simul():
    while True:
        try:
            dice = int(input("What sided dice would you like to" \
                    " roll? 4, 6 or 12? 0 to not roll: "))
            if dice not in (4, 6, 12, 0):
                raise ValueError()
            break  # valid value, exit the fail loop
         except ValueError:
            print("You must enter either 4, 6, 12, or 0")

    if dice == 0:
        return 0

    print(randint(1, dice))
    return dice

if __name__ == '__main__':
    while simul() != 0:
        pass

edited Dec 30 '14 at 20:13

zopieux

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answered Dec 30 '14 at 17:54

Joseph Hansen

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ok, i see what you have done here. Very silly of me to not include the try and except function in the code. Thank you so much for this!!! – pythoner888 Dec 30 '14 at 18:06
2

You should try to avoid recursion when possible, because the stack will grow everytime there's an error. In this case, you could trivially use a loop instead. – Antoine Pietri Dec 30 '14 at 18:14
What @serialk said. Another way to improve the code (and address the stack issue) would be to code other ways to escape the program besides just entering 0, such as a timer or a counter that counts the number of times input is received. – Joseph Hansen Dec 30 '14 at 18:19
FYI, I updated my answer. It's been a while since I coded in Python. My original answer used `raw_input()` and `print` where as Python 3.0+ says the correct syntax is `input()` and `print()`. – Joseph Hansen Dec 30 '14 at 18:21
I updated your code to something a bit cleaner, using the neat `in foo` syntax for validation and an outer infinite loop as the "main" program. As said, using recursion would eventually crash your program. – zopieux Dec 31 '14 at 14:16
@Zopieux Thanks. The changes are helpful. I just meant to be similar to the OP. – Joseph Hansen Dec 31 '14 at 23:14

score 1 · Answer 3 · answered Dec 30 '14 at 18:06

I would encapsulate the "constrained input" functionality into a separate, reusable function:

def constrained_int_input(prompt, accepted, toexit=0):
    msg = '%s? Choose %s (or %s to exit)' % (
       prompt, ', '.join(str(x) for x in sorted(accepted)), toexit)
   while True:
       raw = raw_input(msg)
       try:
           entered = int(raw)
       except ValueError:
           print('Enter a number, not %r' % raw)
           continued
       if entered == toexit or entered in accepted:
           return entered
       print('Invalid number: %r -- please enter a valid one' % entered)

Now you can call e.g

dice = constrained_int_input('What sided dice would you like to roll', (4,6,12))

whenever required and be sure that dice will end up with one of the accepted integers, including possibly the to-exit value of 0.

score 0 · Answer 4 · answered Dec 30 '14 at 17:56

while True:
    x=input("4,6 or 12? 0 to not roll: ")
    if x.isalpha():
        print ("only numbers.")
        continue
    elif int(x)==0:
        print ("quiting..")
        break
    elif int(x)!=4 and int(x)!=6 and int(x)!=12:
        print ("Not allowed.")
    else:
        print (random.randint(1,int(x)))

Here is another method, use isalpha().

Only allowing Integer input in python 3.3.2

4 Answers4