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

[Tomas Kalibera]

Yes, the performance overhead of fixing this at R level would be too 
large and it would complicate the code significantly. The result of 
binary operations involving NA and NaN is hardware dependent (the 
propagation of NaN payload) - on some hardware, it actually works the 
way we would like - NA is returned - but on some hardware you get NaN or 
sometimes NA and sometimes NaN. Also there are C compiler optimizations 
re-ordering code, as mentioned in ?NaN. Then there are also external 
numerical libraries that do not distinguish NA from NaN (NA is an R 
concept). So I am afraid this is unfixable. The disclaimer mentioned by 
Duncan is in ?NaN/?NA, which I think is ok - there are so many numerical 
functions through which one might run into these problems that it would 
be infeasible to document them all. Some functions in fact will preserve 
NA, and we would not let NA turn into NaN unnecessarily, but the 
disclaimer says it is something not to depend on.

Tomas

On 07/03/2018 11:12 AM, Jan Gorecki wrote:
> Thank you for interesting examples.
> I would find useful to document this behavior also in `?mean`, while `+`
> operator is also affected, the `sum` function is not.
> For mean, NA / NaN could be handled in loop in summary.c. I assume that
> performance penalty of fix is the reason why this inconsistency still
> exists.
> Jan
>
> On Mon, Jul 2, 2018 at 8:28 PM, Barry Rowlingson <
> b.rowlingson using lancaster.ac.uk> wrote:
>
>> 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
>>>>
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