similar to: simple list question (question withdrawn)

I'd appreciate some tips, I'm making a simple mistake trying to extract elements from a list of fit objects. The following command works, coef( cph.list[[1]] ) ,.... coef( cph.list[[3]] ) and so forth. These extract regression coefficients from the appropriate list element. When I try to run this inside a for loop, or an lapply (lapply(cph.list,coef)) I get this error

Hello, all R-users! I am working on fitting a survival analysis model using the coxph function for Cox proportional hazards regression model. Data look like usual: ========================== group block death censor Group1 1 4 1 Group1 1 12 1 ... Group2 30 4 1 Group2 30 4 1 ... Group3 57 16

Hi all, I am trying to get bootstrap resampled estimates of covariates in a Cox model using cph (Design library). Using the following I get the error: > ddist2.abr <- datadist(data2.abr) > options(datadist='ddist2.abr') > cph1.abr <- cph(Surv(strt3.abr,loc3.abr)~cov.a.abr+cov.b.abr, data=data2.abr, x=T, y=T) > boot.cph1 <- bootcov(cph1.abr, B=100, coef.reps=TRUE,

Hi, I have a list of fit objects (fit objects from HMISC functions) I create elements in the list in this way lrm.sumtot <- lrm( ae7bepn ~ trarm + sumtot , data=sd.fix) lrm.list[['lrm.sumtot']] <- lrm.sumtot And I can run (anova(lrm.sumtot)) The following also gives the anova I'd expect zz <- lrm.list[['lrm.sumtot']];anova(zz) And similarly for the summary

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

Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. This is just to make nice LateX tables automatically. I have the coefficients with coef(). How do I do that? Thanks, David Biau. [[alternative HTML version deleted]]

Hi, I am having trouble with the survival library, particualrily the coxph function. the following works coxph(jtree9$cph.call,z,rep(1,dim(z)[1])) Call: coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1])) coef exp(coef) se(coef) z p SM 0.2574 1.294 0.0786 3.274 1.1e-03 Sex -0.1283 0.880 0.1809 -0.709

Dear R-help, I am having a problem with the interpretation of result from validate.cph in the Design package. My purpose is to fit a cox model and validate the Somer's Dxy. I used the hypothetical data given in the help manual with modification to the cox model fit. My research problem is very similar to this example. This is the model without stratification: > library(Design) > f1

Dear R help, I am having a problem with the Design package and my problem is detailed here. I fit a cox model to my data and validate the Somer's Dxy using the Design package. (Because of computation time problem, i only try 10 bootstrap samples for the time being) This is the model without stratification: > library(Design) >

Hi. I am using Windows version of R 1.8.1. Being somewhat new to survival analysis, I am trying to compare cph (Design) with coxph (survival) for use with a survival data set. I was wondering why cph and coxph provide me with different confidence intervals for the hazard ratios for one of the variables. I was wondering if I am doing something wrong? Or if the two functions are calculating hazard

Hi, I am trying to run a conditional logistic model on a nested case-control study using cph() and then estimate survival based on the model. The data came from Prof Bryan Langholz website where he also has the SAS code to this, so I am trying to replicate the SAS results. The data attached. Basically, the variables are: rstime: risk set age rsentry: fake entry time, just before rstime setno:

Dear R users I use Windows XP, R2.5.1 (I have read the posting guide, I have contacted the package maintainer first, it is not homework). In a research project on renal cell carcinoma we want to compute Harrell's c index, with optimism correction, for a multivariate Cox regression and also for some univariate Cox models. For some of these univariate models I have encountered an error

Dear colleagues, I hope someone can help me with my problem. I have fitted a cox model with the following syntax: # cox01def <-cph(Surv(TEVENT,EVENT) ~ ifelse(AGE>50, (AGE-50)^2,0) + BMI + # HDL+DIABETES +HISTCAR2 + log(CREAT)+ as.factor(ALBUMIN)+STENOSIS+IMT,data # = XC, x=T, y=T, surv=T) *1 Furthermore I have estimated my beta's also with a Lasso method - Coxpath ( from

Hi, I encountered a weird problem when using the Design and Hmisc problem. I have 2 data frame called "a" and "b", both have 3 columns: "time", "status" and "scores", a sample of the data frame is like: data frame "a": time status scores 1 21 1 99.61 2 38 0 101.11 3 51 0 100.62 4 48 0 87.52 5 78 0

Hi, I used Dr. Harrell's rms package to make a nomogram. Below is my code for nomogram and calculate total points and probability *in original data set* used for building the nomogram. *My question is how I get the formula for calculating the survival probability for this nomogram. Then I can use this formula to do validation by using other data set. * f1 <- cph(Surv(retime,dfs) ~

Thanks everyone for their responses. My data is organized in a data.table.? My goal is to perform analyses according to some groups.? The results of analysis are objects.? If these objects could be stored as elements of a data.table, this would help downstream summarizing of results. Let me try another example. carsdt <- setDT(copy(mtcars)) carsdt[, unique(cyl) |> length()] #[1] 3

sorry, I had a typo there, it's datadist(b) for the analysis of data frame "b". --- Robert Balshaw <Robert.Balshaw at syreon.com> wrote: > Not sure if this will help, but did you mean to use > datadist(a) for > the analysis of B? > > Rob > > > -----Original Message----- > > From: r-help-bounces at stat.math.ethz.ch > >

I stumbled on the following: > library(stats4) > example(mle) > confint.default(fit2) Error in UseMethod("vcov") : no applicable method for "vcov" In addition: Warning message: In object$coefficients : $ operator not defined for this S4 class, returning NULL > vcov(fit2) lymax lxhalf lymax 0.02857612 -0.04870231 lxhalf -0.04870231 0.11457338

I bumped into the following situation: Browse[1]> coef deg0NA deg4NA deg8NA deg0NP deg4NP deg8NP (Intercept) 462 510 528 492 660 762 Browse[1]> coef[,1] [1] 462 Browse[1]> coef[,1,drop=F] deg0NA (Intercept) 462 where I really wanted neither, but (Intercept) 462 Anyone happen to know a neat way out of the conundrum? I can think of

Dear R Help list, My question is regarding extracting the standard error or Z-score from a cph or coxph call. My Cox model is: - modz=cph(Surv(TSURV,STATUS)~RAGE+DAGE+REG_WTIME_M+CLD_ISCH+POLY_VS, data=kidneyT,method="breslow", x=T, y=T) I've used names(modz) but can't see anything that will let me extract the Z scores for each coefficient or the standard errors in the same