similar to: time grid for "survfit" Survival function outputs

Dear All, There are two things about mle() that I wasn't so sure. 1) can mle() handle vector based parameter? say ll<-function(theta=rep(1,20)){..............} I tried such function, it worked for "optim" but not for "mle". 2) is there a general suggestion for the maximum number of parameters allowed to use in mle() or optim()? Thank you. Regards, MJO

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)

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, 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?

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

> Further I appreciate your new function survmean(). At the moment it > seems to be intended as internal, and not documented in the help. The computations done by print.survfit are now a part of the results returned by summary.survfit. See 'table' in the output list of ?summary.survfit. Both call an internal survmean() function to ensure that any future updates stay in

Hi, I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this: library(survival) data(lung) lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung) # lung.byPS Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung) 1

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

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:

I am using the coxph, survfit and summary.survfit functions to calculate an estimate of predicted survival with confidence interval for future patients based on the survival distribution of an existing cohort of subjects. I am trying to understand the calculation and interpretation of the std.err and confidence intervals printed by the summary.survfit function. Using the default confidence

Hi! I came across R just a few days ago since I was looking for a toolbox for cox-regression. I?ve read "Cox Proportional-Hazards Regression for Survival Data Appendix to An R and S-PLUS Companion to Applied Regression" from John Fox. As described therein plotting survival-functions works well (plot(survfit(model))). But I?d like to do some manipulation with the survival-functions

Hello, I'm trying to estimate a Cox proportional hazard model with time-varying covariates using coxph. The parameter estimates are fine but there is something wrong with the survival curves I get with survfit (results are not plausible). Let me explain why I think something's wrong. To make sure I'm setting up my data correctly to estimate a model with time-varying covariates, I

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I

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 <-

Hello, I create a plot from a coxph object called fit.ads4: plot(survfit(fit.ads4)) plot is located at: https://www.dropbox.com/s/9jswrzid7mp1u62/survfit%20plot.png I also create the following survfit statistics: > print(survfit(fit.ads4),print.rmean=T) Call: survfit(formula = fit.ads4) records n.max n.start events *rmean *se(rmean) median 0.95LCL 0.95UCL 203.0

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

I've a data set with 60000 rows of data representing 6000+ distinct loans. I did a coxph() regression on it (see call below), but a subsequent survfit() call on the coxph object is almost certainly wrong. It gives n=6 when it should be more like 6000+ (I think) > survfit(resultag) Call: survfit.coxph(object = resultag) n events median 0.95LCL 0.95UCL 6 489 Inf

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 =

Dear r-helpers, This is my first time to run survival analysis. Currently, I have a data set which contains two variables, the variable of time to event (or time to censoring) and the variable of censor indicator. For the indicator variable, it was coded as 0 and 1. 0 represents right censor, 1 means event of interest. Now I try to use "survfit" in the package of "survival". I

Hi, I am wondering how the confidence interval for Kaplan-Meier estimator is calculated by survfit(). For example,  > summary(survfit(Surv(time,status)~1,data),times=10) Call: survfit(formula = Surv(rtime10, rstat10) ~ 1, data = mgi)  time n.risk n.event survival std.err lower 95% CI upper 95% CI    10    168      55    0.761  0.0282        0.707        0.818 I am trying to reproduce the