similar to: how to code the censor variable for "survfit"

Terry, I recently noticed the censor argument of survfit. For some analyses it greatly reduces the size of the resulting object, which is a nice feature. However, when combined with the id argument, only 1 prediction is made. Predictions can be made individually but I'd prefer to do them all at once if that change can be made. Chris ##################################### # CODE # create

Hello, I'm using plot.survfit to plot cumulative incidence of an event. Essentially, my code boils down to: cox <-coxph(Surv(EVINF,STATUS) ~ strata(TREAT) + covariates, data=dat) surv <- survfit(cox) plot(surv,mark.time=F,fun="event") Follow-up time extends to 54 weeks, but the last event occurs at week 30, and no more people are censored in between. Is there a

hie when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example n=20 n1=n/2 n2=n/4 a11=4;a12=4 ;a21=4 ;a22=4 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

Hi all, I am getting an error in my code and I don't know what the problem is. I am using R 2.9 on ubuntu. my code is as follows: ## Libraries ## library(survival) library(foreign) ## reading data ## data<-read.dta("http://psfaculty.ucdavis.edu/bsjjones/cabinet.dta") head(data) attach(data) fit1<-survfit(Surv(durat,censor)) and I get the following error >

Hi all! I am trying to extract output information from the survfit function in order to generate a matrix of select output for multiple factors. Specifically, I am interested in extracting the number of events (in the output below: 106, 2, 3). The variable names represented in my function (ee) are shown below, but none of those variables correspond to the column of events as shown in the output.

Hi, when ploting Kaplan-Meier estimate curves as below, the censoring symbols (crosses) to not change thickness along the lines plot(survfit(surv ~ I(x>=cut.off) ),lty=c(1,2), lwd=2) is there any strightforward way to make it happen? thanks robert -- View this message in context: http://r.789695.n4.nabble.com/censoring-symbols-on-survfit-plot-tp3311283p3311283.html Sent from the R help

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

Hi, when I generated a survfit() object, I can get number of patients at risk at various time points by using summary(): fit<-survfit(Surv(time,status)~class,data=mtdata) summary(fit) class=1 time n.risk n.event survival std.err lower 95% CI upper 95% CI 9.9 78 1 0.987 0.0127 0.963 1 41.5 77 1 0.974 0.0179 0.940 1 54.0 76

En el libro EPICALC (pagina 229-230) en el que está el siguiente script, todo nos funciona bien, pero cuando vamos a life table, ya allí no avanza, lo señalamos en el script, por favor quizá se nos haya ido algún detalle, pero fuimos siguiéndolo por el libro paso a paso y no no hemos percatado Todos los de el paquete survival de la ayuda del R funcionan perfectamente

Hello everybody, does anybody know how the function plot.survfit exactly works? I'd like to plot the log of the cummulative hazard against the log time by using plot.survfit(...fun="cloglog") which does not work correctly. The scales are wrong and there is an error message about infinit numbers. It must have something to do with the censored data, doesn't it? #Example:

The answer to this may be obvious, but I was wondering in the rms package and the survfit(), how you can plot the censored time points as ticks. Take for example, library(survival) library(rms) foo <- data.frame(Time=c(1,2,3,4,5,6,10), Status=c(1,1,0,0,1,1,1)) answer <- survfit(Surv(foo$Time, foo$Status==1) ~1) # this shows the censored time points as ticks at Time = 3 and 4 plot(answer)

Hola Jose, y podrias enviar algun link al libro o enviar esa parte escaneada ?? para minimizar la posibilidad de que haya algo en la transcripcion del script ... Slds, eric. On Fri 04 Oct 2013 01:59:33 PM CLT, daniel wrote: > José, > > La función survfit espera como argumento una fórmula. Al pie del help de la > función survfit dice: > > "Versiones anteriores del código

Hi everyone, I am not new to R, but new to running survival models in R. I am trying to create some basic KM curves, using the following code: library(survival) library(KMsurv) (import data etc - basic right censored, with continuously observed time of death) sleepfit <- survfit(Surv(timeb, death), data = sleep) Here timeb is measured is survival in years, death is a 1/0 indicator (1 =

Hi all, I have a little problem. I make an weibull survival analysis using the survival package. It,s OK, them I have the functions. I plot this funcions with curve(). I want to make a plot with the real survival points (proportion of alive x time) and them add the curves to points. I have the time to dead, the censor data and my trataments. To analysis the model is: model1 <-

This works ok: > cox = coxph(surv ~ bucket*(today + accor + both) + activity, data = data) > fit = survfit(cox, newdata=data[1:100,]) but using strata leads to problems: > cox.s = coxph(surv ~ bucket*(today + accor + both) + strata(activity), > data = data) > fit.s = survfit(cox.s, newdata=data[1:100,]) Error in model.frame.default(data = data[1:100, ], formula = ~bucket + :

Hello, I am trying to compute the mdeian of the survival time from the function survfit: > fit <- survfit(Surv(time, status) ~ 1) > fit Call: survfit(formula = Surv(time, status) ~ 1) records n.max n.start events median 0.95LCL 0.95UCL 111 111 111 20 NA NA NA The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing this correct?

Hello, I am having hard time obtaining a value from a function. "fit" is a survival function that produces some results, such as "median", "confidence intervals" etc. But str() function does not list these values. How can I extract these to be able use them? For example, I need "median" value for the group DrugA which is 48. "Print" function does

I have attempted to follow posting guidelines but I have failed to find out what I am doing wrong here. I am trying to use lines.survfit to plot a second curve onto a survival curve produced by plot.survfit. In my case this is to be a progression free survival curve superimposed upon an overall survival curve, but I will illustrate my problem using the example given in the help for

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5