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Dear R useRs, I have a vector with positive and negative numbers: A=c(0,1,2,3,0,4,5) Now if i-th element in vector A is > 0, then i-th element in vector B is a+1 else i-th element in vector b=a (or 0) vector A: 0 1 2 3 0 4 5 vector B: 0 2 3 4 0 5 6 What's the right way to do this. I still have some problems with for and if statements... Cheers, Andrej

Dear useRs, I have two different length vectors: one column (1...m) and one row vector (1...n): 20 40 20 60 5 4 2 Now I have to calculate ratios between column vector elements and each row vector elements: 4 5 10 8 10 20 4 5 20 15 12 30 Thank's in advance for any suggestions, Andrej

Dear R useRs, I have big (23000 rows), vertical bar delimited file: e.g. A00001|Text a,Text b, Text c|345 A00002|Text bla|456 ... .. . Try using A <- read.table('filename.txt', header=FALSE,sep='\|') process stop at line 11975 with warning message: number of items read is not a multiple of the number of columns I have no problems with processing similar file, which is

Dear all, Suppose I plot two normal distributions (A and B) side by side and add vertical line which hipotheticaly represent alpha value; e.g.: x <- seq(-3.5,5, length=1000) y <- dnorm(x) # Plot distribution A plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) # Plot distribution B y2 <- dnorm(x-1.5) lines(y2~x,lwd=2) # Plot vertical line for alpha value

Dear R useRs, is there any simple, build in function to match specific regular expression in data file and write it to a vector. I have the following text file: *NEW RECORD *ID-001 *AB-text *NEW RECORD *ID-002 *AB-text etc. Now I have to match all ID fields and print them to a vector: 001 002 etc. I know that this is very simple with Perl or R-Perl interface, but if possible, I want to do

Dear all, Does anybody have an idea or suggestion how to construct (plot) 4-dimensional hypercube in R. Thanks in advance for any pointers. Regards, Andrej

Der useRs, I'm working on meta analysis using rmeta package. Using code below I plot the forest plot: library(rmeta) data (catheter) a<-meta.MH (n.trt, n.ctrl, col.trt, col.ctrl, data=catheter, names=Name, subset=c(13,6,5,3,7,12,4,11,1,8,10,2)) summary(a) # odds ratio values and confidence intervals metaplot(a$logOR, a$selogOR, nn=a$selogOR^-2,a$names, summn=a$logMH, sumse=a$selogMH,

Dear useRs, I'm working on multiple comparison design on two factor (2 3 levels) ANOVA. Each of the tests I have tried (Tukey, multcomp package) seem to do only with one factor at a time. fm1 <- aov(breaks ~ wool * tension, data = warpbreaks) tHSD <- TukeyHSD(fm1, "tension", ordered = FALSE) $tension diff lwr upr p adj M-L -10.000000 -19.35342

Dear R useRs, I have a problem to add title to the following graphics; Tukey=TukeyHSD(aov(CA~C), "C",ordered=TRUE)) plot (Tukey, main="My first graph") actually, it draw a graph, but it also display: parameter "main" could not be set in high-level plot() function. If I execute: plot(rnorm(1000),main="My first plot"), no error occur. What's the

Dear all, I'm working on ability estimates using Rasch model. Using the "ltm" package, the procedure is quite simple: ## Factor Scores for the Rasch model fit <- rasch(LSAT) factor.scores(fit) What about Partial Credit Model (PCM)? For PCM I use PCM function from eRm package. Is there any similar function like factor.scores to estimate ability scores using PCM model? Best,

Dear all, I try to install the latest R version using checkinstall (v. 1.6.2) on Ubuntu 11.10. After solving all the dependencies (installed using apt-get build-dep r-base) checkinstall fails to build and install R package with the following error (the same commands build and install R-2.13.2 on the same machine without any problem): wget

Dear all, I'm trying to filter out the "number of leaves" (it should be 1 in the example below) from the following string: > string [1] "Java-Object{J48 pruned tree\n------------------\n: 0 (15.0/3.0)\n \nNumber of Leaves : \t1\n\nSize of the tree : \t1\n}" Any idea how to do that as simple as possible? Thanks in advance for any advice. Regards, Andrej

Dear useRs, I use dics.ef function from dprep package to discretize continuous variable using intervals of equal frequencies. Dataset to be discretized include 4 continuous and 2 discrete variables in the following order: Continuous Countinuous Countinuous Discrete Discrete Continuous The problem emerge when I try to discretize the last continuos variable: library(dprep)

Dear all, Using lm models one can calculate predicted residual sum of squares (PRESS) statistic to select appropriate number of prediction variables in lm model. Is there any alternative to PRESS statistic for lda (linear discriminant analysis) model where predicted outcome variable is categorical? Thanks in advance for any suggestion. Best, Andrej

Hi, I draw a surface corresponding to bivariate density of independent variables (rho=0) using persp(). Then I add a contour line (i.e., circle in my case) at a particular density. Below is a minimal example of what I have so far. # Bivariate density dnorm2d <- function(x, y, rho = 0) { xoy = (x^2 - 2 * rho * x * y + y^2)/(2 * (1 - rho^2)) density = exp(-xoy)/(2 * pi * sqrt(1 - rho^2))