[Numpy-discussion] handling of inf

Kilian Koepsell python@koepsell...
Fri Jun 22 03:19:07 CDT 2007


Hi,

I was wondering if the numpy function 'isinf' should return True for  
complex infinity.
I encountered the following behavior that could be considered a bug:

 >>> import numpy as N
 >>> N.isinf(1j*N.inf)
True

 >>> 1j/(N.array(1)-1)
(nannanj)
 >>> N.isinf(1j/(N.array(1)-1))
False

 >>> 1j/(N.array([1])-1)
array([                  nan                 +nanj])
 >>> N.isinf(1j/(N.array([1])-1))
array([False], dtype=bool)

Thanks,
   Kilian

--
Kilian Koepsell
Redwood Center for Theoretical Neuroscience
Helen Wills Neuroscience Institute, UC Berkeley
132 Barker Hall, #3190, Berkeley, CA 94720-3190




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