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]) >>> 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
