similar to: plot.validate.cph

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

Hi! How can I exclude the legends from calibration plots  generated by calibrate.cph regards, Salvo [[alternative HTML version deleted]]

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 R-users,   I  have two questions regarding validation and calibration of Survival regression models.   1.  I am trying to calibrate and validate a cox model using val.surv. here is my code:  f.1<-cph(Surv(time,event)~age, x=T, y=T, data=train)  test1<-test[,"age"]  val.surv(f.1, newdata=data.frame(test1), u=10)    but I get an error message:    Error in val.surv(f.1, newdata

Hello all I am a very new R user I am used to using STATA My problem: I want to build a Cox model and validate this. I have a large number of clinical relevant factors and feel the need to reduce these. Meanwhile I have some clinical variables I deem sufficiently important to force into the model regardless of AIC or p value. This is my present log over commands

Here's a number of items that came up in connection with my course for medical ph.d. students this week: boxplot(): - deals ungracefully with empty groups and factor levels not present in grouping variable. - no indication of what variable is being plotted data.entry() (mostly W95 rseptbeta problems) - Entry of first variable name doesn't take before 2nd

Dear R users, I know, this is the second time i return on this topic. Sorry, but this analysis is of great value for me, and i hope someone can help me. I need to model a time-varying effect in a Cox model. Briefly explained here: http://books.google.com/books?id=9kY4XRuUMUsC&lpg=PP1&hl=it&pg=PA147#v=onepage&q&f=false

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 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

Hello, I am using Dr. Harrell's design package to make a nomogram. I was able to make a beautiful one. However, I want to change 5-year survival probability to 5-year failure probability. I couldn?t get hazard rate from Hazard(f1) because I used cph for the model. Here is my code: f1 <- cph(Surv(retime,dfs) ~ age+her2+t_stage+n_stage+er+grade+cytcyt+Cyt_PCDK2 , data=data11, surv=T,

Hi there, I encountered a weird problem using cph() with Design package: I have 2 datasets, say "dat1" and "dat2", both data frames with 3 columns "time","status" and "scores", all numeric If I run the following: dd<-datadist(dat1) options(datadist='dd') dd time status scores Low:effect 37.0 0 -6.018

Dear UseR, I do not know if this a problem with me, my data or cph/survest in package design. The example below works with a standard data set, but not with my data, but I cannot locate the problem. Note that I am using an older package of survival to avoid a problem with the newly renamed function in survival meeting Design. Dieter # First, check standard example to make sure library(Design)

Hello, I am using Dr. Harrell's rms package to make a nomogram. I was able to make a beautiful one. However, I want to change 5-year survival probability to 5-year failure probability. I couldn?t get hazard rate from Hazard(f1) because I used cph for the model. Here is my code: library(rms) f1 <- cph(Surv(retime,dfs) ~ age+her2+t_stage+n_stage+er+grade+cytcyt+Cyt_PCDK2 , data=data11,

Hello, I am constructing a nomogram using cph and nomogram commands in Dr. Harrell's Design/RMS package. The HR that I obtain for dichotomous and categorical variables are identical to those that I obtain using STATA stcox. However, the inter-quartile HR I obtain for continuous variables is obviously different, since STATA gives me HR for each unit (year, centimeter, etc) like coxph would

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:

Hi, in Design package, a plot of survival probability vs. a covariate can be generated by survplot() on a cph object using the folliowing code: n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age" sex <- factor(sample(c('male','female'), n, TRUE)) cens <- 15*runif(n) h <- .02*exp(.04*(age-50)+.8*(sex=='Female')) dt <-

Can someone help me. I am very new to R. I am fitting a Cox model using Frank Harrell's cph as I want to produce a Nomogram. This is what I have done: Srv<- Surv(time,cens) f.cox<- cph(Srv~ v1+v2+v3+v4, x=T, y=T, surv=T) As soon as I press enter, Windows XP crashes. If I remove surv=T, then it works. I have R version 2.9.0. Is there a way of displaying the parameter estimates (ie beta

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, the calibrate.cph() function in rms package generate calibration curve for Cox model on the same dataset where the model was derived using bootstrapping or cross-validation. If I have the model built on dataset 1, and now I want to produce a calibration curve for this model on an independent dataset 2, how can I do that? Thanks John [[alternative HTML version deleted]]

Could someone explain the summary(cph.object)? The example is in the help file of cph. n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age" sex <- factor(sample(c('Male','Female'), n, rep=TRUE, prob=c(.6, .4))) cens <- 15*runif(n) h <- .02*exp(.04*(age-50)+.8*(sex=='Female')) dt <- -log(runif(n))/h label(dt)