similar to: model frames and update()

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

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

I've never seen this, and have no idea how to reproduce it. For resloution you are going to have to give me a working example of the failure. Also, per the posting guide, what is your sessionInfo()? Terry Therneau On 08/09/2012 04:11 AM, r-help-request at r-project.org wrote: > I have a couple of questions with regards to fitting a coxph model to a data > set in R: > > I have a

Hello, I have a couple of questions with regards to fitting a coxph model to a data set in R: I have a very large dataset and wanted to get the baseline hazard using the basehaz() function in the package : 'survival'. If I use all the covariates then the output from basehaz(fit), where fit is a model fit using coxph(), gives 507 unique values for the time and the corresponding cumulative

Hello, I am running R 2.6.1 on windows xp I am trying to fit a cox proportional hazard model with a shared Gaussian frailty term using coxme My model is specified as: nofit1<-coxme(Surv(Age,cen1new)~ Sex+bo2+bo3,random=~1|isl,data=mydat) With x1-x3 being dummy variables, and isl being the community level variable with 4 levels. Does anyone know if there is a way to get the standard error

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

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, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Dear All, I am using WriteXLS to write tables with multiple sheets with the command: WriteXLS("tables", ExcelFileName = fileName, SheetNames = tableList, perl = "perl", verbose = FALSE, Encoding = c("UTF-8", "latin1"), row.names = TRUE, col.names = TRUE, AdjWidth = TRUE, AutoFilter = FALSE, BoldHeaderRow = FALSE,

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

I am trying to add confidence (error) bars to lattice barcharts (and dotplots, and xyplots). I found this helpful message from Deepayan Sarkar and based teh code below on it: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/50299.html However, I can't get it to work with groups, as illustrated. I am sure I am missing something elementary, but I am unsure what. Using R 2.1.1 on various

Hi, I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just

Hi, I have a column in a data frame looking something like: $sex $language $count male english 0 male english 0 female english 32 male spanish 154 female english 11 female norweigan 7 and so on. What I want to do is to order these in to categories, for instance one category where count>=0 & count<10 and so on.. I want my data to turn out looking something like: male

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)

Hello, is there a R package that provides a log rank trend test for survival data in >=3 treatment groups? Or are there any comparable trend tests for survival data in R? Thanks a lot Markus -- Dipl. Inf. Markus Kreuz Universitaet Leipzig Institut fuer medizinische Informatik, Statistik und Epidemiologie (IMISE) Haertelstr. 16-18 D-04107 Leipzig Tel. +49 341 97 16 276 Fax. +49 341 97 16

Dear R-help - Thanks to those who replied yesterday (Christos H. and Thomas L.) regarding my question on coxph and model formula, the answers worked perfectly. My new question involves the following. I want to run several coxph models (package survival) with the same dataset, but different subsets of that dataset. I have found a way to do this, described below in functions subwrap1 and

Hi, I am just wondering if there is a test available for testing if a linear fit of an independent variable in a Cox regression is enough? Thanks for any suggestions. John Zhang ____________________________________________________________________________________ [[elided Yahoo spam]]

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

Hello! I have something like this: test1 <- data.frame(intx=c(4,3,1,1,2,2,3), status=c(1,1,1,0,1,1,0), x1=c(0,2,1,1,1,0,0), x2=c(1,1,0,0,2,2,0), sex=c(0,0,0,0,1,1,1)) and I can easily fit a cox model: library(survival) coxph(Surv(intx,status) ~ x1 + x2 + strata(sex),test1) However, I want to