similar to: defining custom survreg distributions

Hi all, I am trying to run Weibull PH model in R. Assume in the data set I have x1 a continuous variable and x2 a categorical variable with two classes (0= sick and 1= healthy). I fit the model in the following way. Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull") My questions are 1. Is it Weibull PH model or Weibull AFT model? Call: survreg(formula = Surv(time, delta) ~ x1

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

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

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 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),

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 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 <-

Full_Name: Andrew Manners Version: 2.5.1 OS: windows xp prof 2003 Submission from: (NULL) (130.102.0.177) To whom it may concern, I'm trying to get an ANOVA table within survreg but it always produces NA's in the p-value, regardless of the data set. The data set below comes from Tableman and Kim 2004. I had the same problem on a number of my own data sets. I searched the R site for

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,

---------------------------- Original Message ---------------------------- Subject: Different results from survreg with version 2.6.1 and 2.10.1 From: nathalcs at ulrik.uio.no Date: Fri, March 19, 2010 16:00 To: r-help at r-project.org -------------------------------------------------------------------------- Dear all I'm using survreg command in package survival.

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

Dear list, I would like to simulate individual survival times from a model that has been fitted using the survreg procedure (library survival). Output shown below. My plan is to extract the shape and scale arguments for use with rweibull() since my error terms are assumed to be Weibull, but it does not make any sense. The mean survival time is easy to predict, but I would like to simulate

Dear r-help subscribers, I am working on some survival analysis of some interval censored failure time data in R. I have done similar analysis before using PROC LIFEREG in SAS. In that instance, a gamma survival function was the optimum parametric model for describing the survival and hazard functions. I would like to be able to use a gamma function in R, but apparently the survival package does

Dear r-help subscribers, A couple of weeks ago I sent the following message to the r-help mail list. It hasn't generated any response, and I could really use some help on this. Anyone able to help? Thanks again, Roger Dungan >> I am working on some survival analysis of some interval censored failure time data in R. I have done similar analysis before using PROC LIFEREG in SAS. In

Hi All, I have conducted the following survival analysis which appears to be OK (thanks BRipley for solving my earlier problem). > surv.mod1 <- survreg( Surv(timep1, relall6)~randgrpc, data=Dataset, dist="weibull", scale = 1) > summary(surv.mod1) Call: survreg(formula = Surv(timep1, relall6) ~ randgrpc, data = Dataset, dist = "weibull", scale = 1)

I am using psm to model some parametric survival data, the data is for length of stay in an emergency department. There are several ways a patient's stay in the emergency department can end (discharge, admit, etc..) so I am looking at modeling the effects of several covariates on the various outcomes. Initially I am trying to fit a survival model for each type of outcome using the psm

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 =

What objective function is maximized by survreg with the default Weibull model? I'm getting finite parameters in a case that has the likelihood maximzed at Infinite, so it can't be a simple maximum likelihood. Consider the following: ############################# > set.seed(3) > Stress <- rep(1:3, each=3) > ch.life <- exp(9-3*Stress) > simLife <- rexp(9,

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

Am I correct in assuming that the output below essentially translates to "Males have a mean time that is significantly lower than Females"? Is this the correct way to interpret the fact that the coefficient is negative? Assume the variale sex is treated as a factor with Female =0 and Male=1. survmodel<-survreg(survobj~sex,data=data1, dist="weibull")