similar to: Package "survival" --- Difference of coxph strata with subset?

Hi John, Brilliant solution and the best sort - when you finally solve your problem by yourself. Jim On Thu, Oct 1, 2020 at 2:52 AM array chip <arrayprofile at yahoo.com> wrote: > > Hi Jim, > > I found out why clip() does not work with lines(survfit.object)! > > If you look at code of function survival:::lines.survfit, in th middle of the code: > > do.clip <-

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Dear all, I am using R 3.4.3 on Windows 10. I am preparing some teaching materials and I'm having trouble matching the by-hand version with the R code. I have fitted a Cox model - let's use the ovarian data as an example: library(survival) data(ovarian) ova_mod <- coxph(Surv(futime,fustat)~age+rx,data=ovarian) If I want to make predict survival for a new set of individuals at 100

Hello, I am having some trouble with the 'censoring-adjusted C-index' by Uno et al, in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <-

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158

Hi I am trying to understand how to get the validate() function in Design to work with the subset option. I tried this: ovarian.cph=cph(Surv(futime, fustat) ~ age+factor(ecog.ps)+strat(rx), time.inc=1000, x=T, y=T, data=ovarian) validate(ovarian.cph) #fine when no subset is used, but the following two don't work: > validate(ovarian.cph, subset=ovarian$ecog.ps==2) Error in

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,

Dear R users, I am trying to use "muhaz" and "plot.muhaz" functions in "muhaz" package to estimate and plot hazard funciton. However function "muhaz" always gives error message "Error in Surv(times, delta) : object "times" not found". I could not even run their sample codes in the user's manual as follows: data(ovarian)

Dear UseR, I do not know if this a problem with me, my data or cph/survest in package design. The example below works with a standard data set, but not with my data, but I cannot locate the problem. Note that I am using an older package of survival to avoid a problem with the newly renamed function in survival meeting Design. Dieter # First, check standard example to make sure library(Design)

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.

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

Dear all, I am doing library(survival) fit <- coxph(Surv(futime,fustat) ~ rx, ovarian) plot(survfit(fit,newdata=ovarian),col=c(1,2)) legend("bottomleft", legend=c("rx = 0", "rx = 1"), lty=c(1,2),col=c(1,2)) Is this correct to compare these two groups? Is the 0.31 the p-value that the median f two groups are equal Why lty does not work here? Many thanks

Dear R-users, I apply to your kind attention to know if someone have used the Splus software koq.q (Kent & O'Quigley's measure of dependence for censored data) in R and kindly can help me. I have tried several times to contact the authors Andrej Blejec (andrej.blejec at uni-lj.si) or Janez Stare (janez.stare at mf.uni-lj.si) but unfortunately no one answered me. Following

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 <-

Hello everyone, I tried to use coxph.detail() to get the hazard function. But a warning messge always returns to me, even in the example provided by its help document: > ?coxph.detail > fit <- coxph(Surv(futime,fustat) ~ age + rx + ecog.ps, ovarian, x=TRUE) > fitd <- coxph.detail(fit) Warning message: data length [37] is not a sub-multiple or multiple of the number of rows

Given a cox model: library(Hmisc); library(survival); (library(Design); cox.model=cph(Surv(futime, fustat) ~ age, data=ovarian, surv=T) str(cox.model) What I need is the total estimated time until failure (death), not the probability of failing at a given time (survival probability), or hazard etc, which is what I get from survest and predict for example. I suspect the answer is

Hello: In the example below (or for a censored data) using survfit.coxph, can anyone point me to a link or a pdf as to how the probabilities appearing in bold under "summary(pred$surv)" are calculated? Do these represent acumulative probability distribution in time (not including censored time)? Thanks very much, parmee *fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)*

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <-

hi sorry if this has been discussed before, but I'm wondering why the scaled Schoenfeld residuals do not follow the defining formula for obtaining them from the ordinary Schoenfeld residuals, but are instead offset by the estimated parameter values. e.g. library(survival) attach(ovarian) sv<-Surv(futime,fustat) f1<-coxph(sv~age+ecog.ps) f1