similar to: bug in rpart?

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

Hi everybody! I want to find a closer neighbourins observation. This is my code: ########################## library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

Hi I want to create a nice picture about my result of k -the nearest neighbours algorithm. Here is my easy code: ################################# library(klaR) library(ipred) library(mlbench) data(PimaIndiansDiabetes2) dane=na.omit(PimaIndiansDiabetes2)[,c(2,5,9)] dane[,2]=log(dane[,2]) dane[,1:2]=scale(dane[,1:2]) zbior.uczacy=sample(1:nrow(dane),nrow(dane)/2,F)

I have a generalized linear model to solve. I used package "geepack". When I use the correlation structure "unstructured", I get a messeage that- R GUI front-end has stopped working. Why this happens? What is the solution? The r codes are as follows: a<-read.table("d:/bmt.txt",header=T")

the data is : >table.8.3<-data.frame(expand.grid( marijuana=factor(c("Yes","No"),levels=c("No","Yes")), cigarette=factor(c("Yes","No"),levels=c("No","Yes")), alcohol=factor(c("Yes","No"),levels=c("No","Yes"))), count=c(911,538,44,456,3,43,2,279))

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

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

Hi the list, I am experiencing several issues with profile.mle (and consequently with confint.mle) (stat4 version 2.9.2), and I have to spend a lot of time to find workarounds to what looks like interface bugs. I would be glad to get feedback from experienced users to know if I am really asking too much or if there is room for improvement. * Problem #1 with fixed parameters. I can't

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

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

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,

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

How can I from the summary function, decide which glm (fit1, fit2 or fit3) fits to data best? I don't know what to look after, so I would please explain the important output. > fit1 <- glm(Y~X, family=gaussian(link="identity")) > fit2 <- glm(Y~X, family=gaussian(link="log")) > fit3 <- glm(Y~X, family=Gamma(link="log")) > summary(fit1) Call:

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

Dear list, I am trying to re-analyse something. I do have two time series, one of which (ts.mar) might help explaining the other (ts.anr). In the original analysis, no-one seems to have cared about the data being time-series and they just did OLS. This yielded a strong positive correlation. I want to know if this correlation is still as strong when the autocorrelations are taken into account.

Dear all, I have 3 models (from simple to complex) and I want to compare them in order to see if they fit equally well or not. From the R prompt I am not able to see where I can get this information. Let´s do an example: fit1<- lm(response ~ stimulus + condition + stimulus:condition, data=scrd) #EQUIVALE A lm(response ~ stimulus*condition, data=scrd) fit2<- lm(response ~ stimulus +

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,

Hi all, I was wondering if there is a function out there, or someone has written code for making confidence intervals around model averaged predictions (y~á+âx). The model average estimates are from the dRedging library? It seems a common thing but I can't seem to find one via the search engines Examples of the models are: fit1 <- glm(y~ dbh, family = binomial, data = data) fit2 <-