similar to: expected risk from coxph (survival)

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),

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 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

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'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-experts Execuse me for an easy question, but I need help, sorry for that. >From days I have been working with a large dataset, where operations are needed within a component of dataset. Here is my question: I have big dataset where x1:.....x1000 or so. What I need to do is to work on 4 consequite variables to calculate a statistics and output. So far so good. There are more vector

Hi, I'm trying to print the p-values from the output of a CPH test onto a Kaplan Meier plot. Can this be done? I only really want the p-values from the CPH test to appear but if this can't be done I am willing to have the entire CPH output. This is what I am currently trying: (it doesn't print the CPH output) plot_KM <- function(field) { library(survival) y =

Hi list, I was trying to use "predict.coxph" to calculate martingale residuals on a test data, however, as pointed out before http://tolstoy.newcastle.edu.au/R/e4/help/08/06/13508.html predict(mycox1, newdata, type="expected") is not implemented yet. Dieter suggested to use 'cph' and 'predict.Design', but from my reading so far, I'm not sure they can

rbind() does something strange to dimnames R : Copyright 1997, Robert Gentleman and Ross Ihaka Version 0.50 Beta (April 1, 1997) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type "license()" for details. > test1 <- data.frame(time= c(4, 3,1,1,2,2,3), + status=c(1,NA,1,0,1,1,0), + x= c(0,

Thanks for the reproducable example. I can confirm that it fails on my machine using survival 2-37.5, the next soon-to-be-released version, The issue is with NextMethod, and my assumption that the called routine inherited everything from the parent, including the environment chain. A simple test this AM showed me that the assumption is false. It might have been true for Splus. Working this

Dear all, I'm interested in fitting survival trees with competing risk analysis. The tree should show the cumulative incidence function for each terminal node . I read several paper illustrating this possibility, but to the best of my knowledge no R code are reported. There is any R package for this fit? Thank you very much in advance. Mario Petretta

Dear list, does anyone know of a R-package that has implemented the increasingly popular inclusion of the number of patients at risk below Kaplan-Meier curves like in http://bloodjournal.hematologylibrary.org/content/vol116/issue19/images/large/zh89991058760001.jpeg any hint (or negative answer) is much appreciated. Thanks Thorsten -- Thorsten Raff 2nd Medical Department, University

Dear R-help, I am using R-3.3.2 on Windows 10. I teach on a course which has 4 computer practical sessions related to the development and validation of clinical prediction models. These are currently written for Stata and I am in the process of writing them for use in R too (as I far prefer R to Stata!) I notice that predictions made from a Cox model in Stata are based on un-centred variables,

The title says it all really; I am looking for a function along the lines of cutpt.coxph as described in "Survival Analysis Using S" (Tableman and Kim), Chapter 6. As may be guessed, the function optimises the cutpoint of a continuous variable for cox proportional hazard modelling. I can't find it, or any similar function, on CRAN. Alternatively, perhaps there is a way of extracting

[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)~age, ovarian, subset=rx==1)

I have what I *think* should be a simple problem in R, and hope someone might be able to help me. I'm working with cancer survival data, and would like to calculate adjusted survival figures based on the age of the patient and the tumour classification. A friendly statistician told me I should use Cox proportional hazards to do this, and I've made some progress with using the

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

fit <- coxph(Surv(futime,fustat)~ age +strata(rx), data=ovarian, subset=1:23) curves <- survfit(fit, newdata=ovarian[24:26,]) I don't think this is mentioned in the documentation (I'll have to fix that!), but subscripting works for survfit objects. In this case there are 2 strata and 3 subjects, and curves[1,2] will return the survival curve for strata 1,