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.
Asked
Active
Viewed 1.6k times
15
user7064
- 1,685
- 5
- 23
- 39
-
1Of course: see the survfit() function of the survival package [type help(package="survival")] – Stéphane Laurent Apr 11 '12 at 10:06
-
3@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... – user7064 Apr 11 '12 at 10:16
-
Then use approxfun() – Stéphane Laurent Apr 11 '12 at 13:28
-
can I have an example? – user7064 Apr 11 '12 at 13:29
2 Answers
24
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
Brian Diggs
- 1,021
- 8
- 17
-
4Surreal 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... – Brian Diggs Feb 26 '19 at 22:00
8
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)
bensentropy
- 181
- 1
- 4