similar to: Is profile.mle flexible enough?

System: R 2.3.1 on Windows XP machine. I am building a logistic regression model for a sample of 100 cases in dataframe "d", in which there are 3 binary covariates: x1, x2 and x3. ---------------- > summary(d) y x1 x2 x3 0:54 0:50 0:64 0:78 1:46 1:50 1:36 1:22 > fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit)) >

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")

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

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)) #

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

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

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

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,

Hi, I have a glm gives summary as follows, Estimate Std. Error z value Pr(>|z|) (Intercept) -2.03693352 1.449574526 -1.405194 0.159963578 A 0.01093048 0.006446256 1.695633 0.089955471 N 0.41060119 0.224860819 1.826024 0.067846690 S -0.20651005 0.067698863 -3.050421 0.002285206 then I use confint(k.glm)

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

I can't get confint.glm to work from within a function. Consider the following (using R 1.9.1, Windows 2000): # FIRST: SOMETHING THAT WORKS FROM A COMMAND PROMPT DF <- data.frame(y=.1, N=100) (fit <- glm(y~1, family=binomial, data=DF, weights=DF[,"N"])) Call: glm(formula = y ~ 1, family = binomial, data = DF, weights = DF[, "N"]) Coefficients:

Hi, I have a question about the computation of confidence intervals in the zyp package, in particular using the functions zyp.sen and confint.zyp, or zyp.yuepilon. (1) I'm a bit confused about the confidence intervals given by zyp.sen and confint.zyp. When I request a certain confidence interval in the function, the R output seems to deliver another confidence interval, e.g. when I set

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.