similar to: R newbie...

Dear all, I like to extract row names & column names from the named matrix...... like...... a<-matrix(1:6,2) ro<-c("aa","bb") co<-c("dd","ee","ff") dimnames(a)<-list(ro,co) a > dd ee ff aa 1 3 5 bb 2 4 6 from the above matrix "a" I like to extract rownames separately like

Hello R-users! My (simple?) doubt: How to reverse the sequence of axis Y ?? the diagram below illustrate my idea... (default) | | . | . | . | 3 | 2 | 1 | 0 +---------------------------- 0 1 2 3 ... like I want... 0 | 1 | 2 | 3 | . | . | . | | +---------------------------- 0 1 2 3 ... thanks in advance klebyn

Hi All, This may be trivial in R but I have been trying with out any success. I have a list of 100 elements each having a sub list of different length. I would like to write the list to a ASCII file. I tried with write.table(), after converting my list to a matrix. Now it looks like Robert c("90", "50", "30") John c("91", "20",

Say I have a FOR-loop for computing powers (just a trivial example) for(i in 1:5) { x<-i^2 y<-i^3 } How can I create a data.frame and a 3D plot of (i,x(i),y(i)), i.e. for each iteration Thanks, Serguei Kaniovski -- ___________________________________________________________________ ??sterreichisches Institut f??r Wirtschaftsforschung (WIFO) Name: Serguei Kaniovski

Hello R - users, This may sound simple to may people: I have a list of data as follows type value y 7 y 7 y 8 y 8 y 8 y 9 y 9 y 9 y 9 y 10 y 10 y 10 y 10 y 11 y 11 y 12 y 12 y 14 y 14 y 14 y 15 y 17 y 20 y 20 y 20 y 20 y 25

Hello there, I would like to simulate X --Normal (20, 5) Y-- Normal (40, 10) and the correlation between X and Y is 0.6. How do I do it in R? Thank you very much Lisa Wang Msc. Princess Margaret Hospital Toronto, Ca

Hello, I try to define a global variable. My example: R> a <- "old" R> test <- function () { a <- "new" } R> test() R> a # shoud be "new" This doesn't work. I would like to modify the variable "a" in a procedure. How can I do that. Thank you for helping. Sven Kn侟ppel (Germany-Berlin)

Dear, I am looking for the simplification of a formula to improve the calculation speed of my program. Therefore I want to simplify the following formula: H = sum{i=0..n-1 , [ sum {j=0..m-1 , sqrt ( (Ai - Bj)^2 + (Ci - Dj)^2) } ] } where: A, C = two vectors (with numerical data) of length n B, D = two vectors (with numerical data) of length m sqrt = square root Ai = element of A with index

Dear R-helpers, I apologize for this certainly simple question. I have the following R lines : > m = matrix(1:12 , 3 , 4); > m [,1] [,2] [,3] [,4] [1,] 1 4 7 10 [2,] 2 5 8 11 [3,] 3 6 9 12 > m1 = subset(m , m[,2]>=5); > m1 [1] 2 3 5 6 8 9 11 12 but in fact I would appreciate m1 to be also a matrix, and thus would like to get :

Hello, I would like to learn how to create a pdf and a postscript file from an R graphic. For example, I tried : plot(1) # example pdf("H:/Perso/essai.pdf") postscript("H:/Perso/essai.ps") But the pdf document created is empty. How do these functions work ? Thank you very much. -- David Hajage Interne de santé publique Institut Curie

Hello everyone, I have a problem when loading the library: IlluminaHumanMethylation27k.db I installed the package from zip (from the Bioconductor website), and also installed all the packages it asked for. It went all succesfull: "package 'IlluminaHumanMethylation27k.db' successfully unpacked and MD5 sums checked" But when I am trying to load this library I get an error:

Dear R useRs, Sorry for this foolish question, but I can't find how to escape the * character when using grep : > grep("-", c("/3", "2*3", "4-4")) [1] 3 > grep("/", c("/3", "2*3", "4-4")) [1] 1 > grep("*", c("/3", "2*3", "4-4")) Erreur dans grep("*",

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Dear useRs, I'm trying to create a barplot like so: x=matrix(1:10,2,5) barplot(x,leg=c("left","right"),besid=T) The legend is placed in default position topright, however the data are plotted there too. I tried controlling the legend position by adding x="topleft" but this results in an error that x matches multiple formal arguments. Leaving out the legend

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Hello useRs, I'm trying to use the Roxygen package. Here my code file : #' A packge to check Roxygen's sanity #' @name helloRoxygen-package #' @docType package NA And my R code to generate the package : library(roxygen) package.skeleton("helloRoxygen", code_files = "roxy.r", force = T) roxygenize("helloRoxygen", "helloRoxygen",

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Hello I am writing a replacement method for an S4 class and want to pass an additional argument to "[<-"() along the lines of "["()'s "drop" argument. Specifically, I have an S4 class, call it "foo", with a slot 'x' that is a vector and a slot 'NC' that is a scalar. I want to be able to pass a Boolean argument to the

Dear R users: I have a problem about catch the value from function. I have following two functions (part): fbolus1 <- function() {......... par<-data.frame(Parameter=c("kel","Vd"),Initial=c(0)) check(par) .....} check<-function(par) { if (par[ ,2] <= 0){ cat("\nEnter again (y/n) ?\n\n") ans<-readline() if (ans == "n" ){