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Type 'q()' to quit R. > options(na.action=na.exclude) # preserve missings > options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type > library(survival) > > # > # Simple test of (start, stop] Kaplan-Meier curves, using the test2 data > # set > # > test1 <- data.frame(time= c(9, 3,1,1,6,6,8), + status=c(1,NA,1,0,1,1,0), + x= c(0, 2,1,1,1,0,0)) > test2 <- data.frame(start=c(1, 2, 5, 2, 1, 7, 3, 4, 8, 8), + stop =c(2, 3, 6, 7, 8, 9, 9, 9,14,17), + event=c(1, 1, 1, 1, 1, 1, 1, 0, 0, 0), + x =c(1, 0, 0, 1, 0, 1, 1, 1, 0, 0), + wt = 1:10) > aeq <- function(x,y, ...) all.equal(as.vector(x), as.vector(y), ...) > test2 <- test2[c(1,6,2,7,3,8,4,9,5,10),] # unsorted data is a harder test > > fit1 <- survfit(Surv(start, stop, event) ~1, test2, type='fh2', + error='tsiatis') > fit2 <- survfit(Surv(start, stop, event) ~x, test2, start.time=3, + type='fh2') > > cfit1<- survfit(coxph(Surv(start, stop, event)~1, test2)) > cfit2<- survfit(coxph(Surv(start, stop, event) ~ strata(x), test2, subset=-1)) > > deaths <- (fit1$n.event + fit1$n.censor)>0 > aeq(fit1$time[deaths], cfit1$time) [1] TRUE > aeq(fit1$n.risk[deaths], cfit1$n.risk) [1] TRUE > aeq(fit1$n.event[deaths], cfit1$n.event) [1] TRUE > aeq(fit1$surv[deaths], cfit1$surv) [1] TRUE > aeq(fit1$std.err[deaths], cfit1$std.err) [1] TRUE > > deaths <- (fit2$n.event + fit2$n.censor)>0 > aeq(fit2$time[deaths], cfit2$time) [1] TRUE > aeq(fit2$n.risk[deaths], cfit2$n.risk) [1] TRUE > aeq(fit2$n.event[deaths], cfit2$n.event) [1] TRUE > aeq(fit2$surv[deaths], cfit2$surv) [1] TRUE > > fit3 <- survfit(Surv(start, stop, event) ~1, test2) #Kaplan-Meier > aeq(fit3$n, 10) [1] TRUE > aeq(fit3$time, sort(unique(test2$stop))) [1] TRUE > aeq(fit3$n.risk, c(2,3,5,4,4,5,2,1)) [1] TRUE > aeq(fit3$n.event,c(1,1,1,1,1,2,0,0)) [1] TRUE > aeq(fit3$surv[fit3$n.event>0], c(.5, 1/3, 4/15, 1/5, 3/20, 9/100)) [1] TRUE > temp <- with(fit3, n.event/(n.risk * (n.risk - n.event))) > aeq(fit3$std.err, sqrt(cumsum(temp))) [1] TRUE > > # > # Verify that both surv AND n.risk are right between time points. > # > fit <- survfit(Surv(time, status) ~1, test1) > temp <- summary(fit, time=c(.5,1, 1.5, 6, 7.5, 8, 8.9, 9, 10), extend=TRUE) > > aeq(temp$n.risk, c(6,6,4,4,2,2,1,1,0)) [1] TRUE > aeq(temp$surv, c(1, fit$surv[c(1,1,2,2,3,3,4,4)])) [1] TRUE > aeq(temp$n.event, c(0,1,0,2,0,0,0,1,0)) [1] TRUE > aeq(temp$std.err, c(0, (fit$surv*fit$std.err)[c(1,1,2,2,3,3,4,4)])) [1] TRUE > > > fit <- survfit(Surv(start, stop, event) ~1, test2) > temp <- summary(fit, times=c(.5, 1.5, 2.5, 3, 6.5, 14.5, 16.5)) > aeq(temp$surv, c(1, 1, fit$surv[c(1,2,3,6,6)])) [1] TRUE > > # This next fails. With start-stop data the number at risk at intermediate > # endpoints is not known precisely, since the underlying routine does not report > # time points at which only an addition occured. > if (FALSE) aeq(temp$n.risk, c(0, 2, 3, 3, 4, 1,1)) > > # compute conditional survival > fit1 <- survfit(Surv(start, stop, event)~1, test2, weights=wt) > fit2 <- survfit(Surv(start, stop, event)~1, test2, weights=wt, start.time=5) > > aeq(fit1$surv[2], summary(fit1, time=5)$surv) # verify my subscript [1] TRUE > aeq(fit2$surv, fit1$surv[3:8]/fit1$surv[2]) [1] TRUE > aeq(fit2$std.err^2, fit1$std.err[3:8]^2 - fit1$std.err[2]^2) [1] TRUE > aeq(fit2$cumhaz, fit1$cumhaz[3:8] - fit1$cumhaz[2]) [1] TRUE > aeq(fit2$std.chaz^2, fit1$std.chaz[3:8]^2 - fit1$std.chaz[2]^2) [1] TRUE > > # Now with a Cox model > cfit <- coxph(Surv(start, stop, event)~1, test2, weights=wt) > fit1 <- survfit(cfit) > fit2 <- survfit(cfit, start.time=5) > aeq(fit2$surv, fit1$surv[3:8]/fit1$surv[2]) [1] TRUE > aeq(fit2$std.err^2, fit1$std.err[3:8]^2 - fit1$std.err[2]^2) [1] TRUE > aeq(fit2$cumhaz, fit1$cumhaz[3:8] - fit1$cumhaz[2]) [1] TRUE > aeq(fit2$std.chaz^2, fit1$std.chaz[3:8]^2 - fit1$std.chaz[2]^2) [1] TRUE > > > # bigger data set, with covariates and some tied event times > mfit <- coxph(Surv(age, age+futime/12, death) ~ sex + mspike, mgus2) > dummy <- data.frame(sex='F', mspike=1.3) > > msurv1 <- survfit(mfit, newdata=dummy) > msurv2 <- survfit(mfit, newdata=dummy, start.time=80) > j <- max(which(msurv1$time < 80)) > k <- seq(j+1, length(msurv1$time)) > aeq(msurv2$surv, msurv1$surv[k] / msurv1$surv[j]) [1] TRUE > aeq(msurv2$cumhaz, msurv1$cumhaz[k] - msurv1$cumhaz[j]) [1] TRUE > # standard errors now have a term due to vmat(mfit), so don't factor > # into a simple sum > > proc.time() user system elapsed 0.90 0.09 0.93