search for: mysurv

...creen > fit2 <- coxph(list(Surv(tstart, tstop, bstat) ~ 1, +??????????????????? c(1:4):5 ~ age / common + shared), id= id, istate=bili4, +?????????????? data=pbc2, ties='breslow', x=TRUE) > surv2 <- survfit(fit2, newdata=list(age=50), p0=c(.4, .3, .2, .1, 0)) > test2 <- mysurv(fit2, pbc2$bili4, p0= 4:0/10, fit2, x0 =50) ==31730== Invalid read of size 8 ==31730==??? at 0x298A07: Rf_allocVector3 (memory.c:2861) ==31730==??? by 0x299B2C: Rf_allocVector (Rinlinedfuns.h:595) ==31730==??? by 0x299B2C: R_alloc (memory.c:2330) ==31730==??? by 0x3243C6: do_which (summary.c:1152)...

Hello all, thanks for your time and patience. I'm looking for a method in R to analyse the following data: Time to waking after anaesthetic for medical procedures repeated on the same individual. > str(mysurv) labelled [1:740, 1:2] 20 20 15 20 30+ 40+ 50 30 15 10 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr [1:2] "time" "status" - attr(*, "type")= chr "right" - attr(*, "units")= chr "Day" - attr(*, "t...

I have some data which is censored and I want to determine the median. Its actually cost data for a cohort of patients, many of whom are still on treatment and so are censored. I can do the same sort of analysis for a survival curve and get the median survival... ...but can I just use the survival curve functions to plot an X axis that is $ rather than date? If not is there some other way to

...xph(list(Surv(tstart, tstop, bstat) ~ 1, > + c(1:4):5 ~ age / common + shared), id= id, > istate=bili4, > + data=pbc2, ties='breslow', x=TRUE) > > surv2 <- survfit(fit2, newdata=list(age=50), p0=c(.4, .3, .2, .1, 0)) > > test2 <- mysurv(fit2, pbc2$bili4, p0= 4:0/10, fit2, x0 =50) > ==31730== Invalid read of size 8 > ==31730== at 0x298A07: Rf_allocVector3 (memory.c:2861) > ==31730== by 0x299B2C: Rf_allocVector (Rinlinedfuns.h:595) > ==31730== by 0x299B2C: R_alloc (memory.c:2330) > ==31730== by 0x3243C6: d...

...urvival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized incidence ratio concept found in epidemiology. To answer your technical question: fit <- coxph(Surv(......... mysurv <- survfit(fit, newdata= mydata) This will give a set of predicted curves, one for each observation in mydata. If we assume 2 treatments and 4 ethnicities, this means that there are 8 possible predicted curves. You can certainly take the curves for "trt=1, white" and "trt=2, w...

Dear all: Could anyone please provide some R codes to plot the below survival data to compare two groups (0 vs 1) after 2 simulations (TRL)? need 95% prediction interval on the plot from these 2 trials. I would like to simulate 1000 trials later. Thanks a lot for your great help and consideration! yan TRL ID ECOG BASE PTR8 GROUP POP ST ind 1 1 1 1 2.2636717 0.255634126 1 1 99.4 F 3 1 2 1