similar to: Survival, Kaplan-Meier, left truncation

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

Hi all, I am trying to draw a Kaplan-Meier curve and I found online that Kaplan - Meier estimates are computed with a function called km in the event package. Is there an update for that because when I choose to download packages in R,. there is no package called event, even though I have selected all the repositories. Thanks in advance, Eleni [[alternative HTML version deleted]]

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:

Better would be to use interval censored data. Create your data set so that you have (time1, time2) pairs, each of which describes the interval of time over which the tag was lost. So an animal first captured at time 10 sans tag would be (0,10); with tag at 5 and without at 20 would be (5,20), and last seen with tag at 30 would be (30, NA). Then survit(Surv(time1, time2,

Dear R-users I am trying to make an adjusted Kaplan-Meier curve (using the Survival package) but I am having difficulty with plotting it so that the plot only shows the curves for the adjusted results. My data come from a randomised controlled trial, and I would like the adjusted Kaplan-Meier curve to only show two curves for the adjusted survival: one for those on treatment (Treatment==1)

Dear help news reader, I'm trying to draw a Kaplan-Meier curve and would like to ask the news group for some help Supposing I have study comapring two drugs, "A", and "B" and I recorde the time to get to the clinical endpoint (Time), in my case becommming virus free. I have setup the following frame: Time c Drug 1 5 1 A 2 7 1 B 3 2 1 A 4 10 1

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

What is the best way to report the standard error when publishing Kaplan-Meier plots? In my field (Vascular Surgery), practitioners loosely refer to the "10% error" cutoff as the point at which to stop drawing the KM curve. I am interpreting this as the *standard error of the cumulative hazard*, although I'm having a difficult time finding some guidelines about this (perhaps I am

Helo: After changing "involuntarily" some of the graphics parameters with the command par() (I did not know that changes with this command are permanent), now when I made a plot of the survival Kaplan-Meier function, the Y axis does not start at 1, and the X axis does starts at 0. The commands that I use are: library(survival) BROWN.SPV = Surv(BROWN$TEMPS, BROWN$DEF)

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 |

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:

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

Dear all, I'm using the "survival" package with R 2.4.0 on Mac OS X 10.4.8. I have two core statistics books (one of which is Altman's medical stats book) which suggest showing the number of individuals at risk at different time intervals on the Kaplan-Meier curve. My plot shows two curves that later cross, because of one significant outlier. I have two queries: Is there an

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

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,

Dear all, I have a stat question that may not be related to R, but I would like to have your advice. I have just read a medical paper in which the authors report the 1-p (where p is the cumulative survival probability from the Kaplan Meier curve) as incidence of disease. Specifically, the study followed ~12000 women on drug A and ~20000 women on drug B for 12 months. During that period

This is more of a general question without data. After doing 'survdiff', from the 'survival' package, on strata including four groups (so 4 curves on a Kaplan Meier curve) you get a chi squared p-value whether to reject the null hypothesis or not. Is there a method to followup with pairwise testing on the respective groups? I have searched the library but have come up with

Hello All, ? Am replicating in R an analysis I did earlier using SAS. See this as a test of whether I'm ready to start using R in my day-to-day work. ? Just finished replicating a Kaplan Meier analysis. Everything seems to work out fine except for one thing. The 95% CI around my estimate for the median is substantially larger in R than in SAS. For example, in SAS I have a median of 3.29 with a

Dear R users, i try to calculate the probabilty to survive a given time by using the estimated survival curve by kaplan meier. What is the right way to do that? as far as is see i cannot use the predict-methods from the survival package? library(survival) set.seed(1) time <- cumsum(rexp(1000)/10) status <- rbinom(1000, 1, 0.5) ## kaplan meier estimates fit <- survfit(Surv(time,