On Thu, Apr 1, 2010 at 10:07 PM, Shailendra <shailendra.vi...@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 _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
