Input Types

[Jeff Epler]

Here's a function that can do the job.

def typed_input(prompt="", convert=int, catch=ValueError,
                eprompt="Invalid input."):
    while 1:
        s = raw_input(prompt)
        try:
            return convert(s)
        except catch:
            print eprompt

Usage:
>>> typed_input("Enter an integer: ")
Enter an integer: asdf
Invalid input
Enter an integer: 3
3
>>> typed_input("Enter a number: ", float)
Enter a number: 
Invalid input
Enter a number: 3.14
3.1400000000000001

How does it work?

1. The "while 1" loop is repeated until the "return" statement is
successfully executed.

2. int(s) and float(s) convert a string argument s to the specified
type, or raise the ValueError exception

3.  When there is a ValueError exception, the "error prompt" is printed,
and the while loop returns to the top to try again.

You can also write your own "convert" function to restrict values to a
range, etc:

>>> def int_0_100(s):
...     i = int(s)
...     if i < 0 or i > 100:
...         raise ValueError  # out of range
...     return i
...
>>> typed_input("Enter a number from 0 to 100: ", int_0_100)
Enter a number from 0 to 100: asdf
Invalid input.
Enter a number from 0 to 100: 101
Invalid input.
Enter a number from 0 to 100: 37
37

You can also specify the catch= or eprompt= arguments to change the
exception that is caught or the error prompt string.

Jeff