[Rd] mean(x) != mean(rev(x)) different with x <- c(NA, NaN) for some builds
Henrik Bengtsson
henrik.bengtsson at gmail.com
Sat Apr 1 07:51:24 CEST 2017
On Fri, Mar 31, 2017 at 10:14 PM, Prof Brian Ripley
<ripley at stats.ox.ac.uk> wrote:
> From ?NA
>
> Numerical computations using ‘NA’ will normally result in ‘NA’: a
> possible exception is where ‘NaN’ is also involved, in which case
> either might result.
>
> and ?NaN
>
> 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).
>
> fortunes::fortune(14) applies (yet again).
Thanks; I'm often happy to have contributed to some of the fortune
counters, but not so sure about this one. What's even worse is that
one of my own matrixStats NEWS has an entry go a few years back which
mentions "... incorrectly assumed that the value of prod(c(NaN, NA))
is uniquely defined. However, as documented in help("is.nan"), it may
be NA or NaN depending on R system/platform." I guess the joke is on
me - it's April 1st after all.
But, technically one could test for ISNA(x) for each element before
calculating the intermediate sum, but since that is a quite expensive
test it is not done and sum += x is performed "as is" on NA and NaN
(and -Inf and +Inf). Is that correct?
/Henrik
>
>
> On 01/04/2017 04:50, Henrik Bengtsson wrote:
>>
>> In R 3.3.3, I observe the following on Ubuntu 16.04 (when building
>> from source as well as for the sudo apt r-base build):
>>
>>> x <- c(NA, NaN)
>>> mean(x)
>>
>> [1] NA
>>>
>>> mean(rev(x))
>>
>> [1] NaN
>>
>>> rowMeans(matrix(x, nrow = 1, ncol = 2))
>>
>> [1] NA
>>>
>>> rowMeans(matrix(rev(x), nrow = 1, ncol = 2))
>>
>> [1] NaN
>>
>>> .rowMeans(x, m = 1, n = 2)
>>
>> [1] NA
>>>
>>> .rowMeans(rev(x), m = 1, n = 2)
>>
>> [1] NaN
>>
>>> .rowSums(x, m = 1, n = 2)
>>
>> [1] NA
>>>
>>> .rowSums(rev(x), m = 1, n = 2)
>>
>> [1] NaN
>>
>>> rowSums(matrix(x, nrow = 1, ncol = 2))
>>
>> [1] NA
>>>
>>> rowSums(matrix(rev(x), nrow = 1, ncol = 2))
>>
>> [1] NaN
>>
>> I'd expect NA to trump NaN in all cases (with na.rm = FALSE). sum()
>> does not have this problem and returns NA in both cases (*).
>>
>> For the same R version build from source on RHEL 6.6 system
>> (completely different architecture), I get the expected result (= NA)
>> for all of the above cases, e.g.
>>
>>> x <- c(NA, NaN)
>>> mean(x)
>>
>> [1] NA
>>>
>>> mean(rev(x))
>>
>> [1] NA
>> [...]
>>
>> Before going insane trying to troubleshoot this, I have a vague memory
>> that this, or something related to this, has been discussed
>> previously, but I cannot locate it.
>>
>> Is the above a bug in R, a FAQ, a build error, overzealous compiler
>> optimization, and / or ...?
>>
>> Thanks,
>>
>> Henrik
>
>
>
> --
> Brian D. Ripley, ripley at stats.ox.ac.uk
> Emeritus Professor of Applied Statistics, University of Oxford
>
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