Matthieu Brucher a écrit : > You could use numpy.nansum()/numpy.sum(~numpy.isnan()) or something > like this ? Arf, did not even think that there was a sum for nan, altough I do know & use nanmin() & nanmax(). Sorry for the noise, I'm working on too many things at same time ;-) Thanks. -- http://scipy.org/FredericPetit