similar to: Kaplan Meier - not for dates

I have a dataset which for the sake of simplicity has two endpoints. We would like to test if two different end-points have the same eventual meaning. To try and take an example that people might understand better: Lets assume we had a group of subjects who all received a treatment. The could stop treatment for any reason (side effects, treatment stops working etc). Getting that data is very

Hello R users I have a question with Kaplan-Meier Curve with respect to my research. We have done a retrospective study on fillings in the tooth and their survival in relation to the many influencing factors. We had a long follow-up time (upto 8yrs for some variables). However, we decided to stop the analysis at the 6year follow up time, so that we can have uniform follow-up time for all the

I might be missing something really obvious, but is there an easy way to locate all non-unique values in a data frame? Example mydata <- numeric() mydata$id <- 0:8 mydata$unique <- c(1:5, 1:4) mydata$result <- c(1:3, 1:3, 1:3) > mydata $id [1] 0 1 2 3 4 5 6 7 8 $unique [1] 1 2 3 4 5 1 2 3 4 $result [1] 1 2 3 1 2 3 1 2 3 What I want to to be able to get some form of data output

I've been tweaking code for several days on and off in R, cut and pasting in from a text editor (I just leave them open all the time). I think I got something that was usable but then a powersurge tripped the fuses and unfortunately the machine I was working on doesn't have a UPS. Does R hold the command data in cache some place? I've purposefully not opened R since the crash so

Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to include enough detail. I have a question about survival curves I have been battling off and on for a few months. No one local seems to be able to help, so I turn here. The issue seems to either be how R calculates Kaplan-Meier Plots, or something with the underlying statistic itself that I am misunderstanding. Basically,

Hi, I've seen posts (but no answers) about clearing a file lock on a Samba share. Presumably this kind of thing happens when the program abnormally terminates and doesn't close its open files. I've got a file that smbstatus reports as: 1216 DENY_WRITE 0x20089 RDONLY NONE /shares/share1/Files/december.xls Fri Jan 12 12:51:30 2007 Trying to open this in Excel

Dear all, I have data from 1970 to 1990 for people above age 50. Now I want to calculate survival curves by age starting at age 50 using the Kaplan Meier Estimator. The problem I have is that there are already people in 1970 who are older than 50 years. I guess this is called delayed entry or left truncation (?). I thought the code would be: roland <- survfit(Surv(time=age.enter,

Hi I have a long data set on which I want to do Bland-Altman style plots for each rhythm type Using ggplot2, when I use geom_hline with facet_grid I get an extra set of empty panels. I can't get it to do it with the "Diamonds" data supplied with the package so here is a (much abbreviated) example: > lvexs cvd_basestudy ecd_rhythm fixed_time variable_time 1 CBP05J02

Hi Fello: I am asked to compute the Kaplan-Meier estimator of data with right censoring without using surfit(). Does anyone know how to use R to compute the estimators? The data should input X: vector of right-censored observed time for n individuals, and d: vector of failure time indicators (0=censored individual;1=unconsored individual) and the function should return with t: vector of sorted

I'm hoping I'm missing some (probably fundamental basic process) which might make my life easier! Lets assume I have a 3 column table summarizing results from a trial from three arms (Arm A, B and C). For each arm there will be a number of pieces of information to report. The simplest example might be to compare this to the demographic comparisons often seen in clinical traisl where you

When I try to the code from library(survival) of library(ISwR), the following code survfit(Surv(days,status==1)) that could produce Kaplan-Meier estimates shows the following error "Error in survfit(Surv(days, status == 1)) : Survfit requires a formula or a coxph fit as the first argument" How it can be done in R.2.10 -- View this message in context:

dear all, I want to plot a kaplan Meier plot with the following functions, but I fail to produce the plot I want: library(survival) tim <- (1:50)/6 ind <- runif(50) ind[ind > 0.5] <- 1; ind[ind < 0.5] <- 0; MS <- runif(50) pred <- vector() pred[MS < 0.3] <- 0; pred[MS >= 0.3] <- 1 df <- as.data.frame(cbind(MS, tim, pred, ind)) names(df) <-

There are two ways to view weights. One is to treat them as case weights, i.e., a weight of 3 means that there were actually three identical observations in the primary data, which were collapsed to a single observation in the data frame to save space. This is the assumption of survfit. (Most readers of this list will be too young to remember when computer memory was so small that we had to

I tried to post this a few times last week and it seems to have got stuck somehow so I'm trying from a different email in the hope that works. If somehow this has appeared on the list 20 tiems and I never saw any of them I apologize ;-) I'm basically an R-newbie. But I am VERY computer literate. But this has me stumped... All the examples for using the rmeta package to create a

Hi, I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this: library(survival) data(lung) lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung) # lung.byPS Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung) 1

Is it possible to calculate Kaplan-Meier for left-truncated, right-censored data using survival5? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at

I may be doing this wrong! but I have a function which I have simplified a lot below. I want to pass some 'field names' of a data-frame to the function for it to then do some manipulation of. Here's my code: #build a simple dataset mydataset = data.frame (

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" -

Hello, I have a question about the function lifetab in package KMsurv. The description of the output value surv says "the estimated survival function at the start of the intervals". Are these estimates the ones calculated via Kaplan-Meier probability of survival ? Thanks in advance! -- View this message in context:

Hi I am a newbie to R but have tried a number of ways in R to do this and can't find a good solution. (I could do it out of R in perl or awk but would like to know how to do this in R). I have a large data frame 49 variables and 7000 observations however for simplicity I can express it in the following data frame Base, Image, LVEF, ES_Time A, 1, 4.32, 0.89 A, 2, 4.98, 0.67 A, 3, 3.7, 0.5