similar to: kaplan-meier error

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:

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 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 |

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 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 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]]

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,

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 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

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 have some data which is censored and I want to determine the median. Its actually cost data for a cohort of patients, many of whom are still on treatment and so are censored. I can do the same sort of analysis for a survival curve and get the median survival... ...but can I just use the survival curve functions to plot an X axis that is $ rather than date? If not is there some other way to

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

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'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

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)

I thought this would be trivial, but I can't find a package or function that does this. I'm hoping someone can guide me to one. Imagine a simple case with two survival curves (e.g. treatment & control). I just want to calculate the difference in KM estimates at a specific time point (e.g. 1 year) plus the estimate's 95% CI. The former is straightforward, but the estimates not

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 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

Dear r-helpers, This is my first time to run survival analysis. Currently, I have a data set which contains two variables, the variable of time to event (or time to censoring) and the variable of censor indicator. For the indicator variable, it was coded as 0 and 1. 0 represents right censor, 1 means event of interest. Now I try to use "survfit" in the package of "survival". I

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