similar to: plot of survival probability vs. covariate

Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time <- cumsum(rexp(1000)/10) status <- rbinom(1000, 1, 0.5) ## kaplan meier estimates fit <- survfit(Surv(time,

Dear R-help, I would like to compute the variance for the proportion of treatment effect by a surrogate in a survival model (Lin, Fleming, and De Gruttola 1997 in Statistics in Medicine). The paper mentioned that the covariance matrix matches that of the covariance matrix estimator for the marginal hazard modelling of multiple events data (Wei, Lin, and Weissfeld 1989 JASA), and is implemented

Does anyone know how to make a plot like Fig. 4 of Paik, et al., New England Journal of Medicine, Dec. 30, 2004? Given survival data and a covariate, they plot a curve giving "Rate of Distant Recurrence at 10 Yr (% of patients)" on the y-axis versus the covariate on the x-axis. They also plot curves giving a 95% confidence interval. Thanks very much. -Ben The information in this

First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized

Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is just "1 minus kaplan-Meier survival"? Under what circumstance, you should use cumulative incidence vs KM survival? If the relationship is just CI = 1-survival, then what difference it makes to use one vs. the other? And in R how I can draw a cumulative incidence plot. I know I can make a Kaplan-Meier

I was wondering if someone familiar with survival analysis can help me with the following. I would like to fit a Weibull curve, that may be dependent on a covariate, my dataframe "labdata" that has the fields "cov", "time", and "censor". Do I do the following? wieb<-survreg(Surv(labdata$time, labadata$censor)~labdata$cov,

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,

Hi all, I am trying to detect association between a covariate and a disease outcome using R. This covariate shows time-varying effect, I add a time-covariate interaction item to build Cox model as follows: COX <- coxph(as.formula("Surv(TIME,outcome)~eGFR_BASE+eGFR_BASE:TIME"),ori.data); coef exp(coef) e(coef) z p eGFR_BASE

Hi, I am running a Cox Mixed Effects Hazard model using the library coxme. I am trying to model time to onset (age_sym1) of thought problems (e.g. hearing voices) (sym1). As I have siblings in my dataset, I have decided to account for this by including a random effect for family (famid). My covariate of interest is Mother's diagnosis where a 0 is bipolar, 1 is control, and 2 is major

Hi R experts, I am trying to do a prognostic model validation study, using cancer survival data. There are 2 data sets - 1500 cases used to develop a nomogram, and another of 800 cases used as an independent validation cohort. I have validated the nomogram in the original data (easy with the Design tools), and then want to show that it also has good results with the independent data using 60

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

Hi I have been asked to provide Weibull parameters from a paper using Kaplan Meir survival analysis. This is something I am not familiar with. The survival analysis in R works nicely and is the same as commercial software (only the graphs are superior in R). The Weibull does not and produces an error (see below). Any ideas why this error should occur? My approach may be spurious.

Hello, I'm trying to estimate a cox model with a frailty variable and time-dependent covariate (below there is the statement I use and the error message). It's seems to be impossible, because every time I add the time-dependent covariate the model doesn't converge. Instead, if I estimate the same model without the time-dependent covariate it's converge. I'd like knowing if

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

dear all, I want to plot a kaplan Meier plot with the following functions, but I fail to produce the plot I want: library(survival) tim <- (1:50)/6 ind <- runif(50) ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0; MS <- runif(50) pred <- vector() pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1 df <- as.data.frame(cbind(MS, tim, pred, ind)) names(df) <-

> Dear All, > > I have built a survival cox-model, which includes a covariate * time interaction. (non-proportionality detected) > I am now wondering how could I most easily get survival predictions from my model. > > My model was specified: > coxph(formula = Surv(event_time_mod, event_indicator_mod) ~ Sex + > ageC + HHcat_alt + Main_Branch + Acute_seizure +

Hi I am trying to determine the mean of a Weibull function that has been fit to a data set, adjusted for a categorical covariate , gender (0=male,1=female). Here is my code: library(survival) survdata<-read.csv("data.csv") ##Fit Weibull model to data WeiModel<-survreg(Surv(survdata$Time,survdata$Status)~survdata$gender) summary(WeiModel) P<-pweibull(n,

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 <-

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

Hi, I'm interested in building a Cox PH model for survival modeling, using 2 covariates (x1 and x2). x1 represents a 'baseline' covariate, whereas x2 represents a 'new' covariate, and my goal is to figure out where x2 adds significant predictive information over x1. Ideally, I could get a p-value for doing this. Originally, I thought of doing some kind of likelihood ratio