similar to: Cox model qustion in R

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)

For a fitted Cox model, one can either produce the predicted survival curve for a particular "hypothetical" subject (survfit), or the predicted curve for a particular cohort of subjects (survexp). See chapter 10 of Therneau and Grambsch for a long discussion of the differences between these, and the various pitfalls. By default, survfit produces the curve for a hypothetical

Dear R-users, I am checking the proportional hazard assumption of a cox model for a given covariate, let say Z1, after adjusting for other relavent covariates in the model. To this end, I fitted cox model stratified on the discrete values of Z1 and try to get beslow estimator for the baseline cumulative hazard function (H(t)) in each stratum. As far as i know, if the proportionality assumption

I have what I *think* should be a simple problem in R, and hope someone might be able to help me. I'm working with cancer survival data, and would like to calculate adjusted survival figures based on the age of the patient and the tumour classification. A friendly statistician told me I should use Cox proportional hazards to do this, and I've made some progress with using the

Dear all, Let's assume I have a clinical trial with two treatments and a time to event outcome. I am trying to fit a Cox model with a time dependent treatment effect and then plot the predicted survival curve for one treatment (or both). library(survival) test <- list(time=runif(100,0,10),event=sample(0:1,100,replace=T),trmt=sample(0:1,100,replace=T)) model1 <- coxph(Surv(time,

Dear R-users, I would like to determine the probability of event at specific time using cox model fit. On the development sample data I am able to get the probability of a event at time point(t). I need probability score of a event at specific time, using scoring scoring dataset which will have only covariates and not the response variables. Here is the sample code: n = 1000 beta1 = 2; beta2 =

Dear R-helpers, To get curves for a pseudo cohort other than the one centered at the mean of the covariates, I have been trying to use the newdata argument to survfit with no success. Here is my model statement, the newdata and the ensuing error. What am I doing wrong? > summary(fit) Call: coxph(formula = Surv(Start, Stop, Event, type = "counting") ~ Week + LagAOO + Prior.f +

Hello R users Would somebody know how to estimate survival from a frailty cox model, using the function survfit and the argument newdata ? (or from any other way that could provide individual expected survival with standard error); Is the problem related to how the random term is included in newdata ? kfitm1 <- coxph(Surv(time,status) ~ age + sex + disease + frailty(id,

Good afternoon, I am currently trying to do some work on survival analysis. - I hope to seek your advice re: 2 questions (1 general and 1 specific) (1) I'm trying to do a stratified Cox analysis and subsequently plot(survfit(object)). It seems to work for some strata, but not for others. I have tumor grade, which is a range of 1 - 4. When I divide this range of 1:4 into 2 groups, it

Good morning, Sorry to trouble the list. I have a problem I hope to seek your advice on. Essentially, I am trying to 'validate' a multivariate Cox proportional hazards model built in a training set, by testing it on an external test set. I have performed a survfit using the Cox model to predict survival for the test set, and obtained individual predictions for survival time, with

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

I haven't done any R memory debugging lately, but https://www.mail-archive.com/rcpp-devel at lists.r-forge.r-project.org/msg10289.html shows how I used to have gdb break where valgrind finds a problem so you could examine the details. Also, running your code after running gctorture(TRUE) can help track down memory problems. -Bill On Wed, Feb 7, 2024 at 12:03?PM Therneau, Terry M., Ph.D.

?I've hit a roadblock debugging a new update to the survival package.?? I do debugging in a developement envinment, i.e. I don't create and load a package but rather? source all the .R files and dyn.load an .so file, which makes things a bit easier. ? Running with R -d "valgrind --tool=memcheck --leak-check=full" one of my test files crashes in simple R code a dozen lines

Dear All, I am contemplating centering the covariates in my Cox model to reduce multicollinearity between the predictors and the interaction term and to render a more meaningful interpretation of the regression coefficient. Suppose I have two indicator variables, x1 and x2 which represent age categories (x1 is patients less than 16 while x2 is for patients older than 65). If I use the following

Hi: I have a question about the use of the survfit function after the estimation of a cox proportional hazard model with a frailty term. My goal is to estimate expected survival probabilities while controlling for the group-specific frailty term. First, I estimate a model of the following form: model1 <- coxph(Surv(t0, t, d) ~ x1 + x2 + frailty(id), na.action=na.exclude,

The combination of survfit, coxph, and factors is getting confused. It is not smart enough to match a new data frame that contains a numeric for sitenew to a fit that contained that variable as a factor. (Perhaps it should be smart enough to at least die gracefully -- but it's not). The simple solution is to not use factors. site1 <- 1*(coxsnps$sitenew==1) site2 <-

Dear All, I am using lars package written by Dr. Trevor Hastie, the version is lars_0.9-5 downloaded from cran. When I ran the diabetes example data attached in package, I found that the beta outputs from different machines are different. The difference is only about 10^-11 to 10^-12, some friends suggested that it possibly is a machine precision problem. But I check the machine numerical

I am forwarding a question and response about the use of getelementptr: > Hi. I found an LLVM code example that might be an error. The relevant > URL is: http://llvm.cs.uiuc.edu/docs/LangRef.html#modulestructure > > There is a piece of code which looks like this: > > ; Declare the string constant as a global constant... > %.LC0 = internal constant [13 x sbyte] c"hello

Hello everyone, I want to generate machine code for target thumb, so run with bit code test.bc llc -march thumb test.bc -filetype obj -o test.o It doesn't generate test.o but show a message: "target doesn't support generation of this file type!" What's wrong? Thanks -------------- next part -------------- An HTML attachment was scrubbed... URL:

Hi Heyu Zhu, > I want to generate machine code for target thumb, so run > with bit code test.bc > > llc -march thumb test.bc -filetype obj -o test.o > > It doesn't generate test.o but show a message: > > "target doesn't support generation of this file type!" writing object code directly is not supported yet (though it is being worked on), so for