similar to: need some help about using R for Kaplan-Meier estimator

Displaying 20 results from an estimated 1000 matches similar to: "need some help about using R for Kaplan-Meier estimator"

2008 Jun 28
0
How to get the 5th percentile( with a 95% CI )of the Kaplan-meier estimator?
Hi, all dear R experts, I am really stuck by how to get the 5th percentile( with a 95% CI )of the Kaplan-meier estimator of survival function P(T>t). I have a simulated sample of lifetime Ts, then I set the KM estimator < km.fit<-survfit(Surv(T),type="kaplan-meier",data=Surv(T)) <quantile(km.fit,.05) However, it gave "Error in order...". Does someone have some
2004 Jan 07
2
Survival, Kaplan-Meier, left truncation
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,
2001 Mar 11
2
Kaplan-Meier for left-truncated, right-censored data
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
2011 Jul 13
2
life table and Kaplan-Meier
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:
2003 Jan 16
1
help drawing kaplan-meier plot starting from 0
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
2008 Feb 14
4
Kaplan Meier function
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]]
2012 Oct 24
2
Kaplan Meier Post Hoc?
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
2007 Jul 05
1
Incidence estimated from Kaplan-Meier
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
2012 Nov 26
1
Plotting an adjusted kaplan-meier curve
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)
2012 Jan 05
2
Problem with axes in a plot of Kaplan-Meier
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)
2010 Apr 19
2
Kaplan-Meier survfit problem
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:
2007 Jun 10
0
Question on weighted Kaplan-Meier analysis of case-cohort design
I have a study best described as a retrospective case-cohort design: the cases were all the events in a given time span surveyed, and the controls (event-free during the follow-up period) were selected in 2:1 ratio (2 controls per case). The sampling frequency for the controls was about 0.27, so I used a weight vector consisting of 1 for cases and 1/0.27 for controls for coxph to adjust
2012 Aug 22
1
plotting Kaplan Meier using ggplot2 returns class function error
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
2008 Dec 06
1
Kaplan-Meier function from survfit
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 |
2012 Feb 20
1
Reporting Kaplan-Meier / Cox-Proportional Hazard Standard Error, km.coxph.plot, survfit.object
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
2013 Mar 26
1
Weighted Kaplan-Meier estimates with R
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
2009 Jul 22
1
kaplan-meier error
Hi all, I am getting an error in my code and I don't know what the problem is. I am using R 2.9 on ubuntu. my code is as follows: ## Libraries ## library(survival) library(foreign) ## reading data ## data<-read.dta("http://psfaculty.ucdavis.edu/bsjjones/cabinet.dta") head(data) attach(data) fit1<-survfit(Surv(durat,censor)) and I get the following error >
2006 Dec 09
2
Show number at risk on Kaplan Meier curve
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
2009 Jan 14
2
Kaplan-Meier Plot
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) <-
2012 Apr 13
3
Kaplan Meier analysis: 95% CI wider in R than in SAS
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