# [Numpy-discussion] Is this odd?

Ryan May rmay31@gmail....
Fri Apr 2 09:28:19 CDT 2010

```On Thu, Apr 1, 2010 at 10:07 PM, Shailendra <shailendra.vikas@gmail.com> wrote:
> Hi All,
> Below is some array behaviour which i think is odd
>>>> a=arange(10)
>>>> a
> array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>>> b=nonzero(a<0)
>>>> b
> (array([], dtype=int32),)
>>>> if not b[0]:
> ...     print 'b[0] is false'
> ...
> b[0] is false
>
> Above case the b[0] is empty so it is fine it is considered false
>
>>>> b=nonzero(a<1)
>>>> b
> (array([0]),)
>>>> if not b[0]:
> ...     print 'b[0] is false'
> ...
> b[0] is false
>
> Above case b[0] is a non-empty array. Why should this be consider false.
>
>>>> b=nonzero(a>8)
>>>> b
> (array([9]),)
>>>> if not b[0]:
> ...     print 'b[0] is false'
> ...
>>>>
> Above case b[0] is non-empty and should be consider true.Which it does.
>
> I don't understand why non-empty array should not be considered true
> irrespective to what value they have.
> Also, please suggest the best way to differentiate between an empty
> array and non-empty array( irrespective to what is inside array).

But by using:

if not b[0]:

You're not considering the array as a whole, you're looking at the
first element, which is giving expected results.  As I'm sure you're
aware, however, you can't simply do:

if not b: # Raises exception

So what you need to do is:

if b.any():

or:

if b.all()

Now for determining empty or not, you'll need to look at len(b) or b.shape

Ryan

--
Ryan May