similar to: predict.coxph and predict.survreg

Since I read the list in digest form (and was out ill yesterday) I'm late to the discussion. There are 3 steps for predicting survival, using a Cox model: 1. Fit the data fit <- coxph(Surv(time, status) ~ age + ph.ecog, data=lung) The biggest question to answer here is what covariates you wish to base the prediction on. There is the usual tradeoff between too few (leave out something

Dear all, I'm running a coxph model of the form: coxph(Surv(Start, End, Death.ID) ~ x1 + x2 + a1 + a2 + a3) Within this model, I would like to compare the influence of x1 and x2 on the hazard rate. Specifically I am interested in testing whether the estimated coefficient for x1 is equal (or not) to the estimated coefficient for x2. I was thinking of using a Chow-test for this but the Chow

Dear all, I'm asking this question again as I didn't get a reply last time: I'm doing a survival analysis with time-dependent covariates. Until now, I have used a simple Cox model for this, specifically the coxph function from the survival library. Now, I would like to try out an accelerated failure time model with a parametric specification as implemented for example in the survreg

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 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

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

Hi all, Is there an equivalent to predict(...,type="linear") of a Proportional hazard model for a Cox model instead? For example, the Figure 13.12 in MASS (p384) is produced by: (aids.ps <- survreg(Surv(survtime + 0.9, status) ~ state + T.categ + pspline(age, df=6), data = Aidsp)) zz <- predict(aids.ps, data.frame(state = factor(rep("NSW", 83), levels =

Dear all, I have several character strings with a high number of different levels. unique(x) gives me values in the range of 100-200. This creates problems as I would like to use them as predictors in a coxph model. I therefore would like to convert each of these strings to a new string (x_new). x_new should be equal to x for the top n categories (i.e. the top n levels with the highest

Why do I get different sign of the coefficients of covariates when I run the semi-parametric proportional hazard model (coxph) compared to the parametric proportional hazard model (survreg)? Anyone with experience in extracting information form survreg to make predictions are free to contact me. Cheers, Ullrika [[alternative HTML version deleted]]

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

Dear all, The help file for the generic function vcov states "Classes with methods for this function include: 'lm', 'glm', 'nls', 'lme', 'gls', 'coxph' and 'survreg' (the last two in package 'survival')." Since, I am not able to use vcov.coxph(), I am wondering whether I am missing something (as I suspect..) regards, vito

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 users, I would like to obtain a survival curve from a Cox model that is smooth and does not have zero differences due to no events for those particular days. I have: > sum((diff(surv))==0) [1] 18 So you can see 18 days where the survival curve did not drop due to no events. Is there a way to ask survfit to fit a nice spline for the survival?? Note: I tried survreg and it did not

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

A recent question in r-help made me realize that I should add a drop1 method for coxph and survreg. The default does not handle strata() or cluster() properly. However, for coxph the right options for the "test" argument would be likelihood-ratio, score, and Wald; not chisq and F. All of them reference a chi-square distribution. My thought is use these arguments, and add an

Dear list: I am attempting to use what I thought would be a pretty straightforward practical application of Cox regression. I figure users of the survival package must have come across this problem before, so I would like to ask you how you dealt with it. I have set up an illustrative example and included it at the end of this post. I took a sample of 100 data points from each of two populations

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'"

Dear all, I have a large dataset (N=100,000 with 89 variables per subject). This dataset is stored in a 100.000 x 89 matrix where each row describes one individual and each column one variable. What is the easiest way of selecting a subset of let's say 1.000 individuals out of that whole matrix? Thanks, Michael Michael Haenlein Associate Professor of Marketing ESCP-EAP European School of

Dear all, is there a function similar to extractAIC based on which I can extract the BIC (Bayesian Information Criterion) of a coxph model? I found some functions that provide BIC in other packages, but none of them seems to work with coxph. Thanks, Michael [[alternative HTML version deleted]]

Hi I am estimating the following coxph function with stratification and frailty?where each person had multiple events. m<-coxph(Surv(dtime1,status1)~gender+cage+uplf+strata(enum)+frailty(id),xmodel) ? > head(xmodel) id enum dtime status gender cage uplf 1 1008666 1 2259.1412037 1 MA 0.000 0 2 1008666 2 36.7495023 1 MA 2259.141 0 3 1008666