On Wed, Oct 2, 2013 at 12:51 PM, <[email protected]> wrote:
> 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, ...?
>
nanmean returns nan for empty slices while nansum returns nan in 1.8,
consistent with previous behavior, and will return 0 in 1.9.
The main problem I had was deciding what arg{max, min} should return as the
return value is an integer. I like your suggestion of returning 0.
One further possibility is to add a keyword 'raise' to make the behavior
selectable.
Chuck
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