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[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~a...

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 std.err lower 95% CI upper 95% CI 59 26 1 0.962 0.0377 0.890 1.000 115 25 1 0.923 0.0523 0.826 1.000 156 24...

...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 <- ovarian[1:16,] TE <- ovarian[17:26,] train.fit <- coxph(Surv(futime, fustat) ~ age,x=TRUE, y=TRUE, method="breslow", data=TR) lpnew <- predict(train.fit, newdata=TE) Surv.rsp <- Surv(TR$futime, TR$fustat) Surv.rsp.new <- Surv(TE$futime, TE$fustat) If time is left NULL, the maximum futime in the TE data is assumed: UnoC(Surv.rsp, Surv.rsp.n...

Hola Estoy tratando de correr un survival analysis usando un Cox regression model. Tengo una duda respecto a la organizacion del script. Tengo una variable que es -tamano del individuo- y quiero ver si hay diferencia en sobrevivencia respecto a tamano. Como diseno de campo los tamanos fueron ubicados de forma aleatoria en bloques al azar. Cuado planteo el script tengo algo como:

...hat clip() works if I just do some regular graphing tasks. But as long as I run lines(fit) with "fit" object is a survfit object, this would reset to default plot region. See the ovarian example below: > > > > library(survival) > > ovarian1<-ovarian > > ovarian1$fustat[ovarian$futime>450]<-0 > > ovarian1$futime[ovarian$futime>450]<-450 > > ovarian2<-subset(ovarian,futime>450) > > > > fit1 <- survfit(Surv(futime, fustat) ~ rx, data = ovarian1) > > fit2 <- survfit(Surv(futime, fustat) ~ rx, data = ovarian2) >...

...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, dist='weibull',scale=1) > summary(tmp) Call: survreg(formula = Surv(futime, fustat) ~ ecog.ps + rx, data = ovarian, dist = "weibull", scale = 1) Value Std. Error z p (Intercept) 6.962 1.322 5.267 1.39e-07 ecog.ps...

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 0.0463 3.17 0.0015 rx -0.815 0.443 0.6342 -1.28 0.2000 ecog.ps...

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 days then that is trivial using predict.coxph e.g.: newdata <- data.frame(futime=rep(100,5),fustat=rep(1,5),age=c(45,50,55,60,65),rx=c(1,2,1,2,1)) preds <- predict(ova_mod,newdata,type="expected...

...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) attach(ovarian) fit1 <- muhaz(futime, fustat) it gave the same error message. By the way, the "survival" package is installed and functions well on my computer. Does anyone have the same problem? Thanks! Deming

How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed

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 the predictor, whereas ridge penalty term contains variables `age' and `ph.ecog'. Could someone explain what it means to regularize on parameters which are not part of the model? Based...

Dear all, I am confused with the output of survfit.coxph. Someone said that the survival given by summary(survfit.coxph) is the baseline survival S_0, but some said that is the survival S=S_0^exp{beta*x}. Which one is correct? By the way, if I use "newdata=" in the survfit, does that mean the survival is estimated by the value of covariates in the new data frame? Thank you very much!

Hello there, I am doing analysis on survival data. How do I pick out a probability of survival at a chosen landmark time(for example, 3 years, 4 years) from the result of "survfit"? or any other functions? As I know, the result of 'survfit' only have the probabilities for all event or cencor time. Thank you very much Lisa Wang Princess Margaret Hospital Toronto, Ca tel:

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 [12] in matrix Can anyone suggest why it does not work? I use R 2.1.1 with Windows XP Thanks a lot. Zhen Zhang

in R when carring out the log rank test is the censored variable denoted by 1 or 0 or its of no consequence. thanks --------------------------------- always stay connected to friends. [[alternative HTML version deleted]]

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

...er, 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 of the survival package. Thanks DK. > fit <- coxph(Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta=1),data=ovarian) > a<-summary(fit) Call: coxph(formula = Surv(futime, fustat) ~ ridge(rx, age, ecog.ps, theta = 1), data = ovarian) n= 26 coef se(coef) se2 Chisq DF p ridge(rx) -0.780 0.5862 0.5589 1.77 1 0.1800 ridg...

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 order(c(1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, :...

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 embarrassing s...

...is 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 schres<-resid(f1,type="schoenfeld") schresc<-resid(f1,type="scaledsch") n=sum(fustat) V<-f1$var schresc1<-t(n*V%*%t(schres)) #schresc1 is how the scaled Schoenfeld residuals are defined #in terms of the number of events #variance of...