R help - Dec 2010 - survival package - calculating probability to survive a given time

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, status) ~ 1)
s <- summary(fit)

## 1. possibility to get the probability for surviving 20 units of time
ind <- findInterval(20, s$time)
cbind(s$surv[ind], s$time[ind])

## 2. possibility to get the probability for surviving 20 units of time
ind <- s$time >= 20
sum(ind) / length(ind)

Thanks and best regards

Andreas

On Dec 10, 2010, at 2:07 PM, Andreas Wittmann wrote:
> 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, status) ~ 1)
> s <- summary(fit)
>
> ## 1. possibility to get the probability for surviving 20 units of  
> time
> ind <- findInterval(20, s$time)
> cbind(s$surv[ind], s$time[ind])
See if this helps:

 > head(which(s$surv < 0.5))
[1] 368 369 370 371 372 373
 > plot(fit)
 > abline(h=0.5)
 > abline(v=s$time[368])

>
> ## 2. possibility to get the probability for surviving 20 units of  
> time
> ind <- s$time >= 20
> sum(ind) / length(ind)

-- 
David Winsemius, MD
West Hartford, CT

- ------- included message --------
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?
---- end inclusion --------

The survfit function directly estimates the probability you want, no
predict method is needed.  It is similar to quantile(x, 1:9/10), in that
you would not use predict on that either.

As to getting the value at a specific time, use
	summary(fit, time=20)
This correctly interpolates the step function when "20" is exactly one
of the event times.

Terry Therneau