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

[Duncan Murdoch]

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
> 
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> 
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