similar to: Wrong behavior of lines.survfit for conf.int="only" (PR#13180)

Hallo, I modified a code given by Andrija, a contributor in the list  to achieve two objectives: create 1000 samples from a list of 207 samples with each of the samples cointaining 20 good and 20 bad. THis i have achievedcalcuate AUC each of the 1000 samples, this i get an error. Please see the code below and assist me. > data<-data.frame(id=1:(165+42),main_samp$SCORE,

Dear all, I tried to install rgl package on my linux machine fc5, I got an error. Here I run R as user, $ R > options(repos=c(CRAN="http://cran.at.r-project.org/")) > install.packages("rgl", lib="/home/subianto/local/lib/R/library/site-packages", dependencies=TRUE) trying URL 'http://cran.at.r-project.org/src/contrib/rgl_0.67-2.tar.gz' Content type

Dear all recently I tried to split vector of dates according to some particular date to 2 (more) chunks, but I was not able to perform correct setting. When I want split to 3 chunks it partially works however from help page I supposed to get result without NA. Details: Using both ?right = TRUE? and ?include.lowest = TRUE? will include both ends of the range of dates. dat <-

Jitterbug is just what I want. Bugzilla is great but not for a non-programming (issue-tracking) audience. Can I get jitterbug somewhere please (it seems you have removed it) ? Cheers andy -- andy _______________________________________________ Andy Heath a.k.heath@shu.ac.uk

I am trying to create a frequency distribution and I am a bit confused. Here are the commands I have entered: > data <- read.csv(file="40609_sortedfinal.csv",head=TRUE,sep=",") > NumberOfActionsByStatus = data$STATUS > NumberOfActionsByUser = data$ETS_LOGIN > NumberOfBidOffer = data$BID_OFFER > NumberOfActionsByUser.freq = table(NumberOfActionsByUser) >

When I went to report a (very minor) bug today, I noticed that under the choices for Version, there was no entry for "R 2.15.2" nor "R 2.15.2 patched". I could not find a contact email for the bug tracker itself, so I figured the next best place to send this was r-devel. Perhaps adding new versions to the version choices should also be added to

I am using the coxph, survfit and summary.survfit functions to calculate an estimate of predicted survival with confidence interval for future patients based on the survival distribution of an existing cohort of subjects. I am trying to understand the calculation and interpretation of the std.err and confidence intervals printed by the summary.survfit function. Using the default confidence

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)

The following code: cut(as.POSIXct("2009-11-01 04:00:00", tz="America/Los_Angeles"), "1 day") gives the error: Error in seq.int(0, to - from, by) : 'to' must be finite This is related to November 1st, 2009 being the switchover date from daylight savings time to standard time in the America/Los_Angeles time zone. In particular, in cut.POSIXt, the starting

On 1 Jun 2004, John Summerfield <john@Corridors.wa.edu.au> wrote: > The jitterbug link on http://rsync.samba.org/nobugs.html no longer works. I > suggest it either be fixed or removed. Thanks, fixed. > You make bug-reporting needlessly difficult, I think. I dislike the need to > subscribe to a mailing-list and potentially receive lots of email that > doesn't interest

>> log(data.frame(x=1:4),base=2) > Error: Object "base" not found This one seems to be fixed. Move to Language-fixed. (Note: there seems to be some strangeness with composing replies like this in the web interface to JitterBug. Watch the headers if you want to followup.) -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list --

Hello R users, Good day!! I was wondering if there is a way in R to read in a particular column from a tab-delimited file, edit it and write it back into the file with all other columns intact. When I say edit I mean just replacing all the values in that column. I know to read a particular column from a file using colClasses option in read.delim() function. Is there any such option to write out

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 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

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 =

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, I am wondering how the confidence interval for Kaplan-Meier estimator is calculated by survfit(). For example,  > summary(survfit(Surv(time,status)~1,data),times=10) Call: survfit(formula = Surv(rtime10, rstat10) ~ 1, data = mgi)  time n.risk n.event survival std.err lower 95% CI upper 95% CI    10    168      55    0.761  0.0282        0.707        0.818 I am trying to reproduce the

The answer to this may be obvious, but I was wondering in the rms package and the survfit(), how you can plot the censored time points as ticks. Take for example, library(survival) library(rms) foo <- data.frame(Time=c(1,2,3,4,5,6,10), Status=c(1,1,0,0,1,1,1)) answer <- survfit(Surv(foo$Time, foo$Status==1) ~1) # this shows the censored time points as ticks at Time = 3 and 4 plot(answer)

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |