Estimating a survival probability in R

Based on a sample of $n$ survival times, I would like to estimate the probability of surviving time $t$, for some specific $t$, using the Kaplan-Meier estimator. Is it possible to do this in R? Please, note that $t$ is not necessarily an event time.

• Of course: see the survfit() function of the survival package [type help(package="survival")] Commented Apr 11, 2012 at 10:06
• @Stephane Laurent: The surfit() function outputs the estimated survival at event times. But I'd like to have an automatic procedure to compute that survival at any time t. Thanks... Commented Apr 11, 2012 at 10:16
• Then use approxfun() Commented Apr 11, 2012 at 13:28
• can I have an example? Commented Apr 11, 2012 at 13:29

2 Answers

You can use the output of the survfit function from the survival package and give that to stepfun.

km <- survfit(Surv(time, status)~1, data=veteran)
survest <- stepfun(km$time, c(1, km$surv))


Now survest is a function that can be evaluated at any time.

> survest(0:100)
[1] 1.0000000 0.9854015 0.9781022 0.9708029 0.9635036 0.9635036 0.9635036
[8] 0.9416058 0.9124088 0.9124088 0.8978102 0.8905109 0.8759124 0.8613139
[15] 0.8613139 0.8467153 0.8394161 0.8394161 0.8175182 0.8029197 0.7883212
[22] 0.7737226 0.7664234 0.7664234 0.7518248 0.7299270 0.7299270 0.7225540
[29] 0.7225540 0.7151810 0.7004350 0.6856890 0.6856890 0.6783160 0.6783160
[36] 0.6709430 0.6635700 0.6635700 0.6635700 0.6635700 0.6635700 0.6635700
[43] 0.6561970 0.6488240 0.6414510 0.6340780 0.6340780 0.6340780 0.6267050
[50] 0.6193320 0.6193320 0.5972130 0.5750940 0.5677210 0.5529750 0.5529750
[57] 0.5456020 0.5456020 0.5456020 0.5382290 0.5382290 0.5308560 0.5308560
[64] 0.5234830 0.5234830 0.5234830 0.5234830 0.5234830 0.5234830 0.5234830
[71] 0.5234830 0.5234830 0.5161100 0.5087370 0.5087370 0.5087370 0.5087370
[78] 0.5087370 0.5087370 0.5087370 0.4939910 0.4939910 0.4866180 0.4866180
[85] 0.4791316 0.4791316 0.4791316 0.4716451 0.4716451 0.4716451 0.4640380
[92] 0.4640380 0.4564308 0.4564308 0.4564308 0.4412164 0.4412164 0.4412164
[99] 0.4412164 0.4257351 0.4179945

• Surreal is having a problem, looking for an answer on StackExchange, finding the same question, and realizing you were the one who answered it almost 7 years ago... Commented Feb 26, 2019 at 22:00

A time parameter can be passed to the summary function of the survfit object:

summary(km, times=100)


A vector can also be passed:

summary(km, times=0:100)