options(na.action=na.exclude) # preserve missings options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type library(survival) # # The test data set from Turnbull, JASA 1974, 169-73. # # status 0=right censored # 1=exact # 2=left censored # aeq <- function(x,y, ...) all.equal(as.vector(x), as.vector(y), ...) turnbull <- data.frame( time =c( 1,1,1, 2,2,2, 3,3,3, 4,4,4), status=c( 1,0,2, 1,0,2, 1,0,2, 1,0,2), n =c(12,3,2, 6,2,4, 2,0,2, 3,3,5)) # # Compute the K-M for the Turnbull data # via a slow EM calculation # emsurv <- function(time, status, wt, verbose=T) { left.cen <- (status==2) if (!any(left.cen)) stop("No left censored data!") if (!any(status==1))stop("Must have some exact death times") tempy <- Surv(time[!left.cen], status[!left.cen]) ww <- wt[!left.cen] tempx <- factor(rep(1, sum(!left.cen))) tfit <- survfit(tempy~tempx, weights=ww) if (verbose) cat("Iteration 0, survival=", format(round(tfit$surv[tfit$n.event>0],3)), "\n") stimes <- tfit$time[tfit$n.event>0] ltime <- time[left.cen] lwt <- wt[left.cen] tempx <- factor(rep(1, length(stimes) + sum(!left.cen))) tempy <- Surv(c(time[!left.cen], stimes), c(status[!left.cen], rep(1, length(stimes)))) for (iter in 1:4) { wt2 <- stimes*0 ssurv <- tfit$surv[tfit$n.event>0] sjump <- diff(c(1, ssurv)) for (j in 1:(length(ltime))) { k <- sum(ltime[j]>=stimes) #index of the death time if (k==0) stop("Left censored observation before the first death") wt2[1:k] <- wt2[1:k] + lwt[j]*sjump[1:k] /(ssurv[k]-1) } tfit <- survfit(tempy~tempx, weights=c(ww, wt2)) if (verbose) { cat("Iteration", iter, "survival=", format(round(tfit$surv[tfit$n.event>0],3)), "\n") cat(" weights=", format(round(wt2,3)), "\n") } } survfit(tempy ~ tempx, weights=c(ww, wt2), robust=FALSE) } temp <-emsurv(turnbull$time, turnbull$status, turnbull$n) print(summary(temp)) # First check, use the data from Turnbull, JASA 1974, 169-173. tdata <- data.frame(time =c(1,1,1,2,2,2,3,3,3,4,4,4), status=rep(c(1,0,2),4), n =c(12,3,2,6,2,4,2,0,2,3,3,5)) tfit <- survfit(Surv(time, time, status, type='interval') ~1, tdata, weights=n) all.equal(round(tfit$surv,3), c(.538, .295, .210, .095)) # Second check, compare to a reversed survival curve # This is not as simple a test as one might think, because left and right # censored observations are not treated symmetrically by the routine: # time <= y for left and time> y for right (this is to make the routine # correct for the common situation of panel data). # To get equivalence, make the left censoreds happen just a little bit # earlier. The left-continuous/right-continuous shift is also a bother. # 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)) fit1 <- survfit(Surv(time, status) ~1, test1) temp <- ifelse(test1$status==0, 4.99,5) - test1$time fit2 <- survfit(Surv(temp, status, type='left') ~1, test1) all.equal(round(fit1$surv[1:2],5), round(1-fit2$surv[3:2],5)) rm(tdata, tfit, fit1, temp, fit2) # # Create a data set similar to the one provided by Al Zinsmeister # It is a hard test case for survfit.turnbull # time1 <- c(rep(0,100), rep(1,200), 100, 200, 210, 220, rep(365,100), rep(366,5), 731:741) time2 <- c((1:100)*3, 10+1:100, rep(365:366, c(60,40)), NA, 500, NA, 450, rep(730,90), rep(NA,10), c(528,571,691,730,731), NA, 1095:1099, NA, 1400, 1200, 772, 1461) zfit <- survfit(Surv(time1, time2, type='interval2') ~1) # # There are 100 intervals of the form (0,x) where x is from 3 to 300, # and 200 more of the form (1,x) where x is from 11 to 366. These # lead to a mass point in the interval (1,3), which is placed at 2. # The starting estimate has far too little mass placed here, and it takes # the EM a long time to realize that most of the weight for the first 300 # subjects goes here. With acceleration, it takes 16 iterations, without # it takes >40. (On Al's orginal data, without accel still wasn't there after # 165 iters!) # # The next 4 obs give rise to potential jumps at 100.5, 200.5, 211.5, and # 221. However, the final estimate has no mass at all on any of these. # Assume mass of a,b, and c at 2, 100.5 and 365.5, and consider the # contributions: # 123 obs that overlap a only # 137 obs that overlap a and b # 40 obs that overlap a, b, c # 1 obs that overlap b, c # 108 obs that overlap c (200, 210,200, 365, and 366 starting points) # For some trial values of a,b,c, compare the loglik to that of (a+b),0,c # First one: a^123 (a+b)^137 (a+b+c)^40 (b+c) c^108 # Second: (a+b)^123 (a+b)^137 (a+b+c)^40 c c^108 # Likelhood improves if (1 + b/a)^123 > 1+ b/c, which is true for almost # all a and c. In particular, at the solution a and c are approx .7 and # .18, respectively. # # The program can't see this coming, of course, and so iterates towards a # KM with epsilon sized jumps at 100.5, 200.5, and 211.5. Whether these # intervals should be removed during iteration, as detected, is an open # question for me. # # # True solution: mass points at 2, 365.5, 408, and 756.5, of sizes a, b, c, d # Likelihood: a^260 (a+b)^40 (b+c)^92 (b+c+d)^12 c^5 d^11 # Solution: a=0.6958, b=0.1674, c=0.1079, d=0.0289 tfun <- function(x) { if (length(x) ==3) x <- c(x, .03) x <- x/sum(x) #make probabilities sum to 1 loglik <- 260*log(x[1]) + 40*log(x[1]+x[2]) + 92*log(x[2] + x[3]) + 12*log(x[2]+x[3]+x[4]) + 5*log(x[3]) + 11*log(x[4]) -loglik #find the max, not the min } nfit <- nlminb(start=c(.7,.15, .1), tfun, lower=0, upper=1) nparm <- c(nfit$par, .03) nparm <- nparm / sum(nparm) zparm <- -diff(c(1, zfit$surv[match(c(2, 365.5, 408, 756.5), zfit$time)])) aeq(round(tfun(nparm),4), round(tfun(zparm),4)) # .0001 is the tolerance in survfit.turnbull rm(tfun, nfit, nparm, zparm, time1, time2, zfit)