similar to: For each element in vector do...

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 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, 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 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

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 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 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

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 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

Hi, I have Asterisk 1.4.10 under LMCE (upgrade is not an option) and have this strange error appearing in full log : [Apr 16 14:35:48] NOTICE[10802] chan_sip.c: -- Registration for 'NUMBER at voip.siol' timed out, trying again (Attempt #22) [Apr 16 14:35:48] WARNING[10802] chan_sip.c: No such host: voip.siol [Apr 16 14:35:48] WARNING[10802] chan_sip.c: Probably a DNS error for

Hi, can anyone help me fit betabinomial model to the following dataset where each iD is a cluster in itself , if i use package aod 's betabinom model it gives an estimate of zero to phi(the correlation coeficient ) and if i fix it to the anova type estimate obtained from icc( in package aod) then it says system is exactly singular. And when i try to fit my loglikelihood by

Dear All, I am running a simulation to obtain coverage probability of Wald type confidence intervals for my parameter d in a function of two parameters (mu,d). I am optimizing it using "optim" method "L-BFGS-B" to obtain MLE. As, I want to invert the Hessian matrix to get Standard errors of the two parameter estimates. However, my Hessian matrix at times becomes

Hi How can do this in R. >df 48 1 35 32 80 If df < 30 then replace it with 30 and else if df > 60 replace it with 60. I have a large dataset so I cant afford to identify indexes and then replace. Desired o/p: 48 30 35 32 60 Thanx in advance. Sachin

Dear R-help community, If I have a data.frame df as follows: > df x1 x2 x3 x4 x5 x6 1 5 5 1 1 2 1 2 5 5 5 5 1 5 3 1 5 5 5 5 5 4 5 5 1 4 5 5 5 5 1 5 2 4 1 6 5 1 5 4 5 1 7 5 1 5 4 4 5 8 5 1 1 1 1 5 9 1 5 1 1 2 5 10 5 1 5 4 5 5 11 1 5 5 2 1 1 12 5 5 5 4 4 1 13 1 5 1 4 4 1 14 1 1 5 4 5 5 15 1 5 5 4

I have a matrix and am trying to write a code to 1. identify all neg values along each line 2. for each neg value I need to identify min(j+3) 3. end with this code: eq[i,j]<- ifelse(mat.r[i,j] < (0.5*mat.s[i,j]), mat.all[i,j], 0) This is the code I have so far. I have tried several different methods but I keep getting the same error message that the condition has length >1 and only

Dear All, I luckily found the following feature (or problem) when tried to apply ifelse-function to an ordered data. > test <- c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE) > ifelse(test, 0, 1:4) [1] 0 0 0 4 1 2 3 > It roots into the ifelse-syntax: ans[!test & !nas] <- rep(no, length.out = length(ans))[!test & !nas] Would it be possible to disable this feature in the

I'm trying to categorize a continuous variable (yes, I know that's horrible, but I'm trying to reproduce some exercises from a textbook) and don't really know an efficient way to do this. I have a data frame that looks like: surv_time relapse sex log_WBC rx 1 35 0 1 1.45 0 2 34 0 1 1.47 0 3 32 0 1 2.20 0 4 32

R-help I have a martix with missing values( in which I want the sample size by column) When I : apply(matrix,2,length) I get the length of the vector regardless of missing values. I can't pass an argument to length in apply. Alternatively I could ifelse ( is.na ( matrix [, "columns in matrix " ] ) , 0 , 1) Is there any easier way? Thank you

Dear list, I have two functions created in the same environment, fun1 and fun2. fun2 is called by fun1, but fun2 should use an object which is created in fun1 fun1 <- function(x) { ifelse(somecondition, bb <- "o", bb <- "*") ## mymatrix is created, then myresult <- apply(mymatrix, 1, fun2) } fun2 <- function(idx) { if (bb == "o) { #