[R] prod(NaN, NA) vs. prod(NA, NaN)

[Barry Rowlingson]

I'm having deja-vu of a similar discussion on R-devel:

https://stat.ethz.ch/pipermail/r-devel/2018-July/076377.html

This was the funniest inconsistency I could find:

 > sum(c(NaN,NA))
 [1] NaN
 > sum(NaN,NA)
 [1] NA

THEY'RE IN THE SAME ORDER!!!

The doc in ?NaN has this clause:

     In R, basically all mathematical functions (including basic
     ‘Arithmetic’), are supposed to work properly with ‘+/- Inf’ and
     ‘NaN’ as input or output.

which doesn't define "properly", but you'd think commutativity was a
"proper" property of addition. So although they "are supposed to" they
don't. Naughty mathematical functions!

And then there's...

     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).

Which is at least telling you there is vagueness in the system. But
hey, mathematics is not a precise science... oh wait...

Barry





On Tue, Jul 3, 2018 at 10:09 PM, Rolf Turner <r.turner using auckland.ac.nz> wrote:
>
> On 04/07/18 00:24, Martin Møller Skarbiniks Pedersen wrote:
>
>> Hi,
>>    I am currently using R v3.4.4 and I just discovered this:
>>
>>> prod(NA, NaN) ; prod(NaN, NA)
>> [1] NA
>> [1] NaN
>>
>> ?prod says:
>>      If ‘na.rm’ is ‘FALSE’ an ‘NA’ value in any of the arguments will
>>       cause a value of ‘NA’ to be returned, otherwise ‘NA’ values are
>>       ignored.
>>
>> So according to the manual-page for prod() NA should be returned in both
>> cases?
>>
>>
>> However for sum() is opposite is true:
>>> sum(NA, NaN) ; sum(NaN, NA)
>> [1] NA
>> [1] NA
>>
>> ?sum says:
>>      If ‘na.rm’ is ‘FALSE’ an ‘NA’ or ‘NaN’ value in any of the
>>       arguments will cause a value of ‘NA’ or ‘NaN’ to be returned,
>>       otherwise ‘NA’ and ‘NaN’ values are ignored.
>>
>>
>> Maybe the manual for prod() should say the same as sum() that
>> both NA and NaN can be returned?
>
> But:
>
>  > sum(NA,NaN)
> [1] NA
>  > sum(NaN,NA)
> [1] NA
>
> so sum gives NA "both ways around".  Perhaps a slight inconsistency
> here?  I doubt that it's worth losing any sleep over, however.
>
> Interestingly (???):
>
>  > NaN*NA
> [1] NaN
>  > NA*NaN
> [1] NA
>  > NaN+NA
> [1] NaN
>  > NA+NaN
> [1] NA
>
> So we have an instance of non-commutative arithmetic operations.  And
> sum() is a wee bit inconsistent with "+".
>
> Again I doubt that the implications are all that serious.
>
> cheers,
>
> Rolf Turner
>
> --
> Technical Editor ANZJS
> Department of Statistics
> University of Auckland
> Phone: +64-9-373-7599 ext. 88276
>
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