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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