similar to: Kaplan Meier Post Hoc?

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

Hi list, I want to use the p-value from the survdiff function (package survival) to reuse within a function in a Kaplan-Meier plot. The p-value is somehow not a component of the value list ?! Thanks in advance -- A. Goralczyk G?ttingen, Ger.

Hello, I am trying to set up a loop that can run the survdiff function with the ultimate goal to generate a csv file with the p-values reported. However, whenever I try a loop I get an error such as "invalid type (list) for variable 'survival_data_variables[i]". This is a subset of my data: structure(list(time = c(1.51666666666667, 72, 72, 25.7833333333333, 72, 72, 72, 72, 72,

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

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]]

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

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,

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:

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:

I read through Harrington and Fleming (1982) but it is beyond my statistical comprehension. I have survival data for insects that have a very finite expiration date. I'm trying to test for differences in survival distributions between different groups. I understand that the medical field is most often dealing with censored data and that survival analysis, at least in the package survival,

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

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

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

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)

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)

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

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

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) <-

Hi list I am running this ggplot2 R code: http://www.r-bloggers.com/an-enhanced-kaplan-meier-plot/ with its graphical output looks like this: http://statbandit.files.wordpress.com/2011/03/rstudio-v79384.pdf It gives me this error: ggplot2 doesn't know how to deal with data of class function I looked around seems like it's a common problem, but haven't found a solution. Thanks, Mike

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 |