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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.

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    $\begingroup$ Of course: see the survfit() function of the survival package [type help(package="survival")] $\endgroup$ Commented Apr 11, 2012 at 10:06
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    $\begingroup$ @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... $\endgroup$
    – user7064
    Commented Apr 11, 2012 at 10:16
  • $\begingroup$ Then use approxfun() $\endgroup$ Commented Apr 11, 2012 at 13:28
  • $\begingroup$ can I have an example? $\endgroup$
    – user7064
    Commented Apr 11, 2012 at 13:29

2 Answers 2

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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
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    $\begingroup$ 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... $\endgroup$ Commented Feb 26, 2019 at 22:00
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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)
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