similar to: bug in outer() (PR#755)

Full_Name: Matthias von Davier Version: 1.2.0 OS: linux Submission from: (NULL) (144.81.31.148) pairs(NxK_Matrix,panel=qqplot) produces a message Error in pairs.default(NxK_Matrix,panel=qqplot) : The panel function made a new plot best regards and a happy new year -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read

Dear All, I am trying to solve the optimization problem below, but I am always getting the following error: Error in optim(rep(20, nvar), f, gr, method = "L-BFGS-B", lower = rep(0, : L-BFGS-B needs finite values of 'fn' Any ideas? Thanks in advance, Paul ----------------------------------------------- k <- 10000 b <- 0.3 f <- function(x) { n <- length(x)

R CMD check tells me * checking R code for possible problems ... NOTE agexact.fit.rds: ... may be used in an incorrect context: ?optim(init, agfitfn, ...)? Warning: <anonymous>: ... may be used in an incorrect context: ?optim(init, agfitfn, ...)? Can anyone tell me what this message means? My searches haven't turned up anything useful. This is with R 2.7 and 2.9. The message

Hi, in R-3.0 I get the following error when calling lsfit with more observations than variables, which seems to come from an error in the formatting of the error message (note that this was not happening in 2.15.3): > nobs <- 5; nvar <- 6; lsfit(matrix(runif(nobs*nvar), ncol=nvar), runif(nobs), intercept=FALSE) Error in sprintf(ngettext(nry, "%d response", "%d

Hi, I'm currently using the R package e1071 to train naive bayes classifiers and came across a bug: When the posterior probabilities of all classes are small, the result from the predict.naiveBayes function become NaNs. This is an issue with the treatment of the log-transformed probabilities inside the predict.naiveBayes function. Here is an example to demonstrate the problem (you might need

Dear R community members and R experts I am stuck at a point and I tried with my colleagues and did not get it out. Sorry, I need your help. Here my data (just created to show the example): # generating a dataset just to show how my dataset look like, here I have x variables # x1 .........to X1000 plus ind and y ind <- c(1:100) y <- rnorm(100, 10,2) set.seed(201) P <-

Hallo All, I would like to apply a function to all permutations of variables in a dataframe (except the first). What is the best way to achieve this? I produce the permutations using: nvar <- ncol(dat) - 1 perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] ) Thanks in advance Serguei Test-dataframe, comma-delimited: code,wav,w,area,gdp,def,pop,coast,milspend,agr

I am now getting the occasional complaint about survival routines that are not able to handle big data.?? I looked in the manuals to try and update my understanding of max vector size, max matrix, max data set, etc; but it is either not there or I missed it (the latter more likely).?? Is it still .Machine$integer.max for everything??? Will that change??? Found where? I am going to need to go

hi all i know that one should try and limit the amount of looping in R programs. i have supplied some code below. i am interested in seeing how the code cold be rewritten if we dont use the loops. a brief overview of what is done in the code. ============================================== ============================================== ============================================== 1. the input

Assume a C program invoked by .Call, that returns a list. Near the top of the program we allocate space for all the list elements. (It is my habit to use "xyz2" for the name of the R object and "xyz" for the pointer to its contents.) PROTECT(means2 = allocVector(REALSXP, nvar)); means = REAL(means2); PROTECT(u2 = allocVector(REALSXP, nvar)); u =

The number of recent questions from umn.edu makes me wonder if there's homework involved.... Simpler for your example is to use get and subset. dat <- structure(..... as found below var.to.test <- names(dat)[4:6] #variables of interest nvar <- length(var.to.test) chisq <- double(nvar) for (i in 1:nvar) { tfit <- survdiff(Surv(time, completion==2) ~

Hello Dear R community,  This is my problem.  I have a data set (dataframe) called "mydat". It consist of 3 numerical variable.  They are Centrecode, FSUSN and Round. I want to create unique ID by combining these 3 variables. Follwing commands gives me what I need. mydat1 <- paste(mydat$Centrecode, mydat$FSUSN,mydat$Round,sep="") newds <- data.frame(mydat1)    For a

Hi, among the many (5) methods that I found in the list to do partial correlation in the following two that I had a look I am getting different t-values. Does anyone have any clues on why is that? The source code is below. Thanks. pcor3 <- function (x, test = T, p = 0.05) { nvar <- ncol(x) ndata <- nrow(x) conc <- solve(cor(x)) resid.sd <- 1/sqrt(diag(conc)) pcc <-

Hello, I will use the lca method in the e1071 package. But I get the following error: Error in pas[j, ] <- drop(exp(rep(1, nvar) %*% log(mp))) : number of items to replace is not a multiple of replacement length Does anybody know this error and knows what this means? Kind regards, Tryntsje

Dear R community members I have been struggling on this simple question, but never get appropriate solution. So please help. # my data, though I have a large number of variables var1 <- rnorm(500, 10,4) var2 <- rnorm(500, 20, 8) var3 <- rnorm(500, 30, 18) var4 <- rnorm(500, 40, 20) datafr1 <- data.frame(var1, var2, var3, var4) # my unsuccessful codes nvar <- ncol(datafr1)

Background: I administer a cluster of RedHat EWS 3U4 Linux workstations at a university. I built R 2.1.0 from source: ./configure \ --prefix=/sscc/opt/R-2.1.0 \ --with-blas=no \ 2>&1 \ | tee NUInstall.configure R is now configured for i686-pc-linux-gnu Source directory: . Installation directory: /sscc/opt/R-2.1.0 C compiler:

Hi everyone, I have a question regarding a C function of the "stats" package in R. I tried to understand the ?loglin? basic function of the ?stats? package implemented in R. The implemented function itself runs without any problem (perhaps see sample). When I ran it line by line it stopped at the lines 23-24 of the loglin-function; (the following line): z <- .Call(C_LogLin,

Amigos de la erre. He creado mi primer bucle con for para entrenar unos modelos con GAM. La respuesta es quasipoisson porque estoy trabajando con densidades de peces. Sin embargo, tengo un problema, no se muy bien como añadir el offset a la formula siguiente cuando creo el bucle. GAM.A1 <-gam ((DYO)~s(DMA,k=4)+ s(WOD,k=4)+s(CIN,k=4)+s(DRA,k=4)+s(DBR,k=4)

Dear Thomas & Dear List, the fitting function `coxph.fit' called by `coxph' may fail to estimate the regression coefficients when some values of the design matrix are very large. For example library(survival) ### load example data load(url("http://www.imbe.med.uni-erlangen.de/~hothorn/coxph_fit.Rda")) method <- "efron" ### copied from `coxph.fit' coxfit

I'm a Beginner on R-language and I' have tried to use some packages but the following message have appeared: "Error in .Fortran("pspline", as.integer(n), as.integer(nvar), as.integer(norder), : C/Fortran function name not in load table" I'm using a Windows version and I'd like to know if it is a installation problem or problems with paths? Thanks,