similar to: Printing output on Plot

I am unable to get the basehaz function to apply a proportional hazards model to a new data frame. I replicated my specific situation with the example for coxph in the help, where I changed the x value of the first record from 0 to 1. Is there something incorrect in the syntax that I am using? Thanks in advance! test1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0),

Hi, Can anybody tell me what the message below means and how to overcome it. Thanks, Neil Warning message: X matrix deemed to be singular; variable 2 in: coxph(Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + ......... >

Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)

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

Hi there, After I fitted a cox model, I used vcov to obtain the variance of the parameter estimate. It worked correctly in R 1.9.1. But it failed in R 2.0.0 and the error message is Error in vcov(cox.1) : no applicable method for "vcov" I don't know if it is a bug or there is some update on this function. Thanks! Lei Liu Assistant Professor Division of Biostatistics and

I hope one and all will allow a stats question: When running a cox proportional hazards model ,there are two ways to deal with age, including age as a covariate, or to include age as part of the follow-up time, viz, Age as a covariate: tetest1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0), age= c(0, 2,1,1,1,0,0),

I stumbled onto this working on an update to coxph. The last 6 lines below are the question, the rest create a test data set. tmt585% R R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) # Lines of code from survival/tests/singtest.R > library(survival) Loading required package: splines

I'm testing out some changes to survreg and got the following output, the likes of which I've never seen before: ---------------------------------------------------------------------- R version 2.7.1 (2008-06-23) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it

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

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

I am using the package Design for survival analysis. I want to plot a simple Kaplan-Meier fit of survival vs. age, with age grouped as quantiles. I can do this: survplot(survfit(Surv(time,status) ~ cut(age,3), data=veteran) but I would like to do something like this: survplot(survfit(Surv(time,status) ~ quantile(age,3), data=veteran) #will not work ideally I would like to superimpose

Hi all, I am trying to draw a Kaplan-Meier curve and I found online that Kaplan - Meier estimates are computed with a function called km in the event package. Is there an update for that because when I choose to download packages in R,. there is no package called event, even though I have selected all the repositories. Thanks in advance, Eleni [[alternative HTML version deleted]]

Hello, I have a question about the function lifetab in package KMsurv. The description of the output value surv says "the estimated survival function at the start of the intervals". Are these estimates the ones calculated via Kaplan-Meier probability of survival ? Thanks in advance! -- View this message in context:

One approach is to use the rms package's cph and Mean.cph functions. Mean.cph (cph calls coxph and can compute Kaplan-Meier and other survival estimates) can compute mean restricted life. Frank Dinesh W wrote > I am using survfit to generate a survival curve. My population is such > that my x axis is in days and i have a starting population of say 10,000 > of which say only 2000 are

Dear help news reader, I'm trying to draw a Kaplan-Meier curve and would like to ask the news group for some help Supposing I have study comapring two drugs, "A", and "B" and I recorde the time to get to the clinical endpoint (Time), in my case becommming virus free. I have setup the following frame: Time c Drug 1 5 1 A 2 7 1 B 3 2 1 A 4 10 1

Hi Fello: I am asked to compute the Kaplan-Meier estimator of data with right censoring without using surfit(). Does anyone know how to use R to compute the estimators? The data should input X: vector of right-censored observed time for n individuals, and d: vector of failure time indicators (0=censored individual;1=unconsored individual) and the function should return with t: vector of sorted

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:

Dear all, I have a stat question that may not be related to R, but I would like to have your advice. I have just read a medical paper in which the authors report the 1-p (where p is the cumulative survival probability from the Kaplan Meier curve) as incidence of disease. Specifically, the study followed ~12000 women on drug A and ~20000 women on drug B for 12 months. During that period

Dear R-users I am trying to make an adjusted Kaplan-Meier curve (using the Survival package) but I am having difficulty with plotting it so that the plot only shows the curves for the adjusted results. My data come from a randomised controlled trial, and I would like the adjusted Kaplan-Meier curve to only show two curves for the adjusted survival: one for those on treatment (Treatment==1)

I have a study best described as a retrospective case-cohort design: the cases were all the events in a given time span surveyed, and the controls (event-free during the follow-up period) were selected in 2:1 ratio (2 controls per case). The sampling frequency for the controls was about 0.27, so I used a weight vector consisting of 1 for cases and 1/0.27 for controls for coxph to adjust