[Rd] base::mean not consistent about NA/NaN

[Barry Rowlingson]

And for a starker example of this (documented) inconsistency,
arithmetic addition is not commutative:

 > NA + NaN
 [1] NA
 > NaN + NA
 [1] NaN



On Mon, Jul 2, 2018 at 5:32 PM, Duncan Murdoch <murdoch.duncan using gmail.com> wrote:
> On 02/07/2018 11:25 AM, Jan Gorecki wrote:
>> Hi,
>> base::mean is not consistent in terms of handling NA/NaN.
>> Mean should not depend on order of its arguments while currently it is.
>
> The result of mean() can depend on the order even with regular numbers.
> For example,
>
>  > x <- rep(c(1, 10^(-15)), 1000000)
>  > mean(sort(x)) - 0.5
> [1] 5.551115e-16
>  > mean(rev(sort(x))) - 0.5
> [1] 0
>
>
>>
>>      mean(c(NA, NaN))
>>      #[1] NA
>>      mean(c(NaN, NA))
>>      #[1] NaN
>>
>> I created issue so in case of no replies here status of it can be looked up
>> at:
>> https://bugs.r-project.org/bugzilla/show_bug.cgi?id=17441
>
> The help page for ?NaN says,
>
> "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)."
>
> And ?NA says,
>
> "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). "
>
> So I doubt if this inconsistency will be fixed.
>
> Duncan Murdoch
>
>>
>> Best,
>> Jan
>>
>>       [[alternative HTML version deleted]]
>>
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