similar to: max and pmax of NA and NaN

Dear R users, is the following OK? > max(NA, NaN) [1] NA > max(NaN, NA) [1] NA > pmax(NaN, NA) [1] NA > pmax(NA, NaN) [1] NaN ...or is it a bug? Documentation says that NA has a higher priority over NaN. Best regards, Michal Burda [[alternative HTML version deleted]]

>>>>> Martin Maechler <maechler at stat.math.ethz.ch> >>>>> on Mon, 15 Aug 2016 11:07:43 +0200 writes: >>>>> Suharto Anggono Suharto Anggono <suharto_anggono at yahoo.com> >>>>> on Sun, 14 Aug 2016 03:42:08 +0000 writes: >> useNA <- if (missing(useNA) && !missing(exclude) && !(NA %in%

I am raising this again. With z <- complex(real = c(0,NaN,NaN), imaginary = c(NA,NA,0)) , results of sapply(z, match, table = z) and match(z, z) are different in R 3.4.0 alpha. I think they should be the same. I suggest changing 'cequal' in unique.c such that a complex number that has both NA and NaN matches NA and doesn't match NaN, as such complex number is printed as NA.

useNA <- if (missing(useNA) && !missing(exclude) && !(NA %in% exclude)) "ifany" An example where it change 'table' result for non-factor input, from https://stat.ethz.ch/pipermail/r-help/2005-April/069053.html : x <- c(1,2,3,3,NA) table(as.integer(x), exclude=NaN) I bring the example up, in case that the change in result is not intended.

The quirk as in table(1:3, exclude = 1, useNA = "ifany") is actually somewhat documented, and still in R devel r71104. In R help on 'table', in "Details" section: It is best to supply factors rather than rely on coercion. In particular, ?exclude? will be used in coercion to a factor, and so values (not levels) which appear in ?exclude? before coercion will be mapped to

Looking at the code of function 'table' in R devel r71227, I see that the part "remove NA level if it was added only for excluded in factor(a, exclude=.)" is not quite right. In is.na(a) <- match(a0, c(exclude,NA), nomatch=0L) , I think that what is intended is a[a0 %in% c(exclude,NA)] <- NA . So, it should be is.na(a) <- match(a0, c(exclude,NA),

On 'factor', I meant the case where 'levels' is not specified, where 'unique' is called. > factor(c(complex(real=NaN), complex(imaginary=NaN))) [1] NaN+0i <NA> Levels: NaN+0i Look at <NA> in the result above. Yes, it happens in earlier versions of R, too. On matching both NA and NaN, another consequence is that length(unique(.)) may depend on order.

Dear R-users >prob <- c(0.5,0.4,0.3,0.1,0.0) >cal <- prob * log(prob,base=2) >cal [1] -0.5000000 -0.5287712 -0.5210897 -0.3321928 NaN Is there any way to change NaN to zero ? I did come up with this by applying Ripley's relpy to my previous question cal <-prob*log(pmax(prob,0.00000001),base=2) Any suggestion ? Thank you Taka

Removal of ans[nas] <- NA from the code of function 'ifelse' in R is not committed (yet). Why? -------------------------------------------- On Mon, 28/11/16, Martin Maechler <maechler at stat.math.ethz.ch> wrote: Subject: Re: [Rd] ifelse() woes ... can we agree on a ifelse2() ? Cc: R-devel at r-project.org, maechler at stat.math.ethz.ch Date: Monday, 28 November, 2016, 10:00

I am basically fine with the change. How about using just the following? if(!is.character(exclude)) exclude <- as.vector(exclude, typeof(x)) # may result in NA x <- as.character(x) It looks simpler and is, more or less, equivalent. In factor.Rd, in description of argument 'exclude', "(when \code{x} is a \code{factor} already)" can be removed. A larger

That, for example, complex(real=NaN) and complex(imaginary=NaN) are regarded as equal makes it possible that length(unique(as.character(x))) > length(unique(x)) (current code of function 'factor' doesn't expect it). Yes, an argument for the behavior is that NA and NaN are of one kind. On my system, using 32-bit R for Windows from binary from CRAN, the result of sapply(z, match,

Hi, Please consider the following : x = c(1,3,NA,5) y = c(2,NA,4,1) min(x,y,na.rm=TRUE) # ok [1] 1 max(x,y,na.rm=TRUE) # ok [1] 5 sum(x,y,na.rm=TRUE) # ok [1] 16 pmin(x,y,na.rm=TRUE) # ok [1] 1 3 4 1 pmax(x,y,na.rm=TRUE) # ok [1] 2 3 4 5 psum(x,y,na.rm=TRUE) [1] 3 3 4 6 # expected result Error: could not find function "psum" # actual result

I stand corrected. The part "If set to 'NULL', it implies 'useNA="always"'." is even in the documentation in R 2.8.0. It was my fault not to check carefully. I wonder, why "always" was chosen for 'useNA' for exclude = NULL. Why exclude = NULL is so special? What about another 'exclude' of length zero, like character(0) (not c(),

On 'aggregate data.frame', the URL should be https://stat.ethz.ch/pipermail/r-help/2016-May/438631.html . vector(typeof(ans)) (or vector(storage.mode(ans))) has length zero and can be used to initialize array. Instead of if(missing(default)) , if(identical(default, NA)) could be used. The documentation could then say, for example: "If default = NA (the default), NA of appropriate

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. 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 Best, Jan [[alternative HTML version deleted]]

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?: >

By "an argument named 'use.names' is included for concatenation", I meant something like this, that someone might try. > c(as.Date("2016-01-01"), use.names=FALSE) use.names "2016-01-01" "1970-01-01" See, 'use.names' is in the output. That's precisely because 'c.Date' doesn't have 'use.names', so

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

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 at 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

Here is a patch to function 'stopifnot' that adds 'evaluated' argument and makes 'exprs' argument in 'stopifnot' like 'exprs' argument in 'withAutoprint'. --- stop.R 2019-05-30 14:01:15.282197286 +0000 +++ stop_new.R 2019-05-30 14:01:51.372187466 +0000 @@ -31,7 +31,7 @@ .Internal(stop(call., .makeMessage(..., domain = domain))) }