similar to: rpart for CART with weights/priors

Hi, I am working with R version 2.1.0, and I seem to have run into what looks like a bug. I get the same error message when I run R on Windows as well as when I run it on Linux. When I call anova to do a LR test from inside a function, I get an error. The same call works outside of a function. It appears to not find the right environment when called from inside a function. I have provided

Dear R-users, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

fit1<-rq(formula=op~inp1+inp2+inp3+inp4+inp5+inp6+inp7+inp8+inp9,tau=0.15,data=wbc) fit2<-rq(formula=op~inp1+inp2+inp3+inp4+inp5+inp6+inp7+inp8+inp9,tau=0.5,data=wbc) fit3<-rq(formula=op~inp1+inp2+inp3+inp4+inp5+inp6+inp7+inp8+inp9,tau=0.15,data=wbc) fit4<-rq(formula=op~inp1+inp2+inp3+inp4+inp5+inp6+inp7+inp8+inp9,tau=0.15,data=wbc)

On Wed, December 22, 2004 12:11 am, r.ghezzo at staff.mcgill.ca said: > ----- Forwarded message from r.ghezzo at staff.mcgill.ca ----- > Date: Mon, 20 Dec 2004 15:45:11 -0500 > From: r.ghezzo at staff.mcgill.ca > Reply-To: r.ghezzo at staff.mcgill.ca > Subject: [R] problems with limma > To: r-help at stat.math.ethz.ch > > I try to send this message To Gordon

I try to send this message To Gordon Smyth at smyth at vehi,edu.au but it bounced back, so here it is to r-help I am trying to use limma, just downloaded it from CRAN. I use R 2.0.1 on Win XP see the following: > library(RODBC) > chan1 <- odbcConnectExcel("D:/Data/mgc/Chips/Chips4.xls") > dd <- sqlFetch(chan1,"Raw") # all data 12000 > # > nzw <-

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Hello R users,  I am trying to obtain a direct adjusted survival curve. I am sending my whole code (see below). It's basically the larynx cancer data with Stage 1-4. I am using the cox model using coxph option, see the fit3 coxph. When I use the avg.surv option on fit3, I get the following error: "fits<-avg.surv(fit3, var.name="stage.fac", var.values=c(1,2,3,4), data=larynx)

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

Well, I would suggest using the code already in place in the survival package. Here is my code for your problem. I'm using a copy of the larynx data as found from the web resources for the Klein and Moeschberger book. larynx <- read.table("larynx.dat", skip=12, col.names=c("stage", "time", "age", "year",

Hello. One of my teaching assistants was experimenting and encountered unexpected behaviour with the lrtest function in the rms package. It appears that when you have a pair of non-nested models that employ an RCS, the error checking for non-nested models appears not to work. Here is a reproducible example. > library(rms) Loading required package: Hmisc Loading required package: lattice

Folks, I am trying to get a loop to run which increments the object name as part of the loop. Here "fit1" "fit2" "fit3" and "fit4" are linear regression models that I have created. > for (ii in c(1:4)){ + SSE[ii]=rbind(anova(fit[ii])$"Sum Sq") + dfe[ii]=rbind(summary(fit[ii])$df) + } Error in anova(fit[ii]) : object 'fit' not found

Dear all, I'm getting an error in one of the stock examples in the 'mvpart' package. I tried: require(mvpart) data(spider) fit3 <- rpart(gdist(spider[,1:12],meth="bray",full=TRUE,sq=TRUE)~water+twigs+reft+herbs+moss+sand,spider,method="dist") #directly from ?rpart summary(fit3) ...which returned the following: Error in apply(formatg(yval, digits - 3), 1,

Dear R users: I'm generating the following survival data: set.seed(123) n=200 #sample size x=rbinom(n,size=1,prob=.5) #binomial treatment v=rgamma(n,shape=1,scale=1) #gamma frailty w=rweibull(n,shape=1,scale=1) #Weibull deviates b=-log(2) #treatment's slope t=exp( -x*b -log(v) + log(w) ) #failure times c=rep(1,n) #uncensored indicator id=seq(1:n) #individual frailty indicator

I want to organize outputs from several regressions into a handy table. When I try the following, each of my "fit_s" is replaces instead of read. Is there a way to read from the regression summaries that does not require writing separate lines of code for each? -Ben Osborne > fit1<-lm(dBA.spp16$sp2.dBA.ha~dBA.spp16$sp1.dBA.ha) >

Hello all, I wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

Greetings, I am using rpart for classification with "class" method. The test data is the Indian diabetes data from package mlbench. I fitted a classification tree firstly using the original data, and then exchanged the order of Body mass and Plasma glucose which are the strongest/important variables in the growing phase. The second tree is a little different from the first one. The

Dear all, Consider a completely randomized block design (let's use data(Oats) irrespoctive of the split-plot design it was arranged in). Look: library(nlme) fit <- lme(yield ~ nitro, Oats, random = ~1|Block, method="ML") fit2 <- lm(yield ~ nitro + Block, Oats) anova(fit, fit2) gives this: Model df AIC BIC logLik Test L.Ratio p-value fit 1 4 624.3245

Greetings, I checked the Indian diabetes data again and get one tree for the data with reordered columns and another tree for the original data. I compared these two trees, the split points for these two trees are exactly the same but the fitted classes are not the same for some cases. And the misclassification errors are different too. I know how CART deal with ties --- even we are using the

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp