search for: survreg

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, 70168000, 69120000, 68798000, 72615000, 133051000, 38384000, 15204000, 1558000, 14181000), stress=c(90, 100, 110, 90, 100, 80, 70, 60, 56, 62, 62, 59, 56...

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, weight...

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,...

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian, dist='...

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 'flexsurv' offer functions similar to survreg with gompertz support. However, due to the...

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....

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 by the change in sign between the two model...

...ments that I am trying to fit a model to. I also have uncertainties for these measurements. Some of the measurements are not well detected, so I'd like to use a limit instead of the actual measurement. (I am always dealing with upper limits, i.e. left censored data.) I have successfully run survreg using the combination of well detected measurements and limits, but I would like to include the measurement uncertainty (for the well detected measurements) in the fitting. As far as I can tell, survreg doesn't support this. Does anyone have a suggestion for how to accomplish this? Thanks,...

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 - t...

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 1.40297214 1 4 0.0000000 0.09112871 1 5 0.0000000 0.75115280 1 6 0.0000...

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 # OK fit2 <- survreg(Surv(time, st...

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 for the intercept term(in survreg)=log(scale) (in aftreg) that are similar but with a worse precision of 10^-1 or 10^-2. Any i...

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...

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 solution to this problem? Sincerely, Jerome Asselin libra...

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 to create that correctly. I've tried to inherit it from the "extreme" value dist, analogous...

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 dfico is -0.002 instead of 0.005. Al...

Full_Name: Tim Cohn Version: 1.6.1 OS: Macintosh OS X Submission from: (NULL) (130.11.34.250) The Mac version of survreg does not handle left-censored data correctly (at least the results are not what I get doing it other ways, and they are not the same as I get running R 1.6.1 in Windows 98se; the Windows 98 results are correct). On the windows version of R 1.6.1. > survreg(Surv(c(0,1,-1),c(T,T,F),type='l...

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 int...

...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 coefficient -0.4035245: is it feasible or meaningful to translate it to HR? Thanks in advance, Vincent Vinh-Hung Radiation Oncology, Geneva University Hospitals On Sun, Nov 14, 2010 at 6:51 AM, Biau David <djmbiau at yahoo.fr> wrote: > Dear R help list, > &g...

Does anyone know if the survreg function in the survival package can fit numerical covariates ? When I fit a survival model of the form survreg( Surv(time,censored) ~ x ) then x is always treated as a factor even if it is numeric (and even if I try to force it to be numeric using as.numeric(x). Thus, in the particular examp...