similar to: How to keep the same class?

Windows Vista R 2.10.1 Does the following use of groupedData and lme produce an analysis with both random intercept and slope, or only random slope? zz<-groupedData(y~time | Subject,data=data.frame(data), labels = list( x = "Time", y = "y" ), units = list( x = "(yr)", y = "(mm)") ) plot(zz)

Hi, I have a question about using lm on matrix, have to admit it is very trivial but I just couldn't find the answer after searched the mailing list and other online tutorial. It would be great if you could help. I have a matrix "trainx" of 492(rows) by 220(columns) that is my x, and trainy is 492 by 1. Also, I have the newdata testx which is 240 (rows) by 220 (columns). Here is

Dear R-helper, How could I count only some variable was exist after running sample (random) function. For example, > testx <- factor(c("Game","Paper","Internet","Time","Money")) > for(i in 1:2) { + x <- sample(testx,replace=TRUE) + print(x) + } [1] Money Money Time Internet Time Levels: Game Internet

All, For my understanding, I wanted to see if I can get glmnet predictions using both the predict function and also by multiplying coefficients by the variable matrix. This is not worked out. Could anyone suggest where I am going wrong? I understand that I may not have the mean/intercept correct, but the scaling is also off, which suggests a bigger mistake. Thanks for your help. Juliet Hannah

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I

Using str() in a function. I am in the early phase of learning R, and I find I spend a lot of time trying to figure out what is actually in objects I have created or read in from a file. I'm trying to make a simple little function to display a couple of things about a object, let's say the summary() and the str(), sequentially, preferably without a bunch of surplus lines between them. I

Dear list, I have a problem on persp() x <- u1data #first coloum in attached data y <- u2data #second coloum in attached data f <- function(x,y){qgev(pnorm(rhoF*qnorm(pnorm((qnorm(y)-rho2*qnorm(x)/sqrt(1-rho2^2)))) +sqrt(1-rhoF^2)*qnorm(0.95)),-0.3935119, 0.4227890, 0.2701648)} z <- outer(x,y,f) persp(x,y,z) The R will display: "Error in persp.default(x, y,

thanks for your attention. temporarily with the following script I'm looking at myself. #!/bin/bash INTERVAL=180 LOG=true function log { if $LOG; then echo $1 fi } while true; do testx=`ps -auxw | grep "smbd" | wc -l` testy=`netstat -ap | grep samba | grep sock | wc -l` if test "$testx" -gt "50" -a "$testy" -gt "50" ;then echo "samba

Hi all I am using the caret package and having difficulty in obtaining the results using regression, I used the glmnet to model and trying to get the coefficients and the model parameters I am trying to use the extractPrediction to obtain a confusion matrix and it seems to be giving me errors. x<-read.csv("x.csv", header=TRUE); y<-read.csv("y.csv", header=TRUE);

Readers, How to solve this unicode input error please? > postscript("~/tmp/test.eps",width=100/25.4,height=100/25.4,horizontal=FALSE,onefile=TRUE,paper="special") > testx<-seq(1:5) > testy<-seq(1:5) > plot(testy~testx) > mtext('text (O?)\n more text',side=3,line=1) Warning messages: 1: In mtext("text (O?)\n more text", side = 3, line =

Hi all, I am trying to create a function that will open connections to all files of one type within the working directory. I've got the function to open the connections, but I am having a bugger of a time trying to get these connections named as objects in the workspace. I am at the point where I can do it outside of the function, but not inside, using assign. I'm sure I'm

Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between

Hi, I am looking at the change in N concentration in plant roots over 4 time points and I have fit a spline to the data using ns and lme: fit10 <- lme( N~ns(day, 3), data = rcn10G) I may want to adjust the model a little bit, but for now, let's assume it's good. I get output for the fixed effects: Fixed: N ~ ns(day, 3) (Intercept) ns(day, 3)1 ns(day, 3)2 ns(day, 3)3

Dear folks-- I always seem to find that I spend more than half my time making sure my input date is in the right form, properly aligned, with no bizarre features. You know the drill: five kinds of missing values, three of them documented. An alpha mistype in one numeric field turns 30,000 numbers into factor levels. SPSS conversion turns 250 factors nicely into R factors, except 3 have levels

Having some problems with using hclust.identify on my data, I revert to trying it out on the example in the help manual. My result is still the same (as with my own data): data(iris) hci <- hclust(dist(iris[,1:4])) plot(hci) testx<-identify.hclust(hci) Error in rect.hclust(x, k = k, x = x$x, cluster = cluster[, k - 1], border = "red") : k must be between 2 and 0

Hi, I'm currently working on the KLEE tool that uses LLVM (2.9 because that's what KLEE currently uses) and in particular I'm having an unexpected problem with the llvm::cl::bits class. I try something like this... enum testx { A,B }; cl::bits<testx> queryLoggingOptions("option",cl::values( clEnumVal(A,"this is a"),

Hello everyone, Considering the following code sample : ---- indexes <- function(vec) { vec <- which(vec==TRUE) return(vec) } mat <- matrix(FALSE, nrow=10, ncol=10) mat[1,3] <- mat[3,1] <- TRUE ---- Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected. Now if I do: ---- mat[1,3] <- mat[3,1] <- FALSE apply(mat, 1, indexes) ---- I would expect a

Hello R-Users, I have recently run into several problems using vgam() in the VGAM package. I am hoping someone might have some solutions... Briefly, I have been trying to fit GAM models for zero-altered negative binomial models. 1. When fitting smoothed parameters (e.g. s(X, df=2)) changing the degrees-of-freedom has no effect on the level of smoothing (e.g. number of knots for the

Hello I have a matlab MAT file that contains one single variable: a. The structure of a is as follows: a.river1.flow (flow values) a.river1.date_flow (date) a.river1.precip (precipitation values) a.river1.date_precip a.river2.flow a.river2.date_flow a.river2.precip a.river2.date_precip I have used readMat to load the variable a in R, however I have no idea how readMat translates a. I managed

Dear folks: Let?s suppose I want a function to print return the name of the object passed to it. > myname <- function(object) {out<-deparse(substitute(object)); out} This works fine on a single object: > O1 <-c(1:4) > myname(O1) [1] "O1" However it does not work if you use lapply to pass it the same object from a list: > O2 <-c(1:4) > object.list <-