i get correct output while compiling with 'python' but not with 'python3'

Question

In the below python code

    var=input("Enter a number between 1 to 10:")
    if (var==1 or var==2 or var==3 or var==4 or var==5):
          print ('the entered number is between 1 to 5')
    elif (var==6 or var==7 or var==8 or var==9 or var==10):
          print ('The entered number is between 5 to 10')
    else:
          print ('Wrong value exiting!!')

When i run it in terminal~$ python name.py

Enter a number between 1 to 10:3

the entered number is between 1 to 5

when i run it in terminal~$ python3 name.py

Enter a number between 1 to 10:3

Wrong value exiting!!

What makes the difference ? & what i have to change in order to get the correct output when i compile with python3`?

is there any simpler way of comparison instead of using "or" for every number ?

Jon Clements · Accepted Answer · 2012-10-20 15:06:21Z

4

In Python 2:

var=input("Enter a number between 1 to 10:")

The above evaluates the expression, so it ends up with var potentially equalling a number. (And as a note should really be avoided)

In Python 3, input is the equiv. of raw_input in Python 2, so it returns a string - if you expect it to be a number you need to do var = int( input("...") ) and be aware of any conversion errors that may occur:

There is the in operator:

if var in (1, 2, 3, 4, 5):
    pass
elif var in (6, 7, 8):
    pass
else:
    pass

Or, if you want a between operation, you can make use of the Python logic system:

if 1 <= var <= 5:
    pass

answered Oct 20, 2012 at 15:06

Jon Clements

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RocketDonkey · Accepted Answer · 2012-10-20 15:11:02Z

2

I believe this is due to your use of input - in Python 3.x, it is the same as raw_input in Python 2.x, meaning that it is turning it into a string as opposed to an integer. As for your comparison, you could try using range:

var=int(input("Enter a number between 1 to 10:"))

if var in range(1, 6):
      print ('the entered number is between 1 to 5')
elif var in range(6,11):
      print ('The entered number is between 5 to 10')
else:
      print ('Wrong value exiting!!')

edited Oct 20, 2012 at 15:11

answered Oct 20, 2012 at 15:05

RocketDonkey

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5 Comments

Paolo Moretti Over a year ago

To get the old behavior of input(), you can use eval(input())

Markus Unterwaditzer Over a year ago

@PaoloMoretti For gods sake, don't give OP ideas!

RocketDonkey Over a year ago

@MarkusUnterwaditzer Haha, after all of the warnings against eval I've seen on this site, I'm afraid to even type it.

Jon Clements Over a year ago

At the very least - ast.literal_eval(input())

Paolo Moretti Over a year ago

@Markus You are rigth, I was just quoting the official py3k release notes. Using eval() is really a bad practice.

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