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`?

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

Answer 1

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