similar to: coxph function (Survival Library) (PR#12638)

Dear all, I am confused with the output of survfit.coxph. Someone said that the survival given by summary(survfit.coxph) is the baseline survival S_0, but some said that is the survival S=S_0^exp{beta*x}. Which one is correct? By the way, if I use "newdata=" in the survfit, does that mean the survival is estimated by the value of covariates in the new data frame? Thank you very much!

[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~age, ovarian, subset=rx==1)

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

Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)

Dear all, I would have some questions on the coxph function for survival analysis, which I use with frailty terms. My model is: mdcox<-coxph(Surv(time,censor)~ gender + age + frailty(area, dist='gauss'), data) I have a very large proportion of censored observations. - If I understand correctly, the function mdcox$frail will return the random effect estimated for each group on the

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

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, I am attempting to plot survival curves estimated by cox proportional hazards regression model. The formula for the model is this: F.cox.weight <- coxph(Surv(Lifespan, Status) ~ MS + Weight + Laid + MS:Laid + Weight:Laid, data = LongF) MS = Mating status (mated/virgin) Weight = adult female weight, continuous covariate Laid = number of eggs laid by each female, continuous covariate I

Dear Mrs Rachel Pearce, I am looking for a function "cutpt-coxph" in R - like you did some years ago. How have you solved the problem? Have you found it or a similar function? thank you, Sincerely, Friederike "The title says it all really; I am looking for a function along the lines of cutpt.coxph as described in "Survival Analysis Using S" (Tableman and Kim), Chapter

Hi All, I'm runnning coxph() on 90 different datasets (in a loop). 1. I'm wondering how can I get "log-likelihood" value from coxph() output. Currently I can only see following: Likelihood ratio test = Wald test = Score (logrank) test = 2. Once I have likelihood value, I would like to extract these from R output (for each data set). Can any one help please, thanks in

I wonder if anyone know why survest (a function in Design package) and standard survfit.coxph (survival) returned different confidence intervals on survival probability estimation (say 5 year). I am trying to estimate the 5-year survival probability on a continuous predictor (e.g. Age in this case). Here is what I did based on an example in "help cph". The 95% confidence intervals

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

Dear all, I am using glmnet (Coxnet) for building a Cox Model and to make actual prediction, i.e. to estimate the survival function S(t,Xn) for a new subject Xn. If I am not mistaken, glmnet (coxnet) returns beta, beta*X and exp(beta*X), which on its own cannot generate S(t,Xn). We miss baseline survival function So(t). Below is my code which takes beta coefficients from glmnet and creates coxph

Dear all, I use the COXPH function in the SURVIVAL package to estimate survival models with a random effect. I would like to know if there is a measure of "model fit and complexity" analogous to the Akaike Information Criterion implemented in COXPH, in order to compare together models including different sets of covariates (models not necessarily nested within each others). Kind

I would like to plot a survival curves of a group with different categories after running a Cox model with frailty() random effects terms. I just could display a survival plot of the covariable?s mean. Here an example: library(survival) fit<-coxph(Surv(time, status) ~ sex+ frailty(litter, dist='gamma', method='em'), rats) summary(fit ) suf<-survfit(fit) plot(suf,

The title says it all really; I am looking for a function along the lines of cutpt.coxph as described in "Survival Analysis Using S" (Tableman and Kim), Chapter 6. As may be guessed, the function optimises the cutpoint of a continuous variable for cox proportional hazard modelling. I can't find it, or any similar function, on CRAN. Alternatively, perhaps there is a way of extracting

fit <- coxph(Surv(futime,fustat)~ age +strata(rx), data=ovarian, subset=1:23) curves <- survfit(fit, newdata=ovarian[24:26,]) I don't think this is mentioned in the documentation (I'll have to fix that!), but subscripting works for survfit objects. In this case there are 2 strata and 3 subjects, and curves[1,2] will return the survival curve for strata 1,

Hello users, I would like to obtain a survival curve from a Cox model that is smooth and does not have zero differences due to no events for those particular days. I have: > sum((diff(surv))==0) [1] 18 So you can see 18 days where the survival curve did not drop due to no events. Is there a way to ask survfit to fit a nice spline for the survival?? Note: I tried survreg and it did not

Hello, I have a couple of questions with regards to fitting a coxph model to a data set in R: I have a very large dataset and wanted to get the baseline hazard using the basehaz() function in the package : 'survival'. If I use all the covariates then the output from basehaz(fit), where fit is a model fit using coxph(), gives 507 unique values for the time and the corresponding cumulative

Hi all, I found that the two different versions of "survival" packages, namely 2.36-5 vs. 2.36-8 or later, give different results for coxph function. Please see below and the data is attached. The second one was done on Linux, but Windows gave the same results. Could you please let me know which one I should trust? Thanks, ...Tao #####============================ R2.13.0,