similar to: worked in R, but not in S-Plus

The following works, you need to include x=TRUE in the call to coxph. Passing the time and status variables as additional arguments is a matter of personal preference. f.coxph.zph<-function(x, timeVar, statusVar) { cox.fit <- coxph(Surv(timeVar, statusVar) ~ x, na.action = na.exclude, method = "breslow", x=TRUE) fit.zph<-cox.zph(cox.fit) fit.zph$table[,3] } time.cox <-

Dear all, I have a question concerning the p-value. When running coxph I get a p-value = 0. :confused: Can this be true? Why aren?t there decimal points? Is there a way to find out the exact p-value? Here is the output: ---------------------------------------------------------------------------------------------------- Call: coxph(formula = Surv(start, stop, status) ~ Albumin_gproL, data = dial,

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

> ... interactions between covariables and time. A model such as "coxph(Surv(ptime, pstat) ~ age + age*ptime, ...." is invalid -- it is not at all what you think. If cph flags this as an error that is a good thing: I should probably add the same message to coxph. > Is is somewhat sensible to use cox.zph() to investigate which variables need time interaction... The cox.zph

Hi, I've been banging my head against the following problem for a while and thought the fine people on r-help might be able to help. I'm using the survival package. I'm studying the survival rate of a population with a preexisting linear-like event rate (there are theoretical reasons to believe it's linear, but of course it's subject to the usual sampling noise) Some of the

Dear All, I am trying to apply the function 'cox.zph' of the library survival, but I receive this error message: not found the object 'residuals.coxph'. I have re-installed the library 'survival' without any change and also a search with RSiteSearch was unsuccessful.. Any suggestion? TIA Giovanni PS: R 2.2.0 Windows XP HE dr. Giovanni Parrinello Section of Medical

  Hi there,   I am beginner in R and I have some basic question. Suppose I run a common procedure such as a t test or cox model like below:   out<-coxph( Surv(tstart,tstop, death1) ~ x1+x1:log(tstop+1) , test1,method=c("breslow"))    Which yields the following result:   Call: coxph(formula = Surv(tstart, tstop, death1) ~ x1 + x1:log(tstop +     1), data = test1, method =

Hello, I just started using R to do epidemiologic simulation research using the Cox proportional hazard model. I have 2 covariates X1 and X2 which I want to model as h(t,X)=h0(t)*exp(b1*X1+b2*X2). I assume independence of X from t. After I simulate Time and Censor data vectors denoting the censoring time and status respectively, I can call the following function to fit the data into the Cox

We occasionally utilize the coxph function in the survival library to fit multinomial logit models. (The breslow method produces the same likelihood function as the multinomial logit). We then utilize the predict function to create summary results for various combinations of covariates. For example:

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

Dear R-help members, Apologies - I am posting on behalf of a colleague, who is a little puzzled as STATA and R seem to be yielding different survival estimates for the same dataset when treating a variable as ordinal. Ordered() is used to represent an ordinal variable) I understand that R's coxph (by default) uses the Efron approximation, whereas STATA uses (by default) the Breslow. but we

I am confused when trying the function survfit. my question is: what does the survival curve given by plot.survfit mean? is it the survival curve with different covariates at different points? or just the baseline survival curve? for example, I run the following code and get the survival curve #### library(survival) fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)

I have a very general question about what the centering option in basehaz does to factors. (basehaz computes the baseline cumulative hazard for a coxph object using the Breslow estimator). Lets say I'm interested in a survival model with two (dichotomous) factors and a continuous covariate. Variable Possible Values Factor1 0 or 1 Factor2 0 or 1

Dear R-users, I am checking the proportional hazard assumption of a cox model for a given covariate, let say Z1, after adjusting for other relavent covariates in the model. To this end, I fitted cox model stratified on the discrete values of Z1 and try to get beslow estimator for the baseline cumulative hazard function (H(t)) in each stratum. As far as i know, if the proportionality assumption

I am calculating cox propotional hazards models with the coxph function from the survival package. My data relates to failure of various types of endovascular interventions. I can successfully obtain the LR, Wald, and Score test p-values from the coxph.object, as well as the hazard ratio as follows: formula.obj = Surv(days, status) ~ type coxph.model = coxph(formula.obj, df) fit =

I'm trying to estimate a cox proportional hazards regression for repeated events (in gap time) with time varying covariates. The dataset consists of just around 6000 observations (lines) (110 events). The (stylized) data look as follows: unit dur0 dur1 eventn event ongoing x 1 0 1 0 0 0 32.23 1 1 2 0 1 1 35.34 1

Dear R Help list, My question is regarding extracting the standard error or Z-score from a cph or coxph call. My Cox model is: - modz=cph(Surv(TSURV,STATUS)~RAGE+DAGE+REG_WTIME_M+CLD_ISCH+POLY_VS, data=kidneyT,method="breslow", x=T, y=T) I've used names(modz) but can't see anything that will let me extract the Z scores for each coefficient or the standard errors in the same

This is a simple R program that I have been trying to run. I keep running into the "singular matrix" error. I end up with no sensible results. Can anyone suggest any changes or a way around this? I am a total rookie when working with R. Thanks, Rasika > library(survival) Loading required package: splines > args(coxph) function (formula, data, weights, subset, na.action, init,

Hello, I am currently trying to simulate data and analyze it using the frailty option in the coxph function. I am working with recurrent event data, using counting process notation. Occasionally, (about 1 in every 100 simulations) I get the following warning: Error in coxph(Surv(start, end, censorind) ~ binary + uniform + frailty(subject, : X matrix deemed to be singular; variable 2 My

Dear R-users, I would like to determine the probability of event at specific time using cox model fit. On the development sample data I am able to get the probability of a event at time point(t). I need probability score of a event at specific time, using scoring scoring dataset which will have only covariates and not the response variables. Here is the sample code: n = 1000 beta1 = 2; beta2 =