similar to: survreg (survival) reports erroneous results for left-censored data (PR#2287)

On Wed, 13 Nov 2002, Jan de Leeuw wrote: > > No problemo. And, in fact, I get the same results in > the R-1.6.0 Carbon version. I don't. Could there be a G3/G4 issue? -thomas > --- Jan > > On Wednesday, November 13, 2002, at 02:05 PM, tim@timcohn.com wrote: > > > Full_Name: Tim Cohn > > Version: 1.6.1 > > OS: Macintosh OS X > > Submission

Thank you for looking into this so quickly. As you correctly surmise, I was using the Carbon version of R-1.6.1 on Mac OS 10.2.2 (Jaguar) when I got the "wrong" answers. One other observation: The right censoring seems to work fine. Thanks again, Tim On Thursday, November 14, 2002, at 11:09 AM, Jan de Leeuw wrote: > I take that back. I now get the "correct" result

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

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

Hello, I'm working with some left censored survival data using accelerated failure time models. I am interested in fitting different distributions to the data but seem to be getting the same results from the model fit using survreg regardless of the assumed distribution. These two codes seem to provide the same results: aft.gaussian <-

Can survreg() handle interval-censored data like the documentation says? I ask because the command: survreg(Surv(start, stop, event) ~ 1, data = heart) fails with the error message Invalid survival type yet the documentation for Surv() states: "Presently, the only methods allowing interval censored data are the parametric models computed by 'survreg'"

Hello, I have interval censored data, censored between (0, 100). I used the tobit function in the AER package which in turn backs on survreg. Actually I'm struggling with the distribution. Data is asymmetrically distributed, so first choice would be a Weibull distribution. Unfortunately the Weibull doesn't allow for zero values in time data, as it requires x > 0. So I tried the

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,

Hey all, I am trying to use the survreg function in R to estimate the mean and standard deviation to come up with the MLE of alpha and lambda for the weibull distribution. I am doing the following: times<-c(10,13,18,19,23,30,36,38,54,56,59,75,93,97,104,107,107,107) censor<-c(1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,1,0,0) survreg(Surv(times,censor),dist='weibull') and I get the following

Dear R-user: I am using survreg(Surv()) for fitting a Tobit model of left-censored longitudinal data. For logarithmic transformation of y data, I am trying use survreg.distributions in the following way: tfit=survreg(Surv(y, y>=-5, type="left")~x + cluster(id), dist="gaussian", data=y.data, scale=0, weights=w) my.gaussian<-survreg.distributions$gaussian

1. The weibull is the only distribution that can be written in both a proportional hazazrds for and an accelerated failure time form. Survreg uses the latter. In an ACF model, we model the time to failure. Positive coefficients are good (longer time to death). In a PH model, we model the death rate. Positive coefficients are bad (higher death rate). You are not the first to be confused

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

Hi all, Sorry if this has been answered already, but I couldn't find it in the archives or general internet. Is it possible to implement the gompertz distribution as survreg.distribution to use with survreg of the survival library? I haven't found anything and recent attempts from my side weren't succefull so far. I know that other packages like 'eha' and

Dear R-list I have been trying to make survreg fit a normal regression model with left truncated data, but unfortunately I am not able to figure out how to do it. The following survreg-call seems to work just fine when the observations are right censored: library(survival) n<-100000 #censored observations x<-rnorm(n) y<-rnorm(n,mean=x) d<-data.frame(x,y) d$ym<-pmin(y,0.5)

Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except

Dear R help list, I am modeling some survival data with coxph and survreg (dist='weibull') using package survival. I have 2 problems: 1) I do not understand how to interpret the regression coefficients in the survreg output and it is not clear, for me, from ?survreg.objects how to. Here is an example of the codes that points out my problem: - data is stc1 - the factor is dichotomous

I am trying to use R to do loglogistic hazard estimation. My plan is to generate a loglogistic hazard sample data and then use survreg to estimate it. If everything is correct, survreg should return the parameters I have used to generate the sample data. I have written the following code to do a time invariant hazard estimation. The output of summary(modloglog) shows the factor loading of

The question was how to get the p-value from the fit below, as an S object sr<-survreg(s~groups, dist="gaussian") Coefficients: (Intercept) groups -0.02138485 0.03868351 Scale= 0.01789372 Loglik(model)= 31.1 Loglik(intercept only)= 25.4 Chisq= 11.39 on 1 degrees of freedom, p= 0.00074 n= 16 ---- In general, good places to start are > names(sr) >

Thanks for sharing the questions and responses! Is it possible to appreciate how much the coefficients matter in one or the other model? Say, using Biau's example, using coxph, as.factor(grade2 == "high")TRUE gives hazard ratio 1.27 (rounded). As clinician I can grasp this HR as 27% relative increase. I can relate with other published results. With survreg the Weibull model gives a

Hello, I?m trying to do some analysis using survreg function. I need to implement there my own distribution with density: lambda*exp(-lambda*y), where y = a1/(1+exp(-a2*x)). a1, a2 are unknown parameters and x >0. I need to get estimates of a1 and a2 (and lambda of course) I?m really not good at programming. Is there any way how to implement this distribution to survreg without