similar to: Extracting approximate Wald test (Chisq) from coxph(..frailty)

Dear all, Does the frailty.object$history[[1]]$theta returns the Variance of random effect? Why is the value different? Here is an example with kidney data: > library(survival) > data(kidney) > frailty.object<-coxph(Surv(time, status)~ age + sex + disease + frailty(id), kidney) > frailty.object Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id), data

I am trying to model survival data with a Weibull distribution using survreg. Units are clustered two apiece, sometimes receiving the same treatment and sometimes opposing treatment.

Dear R-experts, Here below my R code. I get a NaN response for gam with mgcv library. How to solve that problem? Many thanks. ######################################################### library(mgcv) ? y=c(23,24,34,40,42,43,54,34,52,54,23,32,35,45,46,54,34,36,37,48) x1=c(0.1,0.3,0.5,0.7,0.8,0.9,0.1,0.7,0.67,0.98,0.56,0.54,0.34,0.12,0.47,0.52,0.87,0.56,0.71,0.6)

I have the following questions about the variance of the random effects in the survreg() function in the survival package: 1) How can I extract the variance of the random effects after fitting a model? For example: set.seed(1007) x <- runif(100) m <- rnorm(10, mean = 1, sd =2) mu <- rep(m, rep(10,10)) test1 <- data.frame(Time = qsurvreg(x, mean = mu, scale= 0.5, distribution =

Hello R users Would somebody know how to estimate survival from a frailty cox model, using the function survfit and the argument newdata ? (or from any other way that could provide individual expected survival with standard error); Is the problem related to how the random term is included in newdata ? kfitm1 <- coxph(Surv(time,status) ~ age + sex + disease + frailty(id,

Hello, I am using package MuMIn to calculate AIC for a full model with 10 explanatory variables. Thanks in advance in sharing your experience. Q1 In the AIC list of all models, each model is differentiated by model number. Please kindly advise if it is possible to find the corresponding explanatory variable(s) for the model number. Q2 error message I tried to display sub-model with only

Hi all, I have a large dataset with ~8600 observations that I want to compress to weekly means. There are 9 variables (columns), and I have already added a "week" column with 51 weeks. I have been looking at the functions: aggregate, tapply, apply, etc. and I am just not savvy enough with R to figure this out on my own, though I'm sure it's fairly easy. I also have the Dates

When I do: > apc <- glm( D ~ ns( Ax, knots=seq(50,80,10), Bo=c(40,90) ) + + ns( Cx, knots=seq(1880,1940,20), Bo=c(1840,1960) ) + + ns( Px, knots=seq(1960,1980,10), Bo=c(1940,2000) ) + + offset( log( Y ) ), + family=poisson ) > pterm <- predict( apc, type="terms" ) > plink <- predict( apc,

I'm looking for the best way to do the following: run a set of GAM models, and then make predictions with new data. My problem is the size of the gam model object, I would like to strip it down to the bare minimum of information needed to apply the model to new data. For example, if this were a linear model, I would just keep the betas. If this were an ordinary spline fit, I think I

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

Hello, R-folks: Here is a statement I use to make a data frame: iccdata <- data.frame(i=rep(1:10,rep(2,10)),j=rep(1:2,10), x=c(0.35011,0.11989,0.13081,0.09919,0.16000,0.12000,0.00000,0.00000, 0.44023,0.32977,2.67081,2.63919,0.09050,0.03950,0.44019,0.30981,0.59000, 0.57000,4.03000,3.77000)) Then here are the data: > iccdata i j x 1 1 1 0.35011 2 1 2 0.11989 3 2 1 0.13081 4