OR behaviour in python:

Question

I have written the following piece of code, all i want to do is print a yes for if the number passed is a string representation of 1, 0 or 2 and for everything else a false:

Answer 1

The problem is that your code, to Python, reads like this:

if (number is "1") or "0" or "2":

And as any non-empty string evaluates to True, it's always True.

To do what you want to do, a nice syntax is:

if number in {"1", "0", "2"}:

Note my use of a set here - while it doesn't matter too much in this case (with only three values) checking against a set is faster than a list, as a membership test for a set is O(1) instead of O(n).

This is simply a nicer and easier of writing this:

if number == "1" or number == "0" or number == "2":

Which is what you wanted.

Note when making a comparison for value you should always use == not is - is is an identity check (the two values are the same object). Generally you should use is for stuff like is True or is None.

If you wanted to handle this as a number, you could do something like this:

try:
   value = int(number)
except ValueError:
   value = None
if value is not None and 0 <= value <= 2:
    ...

Which could be more useful in situations where you want to compare to a large range of numbers. Note my use of Python's useful comparison chaining (0 <= value <= 2 rather than 0 <= value and value <= 2).