[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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