similar to: Comparing nonlinear, non-nested models

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 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

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 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

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

Hi I have the following formula (I hope it is clear - if no, I can try to do better the next time) h(x, a, b) = integral(0 to pi/2) ( ( integral(D/sin(alpha) to Inf) ( ( f(x, a, b) ) dx ) dalpha ) and I want to do an mle with it. I know how to use mle() and I also know about integrate(). My problem is to give the parameter values a and b to the

Hi- I am trying to fit a log function to my data, with the ultimate goal of finding the second derivative of the function. However, I am stalled on the first step of fitting a curve. When I use the following code: FG2.model<-(nls((CO2~log(a*Time)+b), start=setNames(coef(lm(CO2 ~ log(Time), data=FG2)), c("a", "b")),data=FG2)) I get the following error: Error in

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)) #

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

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 <-

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

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 +

Hello R users & R friends, I just want to ask you if density() can produce a confidence interval, indicating how "certain" the density() line follows the true frequency distribution based on the sample you feed into density(). I've heard of loess.predict(loess(y ~ x), se=TRUE) which gives you a SE estimate of the smoothed scatterplot - but density() kernel smoothing is not the

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 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, I am just wondering if there is a test available for testing if a linear fit of an independent variable in a Cox regression is enough? Thanks for any suggestions. John Zhang ____________________________________________________________________________________ [[elided Yahoo spam]]

Dear list, I am getting weired with fitting data with a 1/x-polynomial. Suggest I have the following data: x <- c(1,2,3,4,5,6,7) y <- c(100,20,4,2,1,.3,.1) I may fit this with a linear model fit1 = lm(y ~ I(x)) Getting plot out of this model I applied library(polynom) pol1 = polynomial(fit1$coefficients) f1 = as.function(pol1) plot(x,y) lines(x, f1(x), col = 2) Clearly, this model

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, I would like to know which is the right way to check the normality assumption for performing ANOVA. How do you check normality for the following example? I did an experiment where people had to evaluate on a 7 point scale, the degree of realism of some stimuli presented in 2 conditions. The problem is that if I check normality with the Shapiro test I get that the data are not