[Numpy-discussion] numpy.sign(numpy.nan)?????

[Stuart Brorson]

Hi guys,

Just a quick note.  I've been playing with NumPy again, looking at
corner cases of function evaluation.  I noticed this:

In [66]: numpy.sign(numpy.nan)
Out[66]: 0.0

IMO, the output should be NaN, not zero.

If you agree, then I'll be happy to file a bug in the NumPy tracker.
Or if somebody feels like pointing me to the place where this is
implemented, I can submit a patch.  (I grepped through the source for
"sign" to see if I could figure it out, but it occurs so frequently
that it will take longer than 5 min to sort it all out.)

Before I did anything, however, I thought I would solicit the opinions
of other folks in the NumPy community about the proper behavior of
numpy.sign(numpy.nan).


Cheers,

Stuart Brorson 
Interactive Supercomputing, inc. 
135 Beaver Street | Waltham | MA | 02452 | USA 
http://www.interactivesupercomputing.com/