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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) > > # Tests of expected survival > aeq <- function(x,y) all.equal(as.vector(x), as.vector(y)) > # > # This makes several scripts easier > # > mdy.Date <- function(m, d, y) { + y <- ifelse(y<100, y+1900, y) + as.Date(paste(m,d,y, sep='/'), "%m/%d/%Y") + } > > # This function takes a single subject and walks down the rate table > # Input: the vector of starting points, futime, and a ratetable > # Output: the full history of walking through said table. Let n= #unique > # rates that were used > # cell = n by #dims of the table: index of the table cell > # days = time spent in cell > # hazard= accumulated hazard = days * rate > # This does not do date or factor conversions -- start has to be numeric > # > ratewalk <- function(start, futime, ratetable=survexp.us) { + if (!is.ratetable(ratetable)) stop("Bad rate table") + ratedim <- dim(ratetable) + nvar <- length(ratedim) + if (length(start) != nvar) stop("Wrong length for start") + if (futime <=0) stop("Invalid futime") + + attR <- attributes(ratetable) + discrete <- (attR$type ==1) #discrete categories + + maxn <- sum(!discrete)*prod(ratedim[!discrete]) #most cells you can hit + cell <- matrix(0, nrow=maxn, ncol=nvar) + days <- hazard <- double(maxn) + + eps <- 1e-8 #Avoid round off error + n <- 0 + while (futime >0) { + n <- n+1 + #what cell am I in? + # Note that at the edges of the rate table, we use the edge: if + # it only goes up the the year 2000, year 2000 is used for any + # dates beyond. This effectively eliminates one boundary + cell[n,discrete] <- start[discrete] + edge <- futime #time to nearest edge, or finish + for (j in which(!discrete)) { + indx <- sum(start[j] >= attR$cutpoints[[j]]-eps) + cell[n, j] <- max(1, indx) + if (indx < ratedim[j]) + edge <- min(edge, (attR$cutpoints[[j]])[indx+1] - start[j]) + } + days[n] <- edge #this many days in the cell + # using a matrix as a subscript is so handy sometimes + hazard[n] <- edge * (as.matrix(ratetable))[cell[n,,drop=F]] + futime <- futime - edge #amount of time yet to account for + start[!discrete] <- start[!discrete] + edge #walk forward in time + } + list(cell=cell[1:n,], days=days[1:n], hazard=hazard[1:n]) + } > > # Simple test of ratewalk: 20 years old, start on 7Sep 1960 > # 116 days at the 1960, 20 year old male rate, through the end of the day > # on 12/31/1960, then 84 days at the 1961 rate. > # The decennial q for 1960 males is .00169. > zz <- ratewalk(c(20.4*365.25, 1, as.Date("1960/09/07")), 200) > all.equal(zz$hazard[1], -(116/365.25)*log(1-.00169)) [1] TRUE > all.equal(zz$days, c(116,84)) [1] TRUE > > > # > # Simple case 1: a single male subject, born 1/1/36 and entered on study 1/2/55 > # > # Compute the 1, 5, 10 and 12 year expected survival > > temp1 <- mdy.Date(1,1,36) > temp2 <- mdy.Date(1,2,55) > exp1 <- survexp(~1, ratetable=survexp.usr,times=c(366, 1827, 3653, 4383), + rmap= list(year=temp2, age=(temp2-temp1), sex=1, race='white')) > > t12 <- as.numeric(temp2-temp1) # difftimes are a PITA > h1 <- ratewalk(c(t12, 1, 1, temp2), 366, survexp.usr) > h2 <- ratewalk(c(t12, 1, 1, temp2), 1827, survexp.usr) > h3 <- ratewalk(c(t12, 1, 1, temp2), 3653, survexp.usr) > h4 <- ratewalk(c(t12, 1, 1, temp2), 4383, survexp.usr) > > aeq(-log(exp1$surv), c(sum(h1$hazard), sum(h2$hazard), sum(h3$hazard), + sum(h4$hazard))) [1] TRUE > > # pyears should give the same result > dummy <- data.frame(time = 4383, + year=temp2, sex = 1, age= temp2-temp1, race="white") > cuts <- tcut(0, c(0, 366, 1827, 3653, 4383)) > exp1c <- pyears(time ~ cuts, data=dummy, ratetable=survexp.usr) > aeq(exp1$surv, exp(-cumsum(exp1c$expected))) [1] TRUE > > > # Just a little harder: > # Born 3/1/25 and entered the study on 6/10/55. The code creates shifted > # dates to align with US rate tables - entry is 59 days earlier (days from > # 1/1/1925 to 3/1/1925). > # > temp1 <- mdy.Date(3,1,25) > temp2 <- mdy.Date(6,10,55) > exp1 <- survexp(~1, ratetable=survexp.usr,times=c(366, 1827, 3653, 4383), + rmap= list(year=temp2, age=(temp2-temp1), sex=2, race='black')) > > tyear <- temp2 - 59 > t12 <- as.numeric(temp2-temp1) > h1 <- ratewalk(c(t12, 2, 2, tyear), 366, survexp.usr) > h2 <- ratewalk(c(t12, 2, 2, tyear), 1827, survexp.usr) > h3 <- ratewalk(c(t12, 2, 2, tyear), 3653, survexp.usr) > h4 <- ratewalk(c(t12, 2, 2, tyear), 4383, survexp.usr) > > aeq(-log(exp1$surv), c(sum(h1$hazard), sum(h2$hazard), sum(h3$hazard), + sum(h4$hazard))) [1] TRUE > > # > # Simple case 2: make sure that the averages are correct, for Ederer method > # > # Compute the 1, 5, 10 and 12 year expected survival > > temp1 <- mdy.Date(1:6,6:11,1890:1895) > temp2 <- mdy.Date(6:1,11:6,c(55:50)) > temp3 <- c(1,2,1,2,1,2) > age <- temp2 - temp1 > > exp1 <- survexp(~1, rmap= list(year=temp2, age=(temp2-temp1), sex=temp3), + times=c(366, 1827, 3653, 4383)) > exp2 <- survexp(~ I(1:6), + rmap= list(year=temp2, age=(temp2-temp1), sex=temp3), + times=c(366, 1827, 3653, 4383)) > exp3 <- exp2$surv > for (i in 1:length(temp1)){ + exp3[,i] <- survexp(~ 1, + rmap = list(year=temp2, age=(temp2-temp1), sex=temp3), + times=c(366, 1827, 3653, 4383), subset=i)$surv + } > > > print(aeq(exp2$surv, exp3)) [1] TRUE > print(all.equal(exp1$surv, apply(exp2$surv, 1, mean))) [1] TRUE > > # They agree, but are they right? > # > for (i in 1:length(temp1)) { + offset <- as.numeric(temp1[i] - mdy.Date(1,1, 1889+i)) + tyear = temp2[i] - offset + haz1 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 366) + haz2 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 1827) + haz3 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 3653) + haz4 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 4383) + print(aeq(-log(exp2$surv[,i]), c(sum(haz1$hazard), sum(haz2$hazard), + sum(haz3$hazard), sum(haz4$hazard)))) + } [1] TRUE [1] TRUE [1] TRUE [1] TRUE [1] TRUE [1] TRUE > > # > # Check that adding more time points doesn't change things > # > exp4 <- survexp(~ I(1:6), + rmap= list(year=temp2, age=(temp2-temp1), sex=temp3), + times=sort(c(366, 1827, 3653, 4383, 30*(1:100)))) > aeq(exp4$surv[match(exp2$time, exp4$time),], exp2$surv) [1] TRUE > > exp4 <- survexp(~1, + rmap = list(year=temp2, age=(temp2-temp1), sex=temp3), + times=sort(c(366, 1827, 3653, 4383, 30*(1:100)))) > aeq(exp1$surv, exp4$surv[match(exp1$time, exp4$time, nomatch=0)]) [1] TRUE > > > # > # Now test Hakulinen's method, assuming an analysis date of 3/1/57 > # > futime <- mdy.Date(3,1,57) - temp2 > xtime <- sort(c(futime, 30, 60, 185, 365)) > > exp1 <- survexp(futime ~ 1, rmap= list(year=temp2, age=(temp2-temp1), sex=1), + times=xtime, conditional=F) > exp2 <- survexp(~ I(1:6), times=futime, + rmap= list(year=temp2, age=(temp2-temp1), sex=1)) > > wt <- rep(1,6) > con <- double(6) > for (i in 1:6) { + con[i] <- sum(exp2$surv[i,i:6])/sum(wt[i:6]) + wt <- exp2$surv[i,] + } > > exp1$surv[match(futime, xtime)] [1] 0.9557362 0.9285840 0.9025661 0.8774220 0.8532489 0.8297416 > aeq(exp1$surv[match(futime, xtime)], cumprod(con)) [1] TRUE > > > # > # Now for the conditional method > # > exp1 <- survexp(futime ~ 1, rmap= list(year=temp2, age=(temp2-temp1), sex=1), + times=xtime, conditional=T) > > cond <- exp2$surv > for (i in 6:2) cond[i,] <- (cond[i,]/cond[i-1,]) #conditional survival > for (i in 1:6) con[i] <- exp(mean(log(cond[i, i:6]))) > > all.equal(exp1$surv[match(futime, xtime)], cumprod(con)) [1] TRUE > cumprod(con) [1] 0.9556656 0.9284398 0.9023612 0.8771798 0.8529944 0.8294940 > > # > # Test out expected survival, when the parent pop is another Cox model > # > test1 <- data.frame(time= c(4, 3,1,1,2,2,3), + status=c(1,NA,1,0,1,1,0), + x= c(0, 2,1,1,1,0,0)) > > fit <- coxph(Surv(time, status) ~x, test1, method='breslow') > > dummy <- data.frame(time=c(.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5), + status=c(1,0,1,0,1,0,1,1,1), x=(-4:4)/2) > > efit <- survexp(time ~ 1, rmap= list(x=x), dummy, ratetable=fit, cohort=F) > > # > # Now, compare to the true answer, which is known to us > # > ss <- exp(fit$coef) > haz <- c( 1/(3*ss+3), 2/(ss+3), 1) #truth at time 0,1,2,4+ > chaz <- cumsum(c(0,haz)) > chaz2 <- chaz[c(1,2,2,3,3,3,3,4,4)] > > risk <- exp(fit$coef*dummy$x) > efit2 <- exp(-risk*chaz2) > > all.equal(as.vector(efit), as.vector(efit2)) #ignore mismatched name attrib [1] TRUE > > # > # Now test the direct-adjusted curve (Ederer) > # > efit <- survexp( ~ 1, dummy, ratetable=fit, se=F) > direct <- survfit(fit, newdata=dummy, censor=FALSE)$surv > > chaz <- chaz[-1] #drop time 0 > d2 <- exp(outer(-chaz, risk)) > all.equal(as.vector(direct), as.vector(d2)) #this tests survfit [1] TRUE > > all.equal(as.vector(efit$surv), as.vector(apply(direct,1,mean))) #direct [1] TRUE > > # Check out the "times" arg of survexp > efit2 <- survexp( ~1, dummy, ratetable=fit, se=F, + times=c(.5, 2, 3.5,6)) > aeq(efit2$surv, c(1, efit$surv[c(2,2,3)])) [1] TRUE > > # > # Now test out the Hakulinen method (Bonsel's method) > # By construction, we have a large correlation between x and censoring > # > # In theory, hak1 and hak2 would be the same. In practice, like a KM and > # F-H, they differ when n is small. > # > efit <- survexp( time ~1, dummy, ratetable=fit, se=F) > > surv <- wt <- rep(1,9) > tt <- c(1,2,4) > hak1 <- hak2 <- NULL > for (i in 1:3) { + wt[dummy$time < tt[i]] <- 0 + hak1 <- c(hak1, exp(-sum(haz[i]*risk*surv*wt)/sum(surv*wt))) + hak2 <- c(hak2, sum(exp(-haz[i]*risk)*surv*wt)/sum(surv*wt)) + surv <- surv * exp(-haz[i]*risk) + } > > all.equal(as.vector(efit$surv), as.vector(cumprod(hak1))) [1] TRUE > > # > # Now do the conditional estimate > # > efit <- survexp( time ~ 1, dummy, ratetable=fit, se=F, + conditional=T) > wt <- rep(1,9) > cond <- NULL > for (i in 1:3) { + wt[dummy$time < tt[i]] <- 0 + cond <- c(cond, exp(-sum(haz[i]*risk*wt)/sum(wt))) + } > > all.equal(as.vector(efit$surv), as.vector(cumprod(cond))) [1] TRUE > > proc.time() user system elapsed 1.53 0.10 1.64