similar to: Bug or a feature that I completely missed?

Dear all, First, I recently had reasons to read the help page of as.vector() and noticed in the example section the following example: x <- c(a = 1, b = 2) is.vector(x) as.vector(x) all.equal(x, as.vector(x)) ## FALSE However, in all versions of R in which I executed this example, the all.equal command returned TRUE which suggest that either the comment in the help file

Hi Jakob, dat1<-read.table(text=" TIME, Value1, Value2 01.08.2011 02:30:00, 4.4, 4.7 01.09.2011 03:00:00, 4.2, 4.3 01.11.2011 01:00:00, 3.5, 4.3 01.12.2011 01:40:00, 3.4, 4.5 01.01.2012 02:00:00, 4.8, 5.3 01.02.2012 02:30:00, 4.9, 5.2 01.08.2012 02:30:00, 4.1, 4.7 01.12.2012 03:00:00, 4.7, 4.3 01.01.2013 01:00:00, 3, 4.3 01.01.2013 01:30:00, 3.8, 4.1 01.01.2013 02:00:00, 3.8,

I need to make a bunch of PDF files of histograms. I tried gatelist = unique(mdf$ArrivalGate) for( gate in gatelist) { outfile = paste("../", airport, "/", airport, "taxiHistogram", gate, ".pdf", sep="") pdf(file = outfile, width = 10, height=8, par(lwd=1)) title=paste("Taxi time for Arrival Gate", gate, "by

Hi, R users, Here is an example. k <- c(1,2,3,4,5) i <- c(0,1,3,2,1) if k=1, then j=0 from i if k=2, then j=0, 1 from i if k=3, then j=0, 1, 2, 3 from i if k=4, then j=0, 1, 2 from i if k=5, then j=0, 1 from i so i'd like to create a list like below. > list k j 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I

Dear all, I would like to suggest changing the pairs.formula command such that a command like pairs(GNP ~ . - Year - GNP.deflator, longley) would behave in a similar fashion as lm(GNP ~ . - Year - GNP.deflator, longley) i.e., make a pairwise scatterplot of GNP and all other variables in the (longley) dataframe except for Year and GNP.deflator. The above command, with the

Hello, I have a list (mode and class are list) in R that is many elements long and of the form: >length(list) [1] 5778 >list[1:4] $ID1 [1] "num1" $ID2 [1] "num2" "num3" $ID3 [1] "num4" $ID4 [1] NA I'd like to convert the $ID2 value to be in one element rather than in two.?? It shows up as c(\"num2\", \"num3\") if I try to use

I want to extra the part of the formula not including the response variable from an lm object. For example if the lm object ABx.lm was created by the call ABx.lm <- lm( y ~ A + B + x, ...) Then ACx.lm is saved as part of a workspace. I wish to extract "~ A + B + x". Later in my code I will fit another linear model of the form z ~ A + B + x for some other response variable z. I

--AZaLVt6Pw+ Content-Type: text/plain; charset=us-ascii Content-Description: message body text Content-Transfer-Encoding: 7bit Dear all, I was looking at "splinefun" and the underlying C code and believe that there is a memory access error in the C routine "spline_eval". Specifically, on line 368 and following the following code appears: if(ul < x[i] || x[i+1] < ul)

Hello, As I am new with R I am completely stuck in resolving a, no doubt, easy problem. I have a dataset with an enormous amount of rows and 17 columns. I need to know per trial and subject number if another variable (tt) exceeds a maximum. If this is true than the last 5000 rows of that specific trial for that subject number needs to be deleted. But I am completely stuck on how to do so.

Hi guys, I was wondering if any one is able to help me on a problem that I was stuck with for a long time. It involves the replacement of character strings with numbers. The character string can take on only 3 possible values, for instance: AA AT TT I would want R to replace AT with 0. Between AA and TT, I want to compare the frequency of either value, and then for the one which occurs more, I

Say I have: > set.seed( 1 ) > m <- matrix( runif(5^2), nrow=5, dimnames = list( c("A","B","C","D","E"), > c("O","P","Q","R","S") ) ) > m O P Q R S A 0.2655087 0.89838968 0.2059746 0.4976992 0.9347052 B 0.3721239 0.94467527 0.1765568

Hi, I'm having trouble with the "which" or the "seq" function, I'm not sure. Here's an example : > lat=seq(1,2,by=0.1) > lat [1] 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 > which(lat==1) [1] 1 > which(lat==1.1) [1] 2 > which(lat==1.2) [1] 3 > which(lat==1.3) [1] 4 > which(lat==1.4) [1] 5 > which(lat==1.5) [1] 6 >

Hi I want to write a little function that takes a matrix X of size m-by-n, and a list L of length "m", whose elements are matrices all of which have the same number of columns but possibly a different number of rows. I then want to get a sort of dumbed-down kronecker product in which X[i,j] is replaced by X[i,j]*L[[j]] where L[[j]] is the j-th of the "m" matrices. For

Hi look at the following session, in which I have a dataframe, and I want to extract the second row, without the first column. Everything works as expected until the last line, where I set the names of x to NULL, and get a non-desired object (I want c(4,3).). Three questions: (1) why is as.vector(a[2,-1]) not a vector? (2) How come setting names to NULL gives me bad weirdness? (3) Can I

I have found a strange scoping problem in Sweave. The following Rnw file doesn't produce the same output in Sweave as it does if I produce an R file using Stangle and execute that: \documentclass[12pt]{article} \begin{document} <<R>>= election <- data.frame(A=1:3, B=9:7, C=rep(0,3)) partytotal <- rep(0, ncol(election)) for (i in 1:ncol(election)) { partytotal[i] <-

Dear R developers, is there a preferred format or strategy for making a patch to contribute to a package that is maintained by R-core? Berwin Turlach and I have written a very minor extension to lmeControl to allow it to pass an argument to nlminb for the maximum number of evaluations of the objective function. I've edited the nlme/R/lme.R and nlme/man/lmeControl.Rd files. I can diff the

Hi, Forgive a very basic question... I need to take two lists-of-lists, and apply a function to each pair of elements in the lists to return a single list... For example l1 <- list(1:5,6:10,2:15) l2 <- list(1:8,4:12,1:19,4:20) I could easily do an lapply across each of them, but is there a function that does a sort-of pairwise-apply across both together? Does anybody know of a good

G'day all, I have daily scripts running to install the patched version of the current R version and the development version of R on my linux box (Ubuntu 18.04.4 LTS). The last development version that was successfully compiled and installed was "R Under development (unstable) (2020-01-25 r77715)" on 27 January. Since then the script always fails as a regression test seems to fail.

hello I tried to use lars() but neither with my own data nor with the sample data it works. I get in both cases the following error prompt: > data(diabetes) > par(mfrow=c(2,2)) > attach(diabetes) > x<-lars(x,y) Error in one %*% x : requires numeric matrix/vector arguments > x<-lars(x,y, type="lasso") Error in one %*% x : requires numeric matrix/vector arguments

G'day all, In Chapter 1.4 (Writing package vignettes) the Writing R Extensions manual states: By default @code{R CMD build} will run @code{Sweave} on all Sweave vignette source files in @file{vignettes}. If @file{Makefile} is found in the vignette source directory, then @code{R CMD build} will try to run @command{make} after the @code{Sweave} runs, otherwise @code{texi2pdf} is run on