similar to: survfit plot question

Hi, When I run the survfit function, I want to get the restricted mean value and the standard error also. I found out using the "print" function to do so, as shown below, print(km.fit,print.rmean=TRUE) Call: survfit(formula = Surv(diff, status) ~ 1, type = "kaplan-meier") records n.max n.start events *rmean *se(rmean) median 200.000

Hello, I have estimated a survival model with six strata: >model.b <- survfit(Surv(time=start.tijd,time2=eind.tijd2,event=va)~strata(product.code) , data=wu.wide) I would like to save the output of >print(model.b,print.n="records",show.rmean=FALSE) in a dataframe so that I can export it later. How do I do this? Note that summary(model.b) gives an error: Error in

Hi, I'm not sure if this is the proper way to ask questions, sorry if not. But here's my problem: I'm trying to do a bootstrap estimate of the mean for some survival data. Is there a way to specifically call upon the rmean value, in order to store it in an object? I've used print(...,print.rmean=T) to print the summary of survfit, but I'm not sure how to access only rmean

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

Hello all, Here goes one of my first functions. I want to make a nonparametric multiple sample comparison with unequal sample sizes (see Zar?s Biostatistical Analysis, 3rd. Ed., pg. 201 Example 10.11, pg. 288 Example 11.10). In the real world, I want to compare samples of fish length captured with different fishing gears. After using the Kruskal-Wallis test I want to check the differences

Dear all, I'm having difficulty getting access to data generated by survfit and print.survfit when they are using with a Cox model (survfit.coxph). I would like to programmatically access the median survival time for each strata together with the 95% confidence interval. I can get it on screen, but can't get to it algorithmically. I found myself examining the source of print.survfit to

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494

Dear Sirs, I want to estimate the survival mean of a few specific teams. I'm trying to calculate it through a Kaplan Meier estimator. For doing so, I load the "survival" package and run the following instructions: "options(survfit.print.mean=TRUE)" allows showing the mean and mean standard error "KM=survfit(Surv(Dias,Censura))"

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Dear r-help, I have a matrix, suppose, 10x10, and I need the matrix 5x5, having in each cell a mean value of the cells from the initial matrix. Please, point me to a function in R, which can help me doing that. Digging the documentation and mail archives didn't give me a result. Thank you. --- Best regards, Vladimir mailto:wl at eimb.ru

Hello, I've executed the following predict.coxph function to enable prediction for new variable values (error is included). *predict(cox_out,newdata=data.frame(Meter3.Value=100.001, Meter4.Value=200.001,Meter5.Value=300.001,Meter10.Value= 400.001,type="expected")) Error in model.frame.default(data = data.frame(Meter3.Value = 100.001, : variable lengths differ (found for

I have a print method for a set of statistical tests, vcdExtra::CMHtest, for which I'd like to have more sensible printing of pvalues, as in print.anova(). [Testing this requires the latest version of vcdExtra, from R-Forge **|install.packages("vcdExtra", repos="http://R-Forge.R-project.org")|** ] With my current print method, I get results like this, but all Prob values

Hello, I need some results from the survival analysis of my data that I do not know whether exist in Survival Package or how to obtain if they do: 1. The Mean survival time 2. The standard error of the mean 3. Point and 95% Lower & Upper Confidence Intervals estimates Any help will be greatly appreciated. Cem [[alternative HTML version

Hi, everyone, I ran the test case in "Introduction to the Bioconductor marrayInput package first. When I ran the session in (read.marrayRaw) I got the following error messages. Other sessions work well. The functions and errors are labled with blue and red respectively. >mraw <- read.Spot(path = datadir, layout = galinfo$layout, gnames = galinfo$gnames, target = swirlTargets) Error

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

Hi All. I've run into a problem with the plinear algorithm in nls that is confusing me. Assume the following reaction time data over 15 trials for a single unit. Trials are coded from 0-14 so that the intercept represents reaction time in the first trial. trl RT 0 1132.0 1 630.5 2 1371.5 3 704.0 4 488.5 5 575.5 6 613.0 7 824.5 8 509.0 9

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?

Terry, I recently noticed the censor argument of survfit. For some analyses it greatly reduces the size of the resulting object, which is a nice feature. However, when combined with the id argument, only 1 prediction is made. Predictions can be made individually but I'd prefer to do them all at once if that change can be made. Chris ##################################### # CODE # create

Dear List, Does anybody no how to compare mean survival times for two (more) groups in R? What test statistics should I use? Thank you very much! Joe [[alternative HTML version deleted]]

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.