similar to: Problems with coxph and survfit in a stratified model with interactions

First, here is your message as it appears on R-help. On 10/14/2012 05:00 AM, r-help-request@r-project.org wrote: > I?m trying to set up proportional hazard model that is stratified with > respect to covariate 1 and has an interaction between covariate 1 and > another variable, covariate 2. Both variables are categorical. In the > following, I try to illustrate the two problems that

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

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 R users, My apologies for the simple question, as I'm starting to learn the concepts behind the Cox PH model. I was just experimenting with the survival and rms packages for this. I'm simply trying to obtain the expected survival time (as opposed to the probability of survival at a given time t). I can't seem to find an option from the "type" argument in the predict

Hello users, I would like to obtain a survival curve from a Cox model that is smooth and does not have zero differences due to no events for those particular days. I have: > sum((diff(surv))==0) [1] 18 So you can see 18 days where the survival curve did not drop due to no events. Is there a way to ask survfit to fit a nice spline for the survival?? Note: I tried survreg and it did not

I can use stepAIC on an lme object in 1.6.2, but I get the following error if I try to do the same in 1.7.0: Error in lme(fixed = resp ~ cov1 + cov2, data = a, random = structure(list( : unused argument(s) (formula ...) Does anybody know why? Here's an example: library(nlme) library(MASS) a <- data.frame( resp=rnorm(250), cov1=rnorm(250), cov2=rnorm(250),

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

the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data. I created a dummy data set to explain my situation: df1 <- data.frame( Observation =

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 <-

Hi I am a new fan of R after getting mad with the graphical functional in SPSS. I have been able to create a nice looking Kaplan Meyer graph using Survplot function. However I have difficulties in turning the y axis to percent instead of the default 0-1 scale. Further I have tried the function yaxt="n" without any results. Any help in this matter will be appreciated. The code is

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!

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.

I have a fairly large model: > length(Y) [1] 3051 > dim(covariates) [1] 3051 211 All of these 211 covariates need to be nested hierarchically within a grouping "class", of which there are 8. I have an accessory vector, " cov2class" that specifies the mapping between covariates and the 8 classes. Now, I understand I can break all this information up into individual

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

Since I read the list in digest form (and was out ill yesterday) I'm late to the discussion. There are 3 steps for predicting survival, using a Cox model: 1. Fit the data fit <- coxph(Surv(time, status) ~ age + ph.ecog, data=lung) The biggest question to answer here is what covariates you wish to base the prediction on. There is the usual tradeoff between too few (leave out something

Hello, I am a new R user and trying to learn how to implement the mahalanobis function to measure the distance between to 2 population centroids. I have used STATISTICA to calculate these differences, but was hoping to learn to do the analysis in R. I have implemented the code as below, but my results are very different from that of STATISTICA, and I believe I may not have interpreted the help

Dear r-helpers, I have a question about multiple testing. Here an example that puzzles me: All matrixes and contrast vectors are presented in treatment contrasts. 1. example: library(multcomp) n<-60; sigma<-20 # n = sample size per group # sigma standard deviation of the residuals cov1 <- matrix(c(3/4,-1/2,-1/2,-1/2,1,0,-1/2,0,1), nrow = 3, ncol=3, byrow=TRUE, dimnames =

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light

Hi all I do not understand why I am getting the following error message. Can anybody help me with this? Thanks in advance. install.packages("cmprsk") library(cmprsk) result1 <-crr(ftime, fstatus, cov1, failcode=1, cencode=0 ) one.pout1 = predict(result1,cov1,X=cbind(1,one.z1,one.z2)) predict.crr(result1,cov1,X=cbind(1,one.z1,one.z2)) Error: could not find function

Hello, I have a question regarding competing risk regression using cmprsk package (function crr()). I am using R2.9.1. How can I do to assess the effect of qualitative predictor (gg) with more than two categories (a,b,c) categorie c is the reference category. See above results, gg is considered like a ordered predictor ! Thank you for your help Jan > # simulated data to test > set.seed(10)