On Fri, May 16, 2008 at 11:23 AM, Stuart Brorson <[EMAIL PROTECTED]> wrote: > 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).
You're probably right. I would like to see what other systems do before changing it, though. The implementation is actually in a #define macro in umathmodule.c.src. Look for _SIGN1 (and _SIGNC if you want to clean up the complex versions, too). -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
