similar to: simple list question

My apology to the list, as it turns out the moment after I hit the send button, I found that the first element in the list was a null. The error message reflects that. > _____________________________________________ > From: Barker, Chris > Sent: Friday, December 12, 2008 6:30 > To: 'r-help@r-project.org' > Subject: simple list question > > > I'd

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,

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 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

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 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

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 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

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

Dear Forum, I have a question about interaction estimate in the Cox model: why the hazard ratio of the interaction is not produced in the summary of the model? (Instead, the estimate of the coefficient is given in the print of the model.) # Example: modINT <-cph( Surv(T_BASE, T_FIN,STATUS)~ NYHA + ASINI + RFP + FE_REC + XX_PR*XX_DISF) print(modINT) coef se(coef) z

On 7 Aug 2000, Peter Dalgaard BSA wrote: > ripley@stats.ox.ac.uk writes: > > > predict.lm has been broken by recent changes to the patched branch. > > > > It fails for all singular fits. An example: <snip> > Sometimes the fix for one bug uncovers another. We have > > > coef.aov > function (object, ...) > { > z <- object$coef >

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

Buenas. Abajo pongo la salida de un modelo de cox , dónde he estratificado por una variable de país (Countryb) y por otra (Q6). Además hay interacción entre la variable mobilityPDurG2 (es una variable 0,1, y 0 es la categoría de referencia) país. La categoría de referencia para país es "united kingdom". Mi duda surge si quiero calcular el hazard ratio para los que tienen un 1

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

Hi, I am attempting to evaluate the prediction error of a coxph model that was built after feature selection with glmnet. In the preprocessing stage I used na.omit (dataset) to remove NAs. I reconstructed all my factor variables into binary variables with dummies (using model.matrix) I then used glmnet lasso to fit a cox model and select the best performing features. Then I fit a coxph model

Dear R-helpers, I'm trying to associate linear coefficients (intercept and slope) to tens of thousands of observations based on a table with benchmark values. #####Example - Value table and their corresponding coefficients (intercept and slope) coef = data.frame(cbind(st=c(1:5),b = runif(5,0.3,5),a = seq(0.5,5,1)))print(coef) #Example of observations to be computedobs = runif(20,1,5)print(obs)

I'm using R on Windows v2.0.1 with the nlme package (v3.1-53) and am finding some unexpected discrepancies in the output of intervals.lme and coef.lme. I've included a toy dataset at the end, but briefly, the data are longitudinal data from couples in marital therapy. Each spouse's relationship satisfaction is measured 4 times; I've fit both linear and quadratic models to the

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 there,   I am beginner in R and I have some basic question. Suppose I run a common procedure such as a t test or cox model like below:   out<-coxph( Surv(tstart,tstop, death1) ~ x1+x1:log(tstop+1) , test1,method=c("breslow"))    Which yields the following result:   Call: coxph(formula = Surv(tstart, tstop, death1) ~ x1 + x1:log(tstop +     1), data = test1, method =

G'day all, I had a look at the GLM code of R (1.4.1) and I believe that there are problems with the function "glm.fit" that may bite in rare circumstances. Note, I have no data set with which I ran into trouble. This report is solely based on having a look at the code. Below I append a listing of the glm.fit function as produced by my system. I have added line numbers so that I