similar to: changes in coxph in "survival" from older version?

Hi all, I cant' wrap my head around an error from the coxph function (package survival). Here's an example: library(survival) n = 100; set.seed(1); time = rexp(n); event = sample(c(0,1), n, replace = TRUE) covar = data.frame(z = rnorm(n)); model = coxph(Surv(time, event)~ . , data = covar) R gives the following error: > model = coxph(Surv(time, event)~ . , data = covar) Error in

I've never seen this, and have no idea how to reproduce it. For resloution you are going to have to give me a working example of the failure. Also, per the posting guide, what is your sessionInfo()? Terry Therneau On 08/09/2012 04:11 AM, r-help-request at r-project.org wrote: > I have a couple of questions with regards to fitting a coxph model to a data > set in R: > > I have a

Hello, I have a couple of questions with regards to fitting a coxph model to a data set in R: I have a very large dataset and wanted to get the baseline hazard using the basehaz() function in the package : 'survival'. If I use all the covariates then the output from basehaz(fit), where fit is a model fit using coxph(), gives 507 unique values for the time and the corresponding cumulative

Hello, I am interested in producing the expected number of events, in a recurring events setting. I am using the Andersen-Gill model, as fit by the function "coxph" in the package "survival." I need to produce expected numbers of events for a cohort, cumulatively, at several fixed times. My ultimate goal is: To fit an AG model to a reference sample, then use that fitted model

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

Dear R users, My apologies for the simple question, as I'm starting to learn the concepts behind the Cox PH model. I was just experimenting with the survival and rms packages for this. I'm simply trying to obtain the expected survival time (as opposed to the probability of survival at a given time t). I can't seem to find an option from the "type" argument in the predict

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

Thanks for the reproducable example. I can confirm that it fails on my machine using survival 2-37.5, the next soon-to-be-released version, The issue is with NextMethod, and my assumption that the called routine inherited everything from the parent, including the environment chain. A simple test this AM showed me that the assumption is false. It might have been true for Splus. Working this

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Hello R users,  I am trying to obtain a direct adjusted survival curve. I am sending my whole code (see below). It's basically the larynx cancer data with Stage 1-4. I am using the cox model using coxph option, see the fit3 coxph. When I use the avg.surv option on fit3, I get the following error: "fits<-avg.surv(fit3, var.name="stage.fac", var.values=c(1,2,3,4), data=larynx)

Well, I would suggest using the code already in place in the survival package. Here is my code for your problem. I'm using a copy of the larynx data as found from the web resources for the Klein and Moeschberger book. larynx <- read.table("larynx.dat", skip=12, col.names=c("stage", "time", "age", "year",

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

Hi, I am experimenting with the function predict() in two versions of R and the R extension package "survival". library(survival) set.seed(123) testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10)) testfm=as.formula('Surv(otime,event)~x') testfun=function(dat,fm) { predict(coxph(fm,data=dat),type='lp',newdata=dat) } # Under R 2.11.1 and

Hi, I am experimenting with the function predict() in two versions of R and the R extension package "survival". library(survival) set.seed(123) testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10)) testfm=as.formula('Surv(otime,event)~x') testfun=function(dat,fm) { predict(coxph(fm,data=dat),type='lp',newdata=dat) } # Under R 2.11.1 and

I wonder if anyone know why survest (a function in Design package) and standard survfit.coxph (survival) returned different confidence intervals on survival probability estimation (say 5 year). I am trying to estimate the 5-year survival probability on a continuous predictor (e.g. Age in this case). Here is what I did based on an example in "help cph". The 95% confidence intervals

Dear Terry, It's not surprising that different modeling functions behave differently in this respect because there's no articulated standard. Please see my response to Martin for my take on the singular.ok argument. For a highly sophisticated user like you, singular.ok=TRUE isn't problematic -- you're not going to fail to notice an NA in the coefficient vector -- but I've

Dear R users: I'm generating the following survival data: set.seed(123) n=200 #sample size x=rbinom(n,size=1,prob=.5) #binomial treatment v=rgamma(n,shape=1,scale=1) #gamma frailty w=rweibull(n,shape=1,scale=1) #Weibull deviates b=-log(2) #treatment's slope t=exp( -x*b -log(v) + log(w) ) #failure times c=rep(1,n) #uncensored indicator id=seq(1:n) #individual frailty indicator

I tried to compare if cch() and coxph() can generate same result for same case cohort data Use the standard data in cch(): nwtco Since in cch contains the cohort size=4028, while ccoh.data size =1154 after selection, but coxph does not contain info of cohort size=4028. The rough estimate between coxph() and cch() is same, but the lower and upper CI and P-value are a little different. Can we

The coxph function in R is not working for me when I use a continuous predictor in the model. Specifically, it fails to converge, even when bumping up the number of max iterations or setting reasonable initial values. The estimated Hazard ratio from the model is incorrect (verified by an AFT model). I've isolated it to the "x1" variable in the example below, which is log-normally

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),