Python - check if array is empty

Question

I have an array of two int's, and I want to check if either is None, so I have this:

print hourArray
if hourArray[0] or hourArray[1] is None:
    print "FAILED???"
else:
    print "array is full"

And even though the print hourArray shows this right before the if statement

[2040, 2640]

It prints FAILED??? even though neither of the elements in the array is None?

Why is this happening?

`if (hourArray[0] is None) or (hourArray[1] is None)` or `if None in [hourArray[0], hourArray[1]]` — Martin Thoma, Jun 23 '15 at 18:25
`if a or b == something` is an incredibly common mistake in Python, see the duplicate. Firstly it should be `if a == something or b == something`. Secondly you should use `==` instead of `is` — Cory Kramer, Jun 23 '15 at 18:25
@CoryKramer Comparing to singletons like `None` should be done with `is` in python, according to [pep8](https://www.python.org/dev/peps/pep-0008/). — jme, Jun 23 '15 at 18:27

Anand S Kumar · Accepted Answer · 2015-06-23T18:27:31.837

2

The issue is that you are checking if (hourArray[0]) or (hourArray[1] is None) , all non-zero integer values are always true.

You should do -

if hourArray[0] is None or hourArray[1] is None:

Example of non-zero integer values being true -

>>> if 1:
...     print('Hello')
...
Hello

edited Jun 23 '15 at 18:27

answered Jun 23 '15 at 18:25

Anand S Kumar

1 Answers1