[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
Graduate Research Assistant
School of Meteorology
University of Oklahoma
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