similar to: Probabilities from survfit.coxph:

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 fit my CDF to sum of exponentials and now I want to take the numerical derivative of this function to obtain probability density.I will really appreciate your help reagrding the error messages I am getting which I don't understand. * * > fitterma <- function(xtime) { a <- -0.09144115 b <- -0.01335756 c <- -2.368057 d <- -0.00600052

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

Hi dear R-users: I have the following problem: I have a list of data.frames (12 variables and 60000 rows, each) I have to merge from an specific point of the list to the end of the list, I am doing so with a for() loop but it is too inefficient and it exhausts memory. How can I convert this for() loop in a function and then use lapply?

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)

Hi It seems that survfit() doesn't accept the argumnet 'error' as below >survfit(fit, error='greenwood') Error in survfit.coxph(fit, error = "greenwood") : unused argument(s) (error ...) Isn't is allowed to do that for a coxph object? Regards Soren Windows XP, SP2 R 2.3.0

--begin inclusion --- I am trying to extract output information from the survfit function in order to generate a matrix of select output for multiple factors. Specifically, I am interested in extracting the number of events (in the output below: 106, 2, 3). The variable names represented in my function (ee) are shown below, but none of those variables correspond to the column of events as shown

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Hi R-helpers, I was trying to put gender='Male' in newdata to create a expected survival curve for a pseudo cohort by using survfit based on Cox regression. My codes are shown below: fit<- coxph(Surv(end, status2)~gender, data=wlwsn1) Summary(fit) coef exp(coef) se(coef) z p genderMale 0.204 1.23 0.0912 2.23 0.025

Hello, How do I get plot.survfit to draw a survival curve with prespecified time periods, say 1 month, 2, months, 3 months, 6 months, 1 year, 2 years etc) instead of the individual times of all events? Sincerely Fredrik Lundgren

Hi, when ploting Kaplan-Meier estimate curves as below, the censoring symbols (crosses) to not change thickness along the lines plot(survfit(surv ~ I(x>=cut.off) ),lty=c(1,2), lwd=2) is there any strightforward way to make it happen? thanks robert -- View this message in context: http://r.789695.n4.nabble.com/censoring-symbols-on-survfit-plot-tp3311283p3311283.html Sent from the R help

Hi all! I am trying to extract output information from the survfit function in order to generate a matrix of select output for multiple factors. Specifically, I am interested in extracting the number of events (in the output below: 106, 2, 3). The variable names represented in my function (ee) are shown below, but none of those variables correspond to the column of events as shown in the output.

Hi, I am trying to do a weighted cox regression, specifically I need a weighted estimate of the baseline survival function. I'm getting an 'unused argument' error when I try to use the weights option in survfit. Looking at the code, it looks as though the documentation may be wrong and you actually can't use weights with survfit at this time where it says if

I am using the package Design for survival analysis. I want to plot a simple Kaplan-Meier fit of survival vs. age, with age grouped as quantiles. I can do this: survplot(survfit(Surv(time,status) ~ cut(age,3), data=veteran) but I would like to do something like this: survplot(survfit(Surv(time,status) ~ quantile(age,3), data=veteran) #will not work ideally I would like to superimpose

I'm having trouble getting fun="cloglog" to work with plot on a survfit object. Here are the data I used for the commands that follow. days status 2 0 2 0 5 1 9 0 14 1 16 0 16 0 17 0 29 1 30 0 37 1 37 0 39 1 44 0 44 0 58 0 60 1 67 1 68 1 82 1 82 1 86 0 86 0 89 1 93 0 97 1 100 0 100 0 100 0 > library(survival) Loading required package: splines > eg1.km <-

You wrote: "It may be best to either write to the package maintainer (me, as you did) or post to the group but not both." This is just a note that I disagree wrt my own packages: I go on vacation or trips, or have other projects so won't always answer Other folks on the list often have good ideas that I'd miss My preferred standard is "ask the list, with a copy

One approach is to use the rms package's cph and Mean.cph functions. Mean.cph (cph calls coxph and can compute Kaplan-Meier and other survival estimates) can compute mean restricted life. Frank Dinesh W wrote > I am using survfit to generate a survival curve. My population is such > that my x axis is in days and i have a starting population of say 10,000 > of which say only 2000 are

fit <- coxph(Surv(futime,fustat)~ age +strata(rx), data=ovarian, subset=1:23) curves <- survfit(fit, newdata=ovarian[24:26,]) I don't think this is mentioned in the documentation (I'll have to fix that!), but subscripting works for survfit objects. In this case there are 2 strata and 3 subjects, and curves[1,2] will return the survival curve for strata 1,

Hello R users Would somebody know how to estimate survival from a frailty cox model, using the function survfit and the argument newdata ? (or from any other way that could provide individual expected survival with standard error); Is the problem related to how the random term is included in newdata ? kfitm1 <- coxph(Surv(time,status) ~ age + sex + disease + frailty(id,

Hi, I am working on fitting a proportional hazard model to predict the probability of default for mortgage loans. I have a question regarding survfit function. My historical data set is a pool of loans with monthly observed default status for the next 24 months. The data is left truncated (delayed entry to observation window after the loan is opened) and right censored. I would like to