On Wed, Oct 2, 2013 at 2:49 PM, <[email protected]> wrote: > On Wed, Oct 2, 2013 at 2:05 PM, Stéfan van der Walt <[email protected]> wrote: >> On 2 Oct 2013 19:14, "Benjamin Root" <[email protected]> wrote: >>> >>> And it is logically consistent, I think. a[nanargmax(a)] == nanmax(a) >>> (ignoring the silly detail that you can't do an equality on nans). >> >> Why do you call this a silly detail? It seems to me a fundamental flaw to >> this approach. > > a nan is a nan is a NaN > >>>> np.testing.assert_equal([0, np.nan], [0, np.nan]) >>>>
and the functions have "nan" in their names nan in - NaN out what about nanmean, nansum, ...? Josef > > Josef > >> >> Stéfan >> >> >> _______________________________________________ >> NumPy-Discussion mailing list >> [email protected] >> http://mail.scipy.org/mailman/listinfo/numpy-discussion >> _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
