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).