search for: surve

Dear All, a Surv object I put in a data frame behaves somehow unexpected (see example). If I do a Cox regression on the original Surv object it works. If I put it in a data.frame and do the regression on the data frame it does not work. Seemingly it has to do with the class attribute, because if I change the class attribute to let "Surv" appeare first, again it works. Is this known?

Does Surv in psm handle interval2 data? The argument list seems to indicate it does but I get an error. Thanks, Chris # code library('survival') left <- c(1, 3, 5, NA) right <-c(2, 3, NA, 4) Surv(left, right, type='interval2') survreg(Surv(left, right, type='interval2') ~ 1) library('rms') Surv(left, right, type='interval2') # error args(Surv)

Hello, I am having some trouble with the 'censoring-adjusted C-index' by Uno et al, in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <-

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

Hello, I am using Dr. Harrell's design package to make a nomogram. I was able to make a beautiful one without stratifying, however, I will need to stratify to meet PH assumptions. This is where I go wrong, but I'm not sure where. Non-Stratified Nomogram:

Full_Name: Victor Moreno Version: 2.2.0 OS: windows Submission from: (NULL) (213.151.99.160) summary of an object class Surv (package survival) hangs and closes R 2.2.0 I would like to have Surv objects in a data frame to build models easier: t<-rexp(100) s<-rbinom(100,1,.3) surv<-Surv(t,s) x1<-rnorm(100,10,2) x2<-rnorm(100,10,2) dat<-data.frame(surv,x1,x2)

hie Im trying to calculate the power of the logrank test for different values of rho .I was just wandering whether the following programme would do it. any suggestions are welcome s=50 number=1 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count7=0; count8=0;count9=0 while(s!=0){ n=20 n1=n/2

Hi, I'm trying to do a simple survival analysis on some data, and I'm having the following problem (here's my code and the error message): out <- Surv(fup,event=status) Error in Surv(fup, event = status) : argument "time2" is missing, with no default >From reading the documentation, it seems that I should be able to simply write: Surv(time1, event) if my data is

Dear All, since years I am struggling with Surv objects in data.frames. The following seems to have to do with it. See below the modified example from the help page of survSplit. The original works, as expected. If, however, a Surv object is added to the data.frame, each record gets doubled. Is there some solution other than avoiding Surv objects in data.frames? Thanks, Heinz

1. A Surv object is a matrix with some extra attributes. The as.matrix.Surv function removes the extras but otherwise leaves it as is. 2. The last several versions of the survival library were accidentally missing the S3method('as.matrix', 'Surv') line from their NAMESPACE file. (Instead it's position is held by a duplicate of the line just above it in the NAMESPACE file,

Dear R People: I used the "Surv" function to produce the following object: >a <- Surv(1:4,2:5,c(0,1,1,0)) a [1] (1,2+] (2,3 ] (3,4 ] (4,5+] >str(a) Error in "[.Surv"(object, 1:ile) : subscript out of bounds > Why does str(a) give an error, please? Or did I do something wrong? Thanks in advance. R Version 1.9.1 Windows Sincerely, Laura Holt mailto: lauraholt_983

Dear everyone, My question involves the use of the survival object. We can have Surv(time,time2,event, type=, origin = 0) (1) As detailed on p.65 of: http://cran.r-project.org/web/packages/survival/survival.pdf My data (used in my study) is 'right censored' i.e. my variable corresponding to 'event' indicates whether a person is alive (0) or dead (1) at date last seen

Dear All, there seems to be some strange influence of the Design package on data.frame. If I build a data.frame containing a Surv object without loading the package Design, the data frame is usable to coxph. If instead I just load Design and build a data.frame afterwards, the naming of the Surv object is different and it does not work with coxph. (In my real application I loaded Design to use the

Good day R help list!!! I've been trying to do Bootstrap in R on Censored data. I encountered WARNING/ERROR messages which I could not find explanation. I've been searching on the literature for two days now and still can't find answers. I hope there's anyone out there who can help me with these two questions: 1. If the "Loglik converged before variable..." message

I would like to produce a complimentary log-log survival plot with only the points appearing on the graph. I am using the code below, taken from the plot.survfit page of help for the the survival package (version 2.35-7). I am running in R 2.10.0 on Windows XP, and the list of packages following the error is loaded. Is there some specific 'type= ' syntax, or an additional parameter that

my program given below can some one make it presentable. I trying to simulate survival data and calculate the power. I think i could have done better. s=10 number=0 count1=0;count2=0;count3=0;count4=0;count5=0;count6=0;count7=0;count8=0; count9=0; count11=0;count22=0;count33=0;count44=0;count55=0;count66=0;count77=0; count88=0;count99=0; while(s!=0){ n=100 n1=n/2 n2=n/4

hie when i use plot.survfit to plot more than one graph why I only see the last graph how do i see the other graphs.for example n=20 n1=n/2 n2=n/4 a11=4;a12=4 ;a21=4 ;a22=4 t1<-array(1,c(n1)) t2<-array(2,c(n1)) treatgrp=matrix(c(t1,t2))

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

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) <-

Hello R-users. I believe that the way basehaz (in the survival package) compute the baseline hazard function is false. I come to question this function when it gives me hazard probabilities greater than 1. Looking at the code I think I've localised the error : hazard probability is computed as : H <- -log(surv) but it seems to me that hazard probabilities is rather an instantaneous