similar to: Calling survfit within a function

The new code seems to have broken various things. Autoloading of libraries doesn't seem to work: > library(survival4) Autoloading required library: splines Error in pos.to.env(pos) : invalid "pos" argument > search() [1] ".GlobalEnv" "library:survival4" "library:date" [4] "library:base" The coxph function the

The new code seems to have broken various things. Autoloading of libraries doesn't seem to work: > library(survival4) Autoloading required library: splines Error in pos.to.env(pos) : invalid "pos" argument > search() [1] ".GlobalEnv" "library:survival4" "library:date" [4] "library:base" The coxph function the

I installed the latest version of R (Version 0.50 Alpha-4) and it breaks survival4 as follows (it worked fine under Version 0.50 Alpha-3) R> library(survival4) Autoloading required library: splines R> survfit(Surv(runif(25),rbinom(25,1,0.4)) ~ 1) Error in eval(frame, data)[subset, , drop = FALSE] : incorrect number of dimensions Thanks E. S. Venkatraman

Hi, I'm using the rsept for WINDOWS 95 (with modified men?, and "fixed" survival4). See the Ennapadam Venkatraman's example made in date 35/07/97 19.02 Ennapadam Venkatraman wrote: > library(splines) > library(survival4) > xx <- NULL > xx$ftim <- exp(-runif(100)) > xx$ctim <- 2*runif(100) > xx$sts <- 1*(xx$ftim <= xx$ctim) > xx$otim <-

Is this a feature/bug or am I missing something: R : Copyright 1999, The R Development Core Team Version 0.63.2 (January 12, 1999) ...... > ls() character(0) > x <- NULL > x$x1 <- rnorm(10) > x$x2 <- rnorm(10) > mode(x) [1] "pairlist" > as.data.frame(x) Error: can't coerce pairlist into a

I reported bugs in "rhyper" and the TASKS.OLD file claims that they have been fixed. But R continues to get stuck (the prompt never comes back) if I use rhyper with N1, N2, n large, for example "rhyper(5,200,250,60)". E. S. Venkatraman (venkat at biosta.mskcc.org) =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= r-help mailing list -- Read

ls.print produces error that I don't seem to be able to trace. Output of the commands as follows: (hyeung is a 24x2 matrix of data) ------------------------------------------------- > summary(hyeung) x.1 x.2 Min. : 28.0 Min. : 10.0 1st Qu.: 72.0 1st Qu.: 87.5 Median : 86.5 Median : 92.5 Mean : 81.0 Mean : 82.5 3rd Qu.: 97.0 3rd Qu.:100.0 Max.

ls.print produces error that I don't seem to be able to trace. Output of the commands as follows: (hyeung is a 24x2 matrix of data) ------------------------------------------------- > summary(hyeung) x.1 x.2 Min. : 28.0 Min. : 10.0 1st Qu.: 72.0 1st Qu.: 87.5 Median : 86.5 Median : 92.5 Mean : 81.0 Mean : 82.5 3rd Qu.: 97.0 3rd Qu.:100.0 Max.

================================================= This is an automated summary of the status of the R-bugs repository. Note that this may be neither complete nor perfectly correct at any given instance: Not all bugs are reported, and some reported bugs may have been fixed, but the repository not yet updated. Some bug fixes are difficult to verify because they pertain to specific hardware or

A cut and paste typo has crept in and is rendering all values returned for the hypergeometric distribution incorrect. The problem is in src/main/arithmetic.c in the function "math4". The lines PROTECT(sy = allocVector(REALSXP, n)); a = REAL(sa); b = REAL(sb); c = REAL(sc); d = REAL(sc); /* <-- change this line */ y = REAL(sy); should

Windows 9*/NT version, on an NT box R 0.63.0 Beta (Nov 13, 1998) > library(survival4) > test <- function(x,d) { survfit(Surv(x,d) ~1) } > test( rexp(50), sample(c(0,1), size=50, replace=T) ) Error: Object "x" not found > Other functions like in the ctest library works out OK, the x and d got passed to the inner function. And here it did not got passed to

At 10:37 PM 5/1/00 -0400, E. S. Venkatraman wrote: >I have the following problem. I have survival data (time, status) along >with several covariates (X1, X2,..., Xn). I want to fit a Cox model for >each of the covariate (univariately) and obtain the fitted probability of >survival at a fixed time point t0 and covariate value Xi0. I tried to do >this in a for loop where the index

================================================= This is an automated summary of the status of the R-bugs repository. Note that this may be neither complete nor perfectly correct at any given instance: Not all bugs are reported, and some reported bugs may have been fixed, but the repository not yet updated. Some bug fixes are difficult to verify because they pertain to specific hardware or

Hi I always have following error message when I try to read a big ascii-file: > inzp<-read.data() > Error: heap memory (1953 Kb) exhausted [needed 0 Kb more] read.data() is a small function that reads the ascii-file. When I cut the ascii-file to a small one I don't have this problem. Can I extend this 'heap memory' for reading big data files? How? I'm working with R on a

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)

Hello, I create a plot from a coxph object called fit.ads4: plot(survfit(fit.ads4)) plot is located at: https://www.dropbox.com/s/9jswrzid7mp1u62/survfit%20plot.png I also create the following survfit statistics: > print(survfit(fit.ads4),print.rmean=T) Call: survfit(formula = fit.ads4) records n.max n.start events *rmean *se(rmean) median 0.95LCL 0.95UCL 203.0

I've a data set with 60000 rows of data representing 6000+ distinct loans. I did a coxph() regression on it (see call below), but a subsequent survfit() call on the coxph object is almost certainly wrong. It gives n=6 when it should be more like 6000+ (I think) > survfit(resultag) Call: survfit.coxph(object = resultag) n events median 0.95LCL 0.95UCL 6 489 Inf

Hello, I am trying to compute the mdeian of the survival time from the function survfit: > fit <- survfit(Surv(time, status) ~ 1) > fit Call: survfit(formula = Surv(time, status) ~ 1) records n.max n.start events median 0.95LCL 0.95UCL 111 111 111 20 NA NA NA The results is NA? the fit$surv gives values between 1 and 0.749! Am I doing this correct?

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)

Hi everyone, I am not new to R, but new to running survival models in R. I am trying to create some basic KM curves, using the following code: library(survival) library(KMsurv) (import data etc - basic right censored, with continuously observed time of death) sleepfit <- survfit(Surv(timeb, death), data = sleep) Here timeb is measured is survival in years, death is a 1/0 indicator (1 =