> On 20 Jan 2018, at 07:53 , Suharto Anggono Suharto Anggono via R-help
<r-help at r-project.org> wrote:
>
> Extremes.Rd, that documents 'max' and 'pmax', has this in
"Details" section, in the paragraph before the last.
> By definition the min/max of a numeric vector containing an NaN is NaN,
except that the min/max of any vector containing an NA is NA even if it also
contains an NaN.
...but how do you infer that this applies to pmin/pmax? You may want it to, but
it specifically talks about the non-parallel min/max.
-pd
>
> ------------------
>>>>>> Michal Burda <michal.burda at centrum.cz>
>>>>>> on Mon, 15 Jan 2018 12:04:13 +0100 writes:
>
>> Dear R users, is the following OK?
>
>>> max(NA, NaN)
>> [1] NA
>>> max(NaN, NA)
>> [1] NA
>>> pmax(NaN, NA)
>> [1] NA
>>> pmax(NA, NaN)
>> [1] NaN
>
>> ...or is it a bug?
>
>> Documentation says that NA has a higher priority over NaN.
>
> which documentation ??
> [That would be quite a bit misleading I think. So, it should be amended
...]
>
>> Best regards, Michal Burda
>
>
> R's help pages are *THE* reference documentation and they have
> (for a long time, I think) had :
>
> ?NaN has in its 3rd 'Note:'
>
> Computations involving ?NaN? will return ?NaN? or perhaps ?NA?:
> which of those two is not guaranteed and may depend on the R
> platform (since compilers may re-order computations).
>
> Similarly, ?NA contains, in its 'Details':
>
> Numerical computations using ?NA? will normally result in ?NA?: a
> possible exception is where ?NaN? is also involved, in which case
> either might result (which may depend on the R platform). ........
>
> -----
>
> Yes, it is a bit unfortunate that this is platform dependent; if
> we wanted to make this entirely consistent (as desired in a
> perfect world), I'm almost sure R would become slower because
> we'd have to do add some explicit "book keeping" / checking
> instead of relying on the underlying C library code.
>
> Note that for these reasons, often NaN and NA should not be
> differentiated, and that's reason why using is.na(*) is
> typically sufficient and "best" -- it gives TRUE for both NA and
NaN.
>
>
> Martin Maechler
> ETH Zurich
>
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