similar to: Interpreting the example given by Prof Frank Harrell in {Design} validate.cph

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) >

Dear R-help, This is an example in the {Hmisc} manual under rcorr.cens function: > set.seed(1) > x <- round(rnorm(200)) > y <- rnorm(200) > round(rcorr.cens(x, y, outx=F),4) C Index Dxy S.D. n missing uncensored Relevant Pairs Concordant Uncertain 0.4831 -0.0338 0.0462 200.0000

I am interested Andrea is whether you ever established why your R2 was 1. I have had a similar situation previously. My main issue though, which I'd be v grateful for advice on, is why I am obtaining such negative values -0.3 for Somers Dxy using validate.cph from the Design package given my value of Nagelkerke R2 is not so low 13.2%. I have this output when fitting 6 variables all with

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

Dear list, I am relatively new to ordinal models and have been working through the example given by Frank Harrell in the predict.lrm {Design} help All of this makes sense to me, except for the responses, i,e how do i interpret them? i would be extremely grateful if someone could explain the results? First i establish the date and model - > y <- factor(sample(1:3, 400, TRUE), 1:3,

Dr. Frank E. Harrell, Jr., Professor and Chair of the Department of Biostatistics at Vanderbilt University is giving a one-day workshop on Regression Modeling Strategies on Friday, April 29, 2005. Analyses of the example datasets use R/S-Plus and make extensive use of the Hmisc library written by Professor Harrell.The workshop is sponsored by the Southern California Chapter of the American

Does anyone have a valid email address for Frank Harrell of Hmisc fame? I've tried getting in touch with him at both fharrell at virginia.edu and f.harrell at vanderbilt.edu, but messages to either of those addresses get bounced. Frank, if you're reading this, please email me from an account that will accept a reply. Jeff Hallman

Dear All, I am trying to calculate a 95% confidence interval for the difference in two c statistics (or equivalently D statistics). In Stata I gather that this can be done using the lincom command. Is there anything similar in R? As you can see below I have two datasets (that are actually two independent subsets of the same data) and the respective c statistics for the variables in both cases.

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

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

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

Dear all: Could anyone please provide some R codes to plot the below survival data to compare two groups (0 vs 1) after 2 simulations (TRL)? need 95% prediction interval on the plot from these 2 trials. I would like to simulate 1000 trials later. Thanks a lot for your great help and consideration! yan TRL ID ECOG BASE PTR8 GROUP POP ST ind 1 1 1 1 2.2636717 0.255634126 1 1 99.4 F 3 1 2 1

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

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

Hello R Community, I'm continuing to work through logistic regression (thanks for all the help on score test) and have come up against a new opposition. I'm trying to compute Somers Dyx as some suggest this is the preferred method to Somers Dxy (Demaris, 1992). I have searchered the [R] archieves to no avail for a function or code to compute Dyx (not Dxy). The overview of Hmisc has

Hi to everybody! I´m just a beginner in R, and I´m trying to replace values in a distance matrix with a concret condition: replace all values (elements) lower than 4.5 with value=18. I´ve tried this, but it doesn´t work... Dxy would be my 117 x 117 euclidean distance matrix M18 and M4.5 would be 117 x 117 matrices: M18<-matrix(rep(18,13689),nrow=117)

Calum had a long question about drawing survival curves after fitting a Weibull model, using pweibull, which I have not reproduced. It is easier to get survival curves using the predict function. Here is a simple example: > library(survival) > tfit <- survreg(Surv(time, status) ~ factor(ph.ecog), data=lung) > table(lung$ph.ecog) 0 1 2 3 <NA> 63 113 50 1

Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with the "old" comorbidity measure, Charlson's Comorbidity Index. I have nearly 10,000

I am reposting my message from April 8th because I never received a response to the original post: Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with

Hi I am trying to understand how to get the validate() function in Design to work with the subset option. I tried this: ovarian.cph=cph(Surv(futime, fustat) ~ age+factor(ecog.ps)+strat(rx), time.inc=1000, x=T, y=T, data=ovarian) validate(ovarian.cph) #fine when no subset is used, but the following two don't work: > validate(ovarian.cph, subset=ovarian$ecog.ps==2) Error in