similar to: Parametric Survival Models with Left Truncation, survreg

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,

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 =

I am using survreg from the survival package to run a left censored tobit model on “non-survival” data. I have to four questions that I hope someone can help me with: 1) Is there anything I should take into consideration when using frailty() to estimate random intercepts? 2) Is there anyway of extracting the estimated intercepts produced by survreg when using frailty()? 3) Can someone point

> I was wondering why survreg (in survival package) can not handle > three-way interactions. I have an AFT ..... You have given us no data to diagnose your problem. What do you mean by "cannot handle" -- does the package print a message "no 3 way interactions", gives wrong answers, your laptop catches on fire when you run it, ....? Also, make sure you read

Dear All, I would like to fit some parametric survival models using left truncated, right censored data in R. However I am having problems finding a function to fit parametric survival models which can handle left truncated data. I have tested both the survreg function in package survival: fit1 <- survreg(Surv(start, stop, status) ~ X + Y + Z, data=data1) and the psm function in package

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),

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 log-density of independent Gaussian variables, as one would expect: >

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

Dear R-Helpers, I am hoping to perform survival analyses using the "ex-Gaussian" distribution. I understand that the ex-Gaussian is a convolution of exponential and Gaussian distributions for survival data. I checked the "survreg.distributions" help and saw that it is possible to mix pre-defined distributions. Am I correct to think that the following code makes the

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 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-users, I need to use the aftreg function in package 'eha' to estimate failure times for left truncated survival data. Apparently, survreg still cannot fit such models. Both functions should be fitting the accelerated failure time (Weibull) model. However, as G?ran Brostr?m points out in the help file for aftreg, the parameterisation is different giving rise to different

Dear R users: I am using survreg for modeling left censored longitudinal data. When I am using the following code for fitting the tobit model I am getting some output with an warning message(highlighted with red color): > survreg(Surv(y, y>=0, type='left')~x + frailty(id), cytokine.data, weight=w, dist='gaussian', scale=1) Call: survreg(formula = Surv(y, y >= 0, type

Dear R-users, I know that a few people have asked whether survreg handles left-truncation data and the reply that i have seen so far is that it does. However, when I try to use survreg on left-truncated data, I got the following error message. > survcs3<-survreg(Surv(start,end,status)~AG, data=DPONEcs3, dist="exponential") Error in survreg(Surv(start, end, status) ~ AG, data =

Dear R-Helpers: I have tried everything I can think of and hope not to appear too foolish when my error is pointed out to me. I have some real data (18 points) that look linear on a log-log plot so I used them for a comparison of lm() and survreg. There are no suspensions. survreg.df <- data.frame(Cycles=c(2009000, 577000, 145000, 376000, 37000, 979000, 17420000, 71065000, 46397000,

Dear R Listers- I am attempting to analyse the survival of seeds in cages (exclosures) that differ in their permeability to rainforest mammals. Because I did not observe the moment of seed disappearance, my data is interval censored. This limits my options for analysis (as I understand it) to survreg, in the survival package. Because I repeated the experiment in 8 sites, I have a random

Hello, I'm quite new to R and want to make a Weibull-regression with the survival package. I know how to build my "Surv"-object and how to make a standard-weibull regression with "survreg". However, I want to fit a translated or 3-parametric weibull dist to account for a failure-free time. I think I would need a new object in survreg.distributions, but I don't know how

Can anybody give me a neat example of interval censored data analysis codes in R? Given that suvreg(Surv(c(1,1,NA,3),c(2,NA,2,3),type="interval2")~1) works why does survreg(Surv(data[,1],data[,2],type="interval2")~1) not work where data is : T.1 T.2 Status 1 0.0000000 0.62873036 1 2 0.0000000 2.07039068 1 3 0.0000000