What is wrong with this Python code? Refuses input [duplicate]

Question

Every time I enter 4, 6 or 12 it doesn't accept it. Why? The code looks fine to me. Please tell me how to correct or what to change.

import random
def roll_the_dice():

    print("Roll The Dice")
    print()
    repeat = True

    while repeat:
        number_of_sides = input("Please select a dice with 4, 6 or 12 sides: ")
        if (number_of_sides in [4,6,12] and len(number_of_sides) == 0 and 
            number_of_sides == int):
            user_score = random.randint(1,number_of_sides)
            print("{0} sided dice thrown, score {1}".format(
                                     number_of_sides,user_score))
            roll_again = input("Do you want to roll the dice again? ")
            roll_again.lower()
            if roll_again == "no":
                print("Have a nice day")
                repeat = False
            elif len(roll_again) == 0:
                print("Error please type 'yes' or 'no'")
                roll_again = input("Do you want to roll the dice again? ")
        else:
            print("You have entered an incorrect value, please try again")
            number_of_sides = input("Please select a dice with 4, 6 or 12 sides: ")

I'm not sure how it accepts any value. number_of_sides cannot simultaneously be one of 4, 6, or 12 and be a string of length 0. In no case can you enter a string that is equal to the type int. — chepner
– chepner, Commented Jun 8, 2013 at 13:31

TerryA · Accepted Answer · 2013-06-08 12:00:46Z

7

In Python 3, when using input(), it returns a string. Thus, you would have something like "4". And "4" is not 4.

So in your script, specifically at the if number_of_sides in [4,6,12], it will always be False, because you are really saying if "4" in [4,6,12] (I'm just doing 4 as an example).

Convert the string to an integer:

>>> int("4")
4

It also looks like you are trying to determine if an input was given. len(...) == 0 is not needed. You can just say if number_of_sides. Because an empty string is False, and if one was entered, then the if-statement will not execute.

Also, number_of_sides == int is not the way to check if an object is an integer. Use isinstance():

>>> isinstance("4", int)
False
>>> isinstance(4, int)
True

Some other tiny things:

.lower() does not sort the string in place, as strings are immutable in python. You might just want to attach .lower() onto the end of the input().

You might also want to use a while loop for your second input. Observe:

roll_again = ''
while True:
    roll_again = input('Do you want to roll the dice again? ')
    if roll_again in ('yes', 'no'):
        break
    print("You have entered an incorrect value, please try again")

if roll_again == "no":
    print("Have a nice day")
    repeat = False
else:
    print("Let's go again!")

edited Jun 8, 2013 at 12:00

answered Jun 8, 2013 at 11:46

TerryA

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12 Comments

Ulrich Eckhardt Over a year ago

Even if it was an integer, calling len() on it doesn't work. Maybe it should be n in ['4', '6', '12'] instead?

Klaus Byskov Pedersen Over a year ago

number_of_sides == int would also always return False.

TerryA Over a year ago

@UlrichEckhardt Or just if int(number_of_sides) in [4, 6, 12], but you are absolutely correct yes.

Ulrich Eckhardt Over a year ago

There is too much wrong that doesn't work. The OP should have made smaller steps and/or reduced the code to the smallest possible piece that doesn't behave as expected. Oh, and saying the exact error and expectations help, too.

Ahmed Abukar Over a year ago

how about if i do int(input("Please select a dice with 4, 6 or 12 sides: "))

|

Burhan Khalid · Accepted Answer · 2013-06-08 12:09:05Z

2

Haidro gave you the reason, but here is a different way to approach your problem:

def get_dice_size():
    dice_size = input('Enter the number of sides on the dice: ')
    while dice_size not in ['4','6','12']:
       print 'Sorry, please enter one of 4, 6 or 12:'
       dice_size = input('Enter the number of sides on the dice: ')
    return int(dice_size)

def main():
   dice_size = get_dice_size()
   repeat = True

   while repeat:
       print('Rolling the dice...')
       user_score = random.randint(1,dice_size)
       print("{0} sided dice thrown, score {1}".format(dice_size,user_score))
       roll_again = input("Do you want to roll the dice again? ")
       if roll_again.lower() == 'yes':
           dice_size = get_dice_size()
       else:
           repeat = False

   print('Thank you for playing!')

edited Jun 8, 2013 at 12:09

answered Jun 8, 2013 at 11:58

Burhan Khalid

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3 Comments

TerryA Over a year ago

OP is using python 3, I'm presuming by the syntax. So print() and not print :)

Burhan Khalid Over a year ago

Well, strictly speaking print() will work in Python 2.7 as well, but you are probably right.

TerryA Over a year ago

Also, this looks really clean. I'm upvoting this.

Mike Müller · Accepted Answer · 2013-06-08 12:53:23Z

0

You should change your script like shown below.

This is the important part:

    try:
        number_of_sides = int(input("Please select a dice with 4, 6 or 12 sides: "))
    except ValueError:
        wrong = True

Convert to int right at input time with int(input("…. If the user enters anything that cannot be converted into an integer Python will raise a ValueError. You can catch this, show a message and go to the start of the loop with continue.

import random


def roll_the_dice():

    print("Roll The Dice")
    print()
    repeat = True

    while repeat:
        wrong = False
        try:
            number_of_sides = int(input("Please select a dice with 4, 6 or 12 sides: "))
        except ValueError:
            wrong = True
        if wrong or number_of_sides not in [4,6,12]:
            print("You have entered an incorrect value, please try again")
            continue
        else:
            user_score = random.randint(1,number_of_sides)
            print("{0} sided dice thrown, score {1}".format(
                                     number_of_sides,user_score))
            roll_again = input("Do you want to roll the dice again? ")
            if roll_again in ('y', 'Y', 'yes', 'Yes'):
                continue
            else:
                print("Have a nice day")
                repeat = False

roll_the_dice()

edited Jun 8, 2013 at 12:53

answered Jun 8, 2013 at 12:05

Mike Müller

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4 Comments

Burhan Khalid Over a year ago

Logic here needs some work. Conder at the first while loop, someone enters "1356"; what will happen?

Mike Müller Over a year ago

1356 is not in [2, 4, 6] so else gets these. Screen output: $ Please select a dice with 4, 6 or 12 sides: 1356 $ You have entered an incorrect value, please try again $ Please select a dice with 4, 6 or 12 sides:

Burhan Khalid Over a year ago

And then after that? It will ask again "Please select a dice with 4, 6 or 12 sides:", even if I enter 4 at the second prompt :)

Ahmed Abukar Over a year ago

@MikeMüller is there anyway of doing with the try: method

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