Survival analysis in R Niels Richard Hansen This note describes a - PDF document

Survival analysis in R Niels Richard Hansen This note describes a few elementary aspects of practical analysis of survival data in R. For further information we refer to the book “Introductory Statistics with R” by Peter Dalgaard and “Dynamic Regression Models for Survival Data” by Torben Martinussen and Thomas Scheike and to the R help files. There is also the classical reference “Statistical Models Based on Counting Processes” by Andersen, Borgan, Gill and Keiding. The analyses that we will do can all be done using the functions in library sur- vival . That library also contains some example data sets. > library(survival) > data(pbc) > pbc[1:10, c("time", "status", "age", "edema", "alb", "bili", + "protime")] time status age edema alb bili protime 1 400 1 58.7652 1 2.60 14.5 12.2 2 4500 0 56.4463 0 4.14 1.1 10.6 3 1012 1 70.0726 1 3.48 1.4 12.0 4 1925 1 54.7406 1 2.54 1.8 10.3 5 1504 0 38.1054 0 3.53 3.4 10.9 6 2503 1 66.2587 0 3.98 0.8 11.0 7 1832 0 55.5346 0 4.09 1.0 9.7 8 2466 1 53.0568 0 4.00 0.3 11.0 9 2400 1 42.5079 0 3.08 3.2 11.0 10 51 1 70.5599 1 2.74 12.6 11.5 The data set contains the main variable time , which is the observed, potentially censored survival time and the status , which indicates if the observation has been censored. In addition there are 17 observed covariates, where we show above five that in previous analyses have been found to be important. We can produce a plot of the survival times with : > nr <- 50 > plot(c(0, pbc$time[1]), c(1, 1), type = "l", ylim = c(0, nr + + 1), xlim = c(0, max(pbc$time[1:nr]) + 10), ylab = "nr", xlab = "Survival tim > for (i in 2:nr) lines(c(0, pbc$time[i]), c(i, i)) > for (i in 1:nr) { 1

+ if (pbc$status[i] == 0) + points(pbc$time[i], i, col = "red", pch = 20) + if (pbc$status[i] == 1) + points(pbc$time[i], i) + } 50 40 30 nr 20 10 0 0 1000 2000 3000 4000 Survival time Estimation of the survival function using the Kaplan-Meier estimator can be done using the survfit function. > pbc.surv <- survfit(Surv(pbc$time, pbc$status)) > pbc.surv Call: survfit(formula = Surv(pbc$time, pbc$status)) n events median 0.95LCL 0.95UCL 418 161 3395 3090 3853 > plot(pbc.surv) 2

1.0 0.8 0.6 0.4 0.2 0.0 0 1000 2000 3000 4000 The result from survfit is an R object of type survfit . Taking summary of a survfit object provides the non-parametric estimate of the survival function including additional information at a long list. The plot is perhaps more infor- mative with the added, pointwise confidence bands. The function Surv applied to the time and status variables for the PBC data is a function that create a survival object. A major point in a data set like the PBC data above is that we record covariate information on the individuals. It is no problem to split the estimation of the survival function according to a factor such as the binary edema variable in the PBC data. > pbc.surv <- survfit(Surv(pbc$time, pbc$status) ~ pbc$edema) > plot(pbc.surv, conf.int = TRUE, col = c("red", "blue")) 3

1.0 0.8 0.6 0.4 0.2 0.0 0 1000 2000 3000 4000 From the plot it seems clear that individuals without edema has a larger chance of a large survival time than those with edema. This can also be testet formally with the so-called log-rank test. > survdiff(Surv(pbc$time, pbc$status) ~ pbc$edema) Call: survdiff(formula = Surv(pbc$time, pbc$status) ~ pbc$edema) N Observed Expected (O-E)^2/E (O-E)^2/V pbc$edema=0 368 121 151.0 5.94 96 pbc$edema=1 50 40 10.0 89.28 96 Chisq= 96 on 1 degrees of freedom, p= 0 But what about including the other covariates also – especially the continuous ones? This can be handled in the framework of the Cox proportional hazards model. > pbc.surv <- coxph(Surv(pbc$time, pbc$status) ~ pbc$edema) > summary(pbc.surv) 4

Call: coxph(formula = Surv(pbc$time, pbc$status) ~ pbc$edema) n= 418 coef exp(coef) se(coef) z p pbc$edema 1.63 5.09 0.185 8.82 0 exp(coef) exp(-coef) lower .95 upper .95 pbc$edema 5.09 0.197 3.54 7.3 Rsquare= 0.129 (max possible= 0.985 ) Likelihood ratio test= 57.7 on 1 df, p=3e-14 Wald test = 77.7 on 1 df, p=0 Score (logrank) test = 96 on 1 df, p=0 We can tell from the summary that the β corresponding to the covariate edema is significantly different from 0. We also see that the hazard rate for those with edema=1 is estimated as roughly 5 times the base-line hazard rate α 0 ( t ). We can include more covariates. > pbc.surv <- coxph(Surv(time, status) ~ edema + age + alb + bili + + protime, data = pbc) > summary(pbc.surv) Call: coxph(formula = Surv(time, status) ~ edema + age + alb + bili + protime, data = pbc) n= 418 coef exp(coef) se(coef) z p edema 0.7222 2.059 0.20764 3.48 5.1e-04 age 0.0365 1.037 0.00811 4.50 6.8e-06 alb -0.9266 0.396 0.20898 -4.43 9.2e-06 bili 0.1246 1.133 0.01271 9.80 0.0e+00 protime 0.1928 1.213 0.05770 3.34 8.3e-04 exp(coef) exp(-coef) lower .95 upper .95 edema 2.059 0.486 1.371 3.093 age 1.037 0.964 1.021 1.054 alb 0.396 2.526 0.263 0.596 bili 1.133 0.883 1.105 1.161 protime 1.213 0.825 1.083 1.358 5

Rsquare= 0.36 (max possible= 0.985 ) Likelihood ratio test= 186 on 5 df, p=0 Wald test = 218 on 5 df, p=0 Score (logrank) test = 309 on 5 df, p=0 It is of course also possible to transform the covariates before using them in the regression. For example; > pbc.surv <- coxph(Surv(time, status) ~ edema + age + log(bili) + + log(alb) + log(protime), data = pbc) > summary(pbc.surv) Call: coxph(formula = Surv(time, status) ~ edema + age + log(bili) + log(alb) + log(protime), data = pbc) n= 418 coef exp(coef) se(coef) z p edema 0.6613 1.937 0.20595 3.21 1.3e-03 age 0.0382 1.039 0.00768 4.97 6.5e-07 log(bili) 0.8975 2.453 0.08271 10.85 0.0e+00 log(alb) -2.4524 0.086 0.65707 -3.73 1.9e-04 log(protime) 2.3458 10.442 0.77425 3.03 2.4e-03 exp(coef) exp(-coef) lower .95 upper .95 edema 1.937 0.5162 1.2938 2.901 age 1.039 0.9625 1.0234 1.055 log(bili) 2.453 0.4076 2.0863 2.885 log(alb) 0.086 11.6167 0.0237 0.312 log(protime) 10.442 0.0958 2.2895 47.624 Rsquare= 0.429 (max possible= 0.985 ) Likelihood ratio test= 234 on 5 df, p=0 Wald test = 233 on 5 df, p=0 Score (logrank) test = 297 on 5 df, p=0 A Theory We consider n individuals, T ∗ 1 , . . . , T ∗ n independent, positive random variables (survival times). 6

Estimation of the distribution by the empirical distribution: n F ( t ) = 1 ˆ � 1( T ∗ i ≤ t ) n i =1 and the survival function S ( t ) = 1 − F ( t ) n F ( t ) = 1 S ( t ) = 1 − ˆ ˆ � 1( T ∗ i > t ) . n i =1 Censoring by positive, random variables C 1 , . . . , C n : We observe only T i = min { T ∗ i , C i } and ∆ i = 1( T ∗ i ≤ C i ) . Thus we observe the pairs ( T 1 , ∆ 1 ) , . . ., ( T n , ∆ n ) . Define the counting process of deaths (non censored events) � N ( t ) = 1( T i ≤ t, ∆ i = 1) . i =1 Define also the process of individuals at risk n � Y ( t ) = 1( t ≤ T i ) . i =1 The jumps for the counting process are given by the process ∆ N ( t ) = N ( t ) − N ( t − ) = N ( t ) − lim ǫ → 0+ N ( t − ǫ ) ∈ { 0 , 1 } , which is simply an indicator process for each jump (death). The Kaplan-Meier estimator of the survival function, based on censored survival times, is � � 1 − ∆ N ( s ) ˆ � S ( t ) = . Y ( s ) s ≤ t The factors are equal to 1 for all s where ∆ N ( s ) = 0. If we denote by τ i the time for the i ’th jump, we can also write the estimator out as � � � � � � 1 1 1 ˆ S ( t ) = 1 − 1 − . . . 1 − Y ( τ 1 ) Y ( τ 2 ) Y ( τ N ( t ) ) Each factor can be interpreted as an estimator of the conditional probability of dying in the time interval ( τ i , τ i +1 ] given survival to time τ i . 7