similar to: NA in table with integer types

Hi -- I have found that by default the function print.glm() uses 3 significant figures when printing out null and model deviances and the aic. Of course, this is not wrong. But if a person fitted two nested models and compared the resulting deviances obtained from print.glm(), the resulting hypothesis test could indeed be wrong because of this rounding. The function summary() applied to a glm

Hi guys. Check this out: > NaN +NA [1] NaN > NA + NaN [1] NA I thought "+" was commutative by definition. What's going on? > R.version _ platform powerpc-apple-darwin6.8 arch powerpc os darwin6.8 system powerpc, darwin6.8 status major 1 minor 9.0 year 2004 month 04 day 12 language R > (Both give NA under linux, so it looks

>>>>> 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%

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

Hi I want to extract a row from a data.frame but I want that object to be a vector . After trying some different ways I end up always with a data.frame or with the wrong vector. Any pointers? x <- data.frame(a = factor(c('a',2,'b')), b = c(4,5,6)) I want to get "a" "4" I tried: as.vector(x[1,]) a b 1 a 4 (resulting in a data.frame even after in my

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),

Hi the list, I am writing a R function that call a C function. The C function needs integers but I do not manage to give a NA integer as argument : --- C code --- void essai(int *t){ Rprintf("\nT0=%i T1=%i T2=%i T3=%i",t[0],t[1],t[2],t[3]); } --- R --- boub <- c(1,2,3,4) .C("pour",as.integer(boub),NAOK=TRUE) # T0=1 T1=2 T2=3 T3=4[[1]] # [1] 1 2 3 4 boub <-

>>>>> 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% exclude)) "ifany" > An example where it change 'table' result for non-factor input, from

Frank Harrell suggested I re-post with information about the version of R Heres's the information: > > version > _ > platform powerpc-apple-darwin6.8 > arch powerpc > os darwin6.8 > system powerpc, darwin6.8 > status > major 2 > minor 0.1 > year 2004 > month 11 > day 15 > language R Here's what happens

Greetings I run into a compilation error when updating to mgcv_1.1-7 in R 2.0.0 on OS X 10.3. Note that other pacakges have compiled nicely. Some details are given below, but in short it looks like it's seeking for /usr/local/lib/powerpc-apple-darwin6.8/3.4.2/ which I don't have. But I do have /usr/lib/gcc/darwin/3.3 i.e a lower version of GCC in a different directory. More

Full_Name: Dennis Wolf Version: 1.9.0 OS: Mac OS 10.3.3 Submission from: (NULL) (160.91.76.23) platform powerpc-apple-darwin6.8 arch powerpc os darwin6.8 system powerpc, darwin6.8 status major 1 minor 9.0 year 2004 month 04 day

I'm having trouble grokking complex NaN's. This first set examples using complex(re=NaN,im=NaN) give what I expect > Re(complex(re=NaN, im=NaN)) [1] NaN > Im(complex(re=NaN, im=NaN)) [1] NaN > Arg(complex(re=NaN, im=NaN)) [1] NaN > Mod(complex(re=NaN, im=NaN)) [1] NaN > abs(complex(re=NaN, im=NaN)) [1] NaN and so do the following > Re(complex(re=1,

Hi, A colleague of mine has spotted me a passage of the doc ?option talking about Inf and NaN check in 'matprod=default' section: https://stat.ethz.ch/R-manual/R-devel/library/base/html/options.html I am wondering if NA should be mentioned too as the check seems to include this "value" too. NA being different from Inf and NaN it is worth mentioning, isn't it? Best,

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]]

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]]

Extremes.Rd, that documents 'max' and 'pmax', has this in "Details" section, in the paragraph before the last. By definition the min/max of a numeric vector containing an NaN is NaN, except that the min/max of any vector containing an NA is NA even if it also contains an NaN. ------------------ >>>>> Michal Burda <michal.burda at centrum.cz>

Hi, I want to recode all Inf and NaN values to NA, but I;m surprised to see the result of the following code. Could anybody enlighten me about this? > df <- data.frame(a=c(NA, NaN, Inf, 1:3)) > df[is.infinite(df) | is.nan(df)] <- NA > df a 1 NA 2 NaN 3 Inf 4 1 5 2 6 3 > Thanks! Cheers!! Albert-Jan

Hi, I want to recode all Inf and NaN values to NA, but I;m surprised to see the result of the following code. Could anybody enlighten me about this? > df <- data.frame(a=c(NA, NaN, Inf, 1:3)) > df[is.infinite(df) | is.nan(df)] <- NA > df a 1 NA 2 NaN 3 Inf 4 1 5 2 6 3 > Thanks! Cheers!! Albert-Jan

Le 17/02/2020 ? 17:50, Tomas Kalibera a ?crit?: > On 2/17/20 5:36 PM, Serguei Sokol wrote: >> Hi, >> >> A colleague of mine has spotted me a passage of the doc ?option >> talking about Inf and NaN check in 'matprod=default' section: >> https://stat.ethz.ch/R-manual/R-devel/library/base/html/options.html >> >> I am wondering if NA should be