similar to: survfit on coxph object with weights

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!

My question is, how to compute hazard function(H(t)) after building the coxph model. I even aware of the terminology that differs from hazard function(H(t)) and the hazard rate(h(t)). Here onward I wish to calculate both. Here what I have done in two different methods; ##########################################################################################

What is the best way to report the standard error when publishing Kaplan-Meier plots? In my field (Vascular Surgery), practitioners loosely refer to the "10% error" cutoff as the point at which to stop drawing the KM curve. I am interpreting this as the *standard error of the cumulative hazard*, although I'm having a difficult time finding some guidelines about this (perhaps I am

I?m trying to set up proportional hazard model that is stratified with respect to covariate 1 and has an interaction between covariate 1 and another variable, covariate 2. Both variables are categorical. In the following, I try to illustrate the two problems that I?ve encountered, using the lung dataset. The first problem is the warning: To me, it seems that there are too many dummies

Hi Dear R-users I have a database with 18000 observations and 20 variables. I am running cox regression on five variables and trying to use survfit to plot the survival based on a specific variable without success. Lets say I have the following coxph: >library(survival) >fit <- coxph(Surv(futime, fustat) ~ age + rx, data = ovarian) >fit what I am trying to do is plot a survival

I would like to be able to construct hazard rates (or unconditional death prob) for many subjects from a given survfit. This will involve adjusting the ( n.event/n.risk) with (coxph object )$linear.predictors I must be having another silly day as I cannot reproduce the linear predictor: fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian) fit$linear.predictors[1] [1] 2.612756

I have a very general question about what the centering option in basehaz does to factors. (basehaz computes the baseline cumulative hazard for a coxph object using the Breslow estimator). Lets say I'm interested in a survival model with two (dichotomous) factors and a continuous covariate. Variable Possible Values Factor1 0 or 1 Factor2 0 or 1

Hello, I'm trying to estimate a Cox proportional hazard model with time-varying covariates using coxph. The parameter estimates are fine but there is something wrong with the survival curves I get with survfit (results are not plausible). Let me explain why I think something's wrong. To make sure I'm setting up my data correctly to estimate a model with time-varying covariates, I

I am fitting a coxph model on a large dataset (approx 100,000 patients), and then trying to estimate the survival curves for several new patients based on the coxph object using survfit. When I run coxph I get the coxph object back fairly quickly however when I try to run survfit it does not come back. I am wondering if their is a more efficient way to get predicted survival curves from a coxph

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

i want to make survival plots for a coxph object using survfit function. mod.phm is an object of coxph class which calculated results using columns X and Y from the DataFrame. Both X and Y are categorical. I want survival plots which shows a single line for each of the categories of X i.e. '4' and 'C'. I am getting the following error: > attach(DataFrame) >

Dear users, In a study with recurrent events: My objective is to get estimates of survival (obtained through a Cox model) by rank of recurrence and by treatment group. With the following code (corresponding to a model with a global effect of the treatment=rx), I get no error and manage to obtain what I want : data<-(bladder)

I am wondering how to estimate the survival curve for a particular case(s) given a coxph model using this example code: #fit a cox proportional hazards model and plot the #predicted survival curve fit <- coxph( Surv(futime,fustat)~resid.ds+strata(rx)+ecog.ps+age,data=ovarian[1:23,]) z <- survfit(fit,newdata=ovarian[24:26,],individual=F) zs <- z$surv zt <-

Hello: In the example below (or for a censored data) using survfit.coxph, can anyone point me to a link or a pdf as to how the probabilities appearing in bold under "summary(pred$surv)" are calculated? Do these represent acumulative probability distribution in time (not including censored time)? Thanks very much, parmee *fit <- coxph(Surv(futime, fustat) ~ age, data = ovarian)*

Hi! I am having a few problems with coxph function, I had the same problem with the use of survfit. Here it is: when calling 'M22<-coxphw(Surv(V1,V2,V4)~V5, data=XTDV, id=XTDV$V3, * prentice*= ~V5, robust=TRUE, censcorr=TRUE)' knowning that my data is: > XTDV[1:10,1:3] V1 V2 V3 V4 V5 1 0 36.39706 36 1 343.2224 2 0 36.39706 943 0 161.5931 3 0 36.39706

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

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

I have a model with quant vars only and the error message does not make sense: (mod1 <- coxph(Surv(time=strt,time2=stp,event=(resp==1))~ +incpost+I(amt/1e5)+rate+strata(termfac), subset=dt<"2010-08-30", data=inc,method="efron")) Call: coxph(formula = Surv(time = strt, time2 = stp, event = (resp == 1)) ~ +incpost + I(amt/1e+05) + rate + strata(termfac),

Hi, The following function does work optimize.W<-function(W,k,G,Groups,cph.call,z){ n<-length(Groups) grp.wt<-rep(0,n) for(i in 1:(length(G))){ ind<-Groups == G[i] if(G[i]!=k){ grp.wt[ind]<-W[i] } elsegrp.wt[ind]<-1 } z<-data.frame(cbind(z,grp.wt=grp.wt)) #needed to make the case weights #part of the data

Thanks for the reproducable example. I can confirm that it fails on my machine using survival 2-37.5, the next soon-to-be-released version, The issue is with NextMethod, and my assumption that the called routine inherited everything from the parent, including the environment chain. A simple test this AM showed me that the assumption is false. It might have been true for Splus. Working this