similar to: frailty models in survreg() -- survival package (PR#2933)

On Tue, 6 May 2003, Jerome Asselin wrote: > > I am confused on how the log-likelihood is calculated in a parametric > survival problem with frailty. I see a contradiction in the frailty() help > file vs. the source code of frailty.gamma(), frailty.gaussian() and > frailty.t(). > > The function frailty.gaussian() appears to calculate the penalty as the > negative

SEE ALSO ORIGINAL POSTING IN PR#2933 On May 6, 2003 03:58 pm, Thomas Lumley wrote: > > Looking at a wider context in the code > > pfun <- function(coef, theta, ndeath) { > if (theta == 0) > list(recenter = 0, penalty = 0, flag = TRUE) > else { > recenter <- log(mean(exp(coef))) > coef <- coef - recenter

I am trying to fit a lognormal distribution on interval-censored data. Some of my intervals have a lower bound of zero. Unfortunately, it seems like survreg() cannot deal with lower bounds of zero, despite the fact that plnorm(0)==0 and pnorm(-Inf)==0 are well defined. Below is a short example to reproduce the problem. Does anyone know why survreg() must behave that way? Is there an alternate

R version: 1.7.1 OS: Red Hat Linux 7.2 In this example, I would expect an error for the overly long variable name. This is always reproducable for me. > formula(paste("y~",paste(rep("x",50000),collapse=""))) Segmentation fault Sincerely, Jerome Asselin -- Jerome Asselin (Jérôme), Statistical Analyst British Columbia Centre for Excellence in HIV/AIDS St.

R version: 1.7.1 OS: Red Hat Linux 7.2 When "covmat" is supplied in princomp(), the output value "center" is all NA's, even though the input matrix was indeed centered. I haven't read anything about this in the help file for princomp(). See code below for an example: pc2$center is all NA's. Jerome Asselin x <- rnorm(6) y <- rnorm(6) X <- cbind(x,y)

I would like to document my findings (with a potential FIX) regarding the issue of calculation of the degrees of freedom with nlme(). The program given at the bottom of this email generates and fit 20 data sets with a mixed-effects LINEAR model, but using the function nlme() instead of lme(). In each case, the correct number of degrees of freedom for the intercept parameter is 12. However, in

Hi, I would like to recommend a minor modification in axis() which I believe can simplify the making of plots for publications. I am trying to define default values for par() in order to make labels bigger and lines thicker, so that the resulting plots look good when resized for publication purposes. I ran into the following problem... axis() does not use par() values as default for

Hi there, I want to justify to right the text of my legend. Consider this short reproducable example. x <- 1:5 y1 <- 1/x y2 <- 2/x plot(rep(x,2),c(y1,y2),type="n",xlab="x",ylab="y") lines(x,y1) lines(x,y2,lty=2) legend(5,2,c("1,000","1,000,000"),lty=1:2,xjust=1,yjust=1)

The problem occurs when the sample odds ratio is Inf, such as in the following example. Given the fact that both upper bounds of the two 95% confidence intervals are Inf, I would have expected that the two lower bounds be equal, but they aren't. x <- matrix(c(9,4,0,2),2,2) x # [,1] [,2] #[1,] 9 0 #[2,] 4 2 rbind("two.sided.95CI"=fisher.test(x)$conf.int,

R version: 1.7.1 OS: Red Hat Linux 7.2 Hi all, The formula object in model.frame() is not retrieved properly when model.frame() is called from within a function and the "subset" argument is supplied. foo <- function(formula,data,subset=NULL) { cat("\n*****Does formula[-3] == ~y ?**** TRUE *****\n") print(formula[-3] == ~y) cat("\n*****Result of model.frame()

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

The combination of survreg + gamma frailty = invalid model, i.e., the example that you quote. I did not realize that this had been added to the survreg help file until very recently. I will try to fix the oversight. Other, more detailed documentation states that Gaussian frailty + AIC is the only valid random effects choice for survreg. Details: frailty(x) with no optional

Dear All, I am attempting to make predictions based on a survreg() model with some censoring and a frailty term, as below: predict works fine on the original data, but not if I specify newdata. # a model with groups as fixed effect model1 <- survreg(Surv(y,cens)~ x1 + x2 + groups, dist = "gaussian") # and with groups as a random effect fr <- frailty(groups,

Hi, I would like to fit parametric survival models to time-to-event data that are left truncated. I have checked the help page for survreg and looked in the R-help archive, and it appears that the R function survreg from the survival library (version 2.16) should allow me to take account of left truncation. However, when I try the command

Hi: I have a few clarification questions about the elements returned by the coxph function used in conjuction with a frailty term. I create the following group variable: group <- NULL group[id<50] <- 1 group[id>=50 & id<100] <- 2 group[id>=100 & id<150] <- 3 group[id>=150 & id<200] <- 4 group[id>=200 & id<250] <- 5 group[id>=250

Hi everybody. I'm trying to fit a weibull survival model with a spline basis for the predictor, using the survival library. I've noticed that it doesn't seem to be possible to use the aic method to choose the degrees of freedom for the spline basis in a parametric regression (although it's fine with the cox model, or if the degrees of freedom are specified directly by the user),

Dear list, I want o fit a shared gamma frailty model with the frailty specification in the survival package. I have partly left-truncated data and time-dependent covariates. Is it possible to combine these two things in the frailty function. Or are the results wrong if I use data in the start-stop-formulation which account for delayed entry? Is the frailty distribution updated in the

It can't be done with the current code. In a nutshell, you are trying to use a feature that I never got around to coding. It's been on my "to do" list, but may never make it to the top. Terry

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,