similar to: Fitting additional variables to model

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

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,

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)

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:

Hi, I am attempting to graph a Kaplan Meier estimate for some claims using the survfit function. However, I was wondering if it is possible to plot a cdf of the kaplan meier rather than the survival function. Here is some of my code: library(survival) Surv(claimj,censorj==0) survfit(Surv(claimj,censorj==0)~1) surv.all<-survfit(Surv(claimj,censorj==0)~1) summary(surv.all) plot(surv.all)

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 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,

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

Hi, I'm trying to print the p-values from the output of a CPH test onto a Kaplan Meier plot. Can this be done? I only really want the p-values from the CPH test to appear but if this can't be done I am willing to have the entire CPH output. This is what I am currently trying: (it doesn't print the CPH output) plot_KM <- function(field) { library(survival) y =

Hello all! I am once again analyzing patient survival data with chronic liver disease. The severity of the liver disease is given by a number which is continuously variable. I have referred to this number as "meld"--model for end stage liver disease--which is the result of a mathematical calculation on underlying laboratory values. So, for example, I can generate a Kaplan-Meier plot

First a statistical issue: The survfit routine will produce predicted survival curves for any requested combination of the covariates in the original model. This is not the same thing as an "adjusted" survival curve. Confusion on this is prevalent, however. True adjustment requires a population average over the confounding factors and is closely related to the standardized

Did you not notice the conf.type = "none" argument to your survfit call and the associated documentation in the survfit help? -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sat, Aug 26, 2017 at 5:18 PM, Adrian Johnson

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 |

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

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 Bert, thank you for suggestion. I am aware of R-help function. I must apologize, my earlier question could lead to assumptions otherwise. As you can see below, I only get Std. error but not lower and Upper CIs. I was wondering if there is another argument or method, could give CIs which I cannot find anywhere. Thanks Adrian > Nsurv <-

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

Hi, I am wondering if anyone can explain to me if cumulative incidence (CI) is just "1 minus kaplan-Meier survival"? Under what circumstance, you should use cumulative incidence vs KM survival? If the relationship is just CI = 1-survival, then what difference it makes to use one vs. the other? And in R how I can draw a cumulative incidence plot. I know I can make a Kaplan-Meier

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) <-

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)