[Numpy-discussion] numpy.average(fooArray, axis=0): dropping nans from calculation

Sven Schreiber svetosch at gmx.net
Fri Jul 14 03:51:44 CDT 2006


Curzio Basso schrieb:
>> Well try it out and see for yourself ;-)
> 
> good point :-)
> 
>> But for sums it doesn't make a difference, right... Note that it's
>> called nan*sum* and not nanwhatever.
> 
> sure, I was still thinking about the first post which was referring to
> the average...
> 
> qrz
> 

Right, having to count the Nans then makes the masked-array solution
more attractive, that's true. So maybe that's a feature request,
complementing the nansum function by a nanaverage?






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