331 lines
11 KiB
Plaintext
331 lines
11 KiB
Plaintext
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R Under development (unstable) (2023-01-09 r83585) -- "Unsuffered Consequences"
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Copyright (C) 2023 The R Foundation for Statistical Computing
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Platform: x86_64-pc-linux-gnu (64-bit)
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R is free software and comes with ABSOLUTELY NO WARRANTY.
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You are welcome to redistribute it under certain conditions.
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Type 'license()' or 'licence()' for distribution details.
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R is a collaborative project with many contributors.
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Type 'contributors()' for more information and
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'citation()' on how to cite R or R packages in publications.
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Type 'demo()' for some demos, 'help()' for on-line help, or
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'help.start()' for an HTML browser interface to help.
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Type 'q()' to quit R.
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> options(na.action=na.exclude) # preserve missings
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> options(contrasts=c('contr.treatment', 'contr.poly')) #ensure constrast type
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> library(survival)
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>
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> # Tests of expected survival
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> aeq <- function(x,y) all.equal(as.vector(x), as.vector(y))
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> #
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> # This makes several scripts easier
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> #
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> mdy.Date <- function(m, d, y) {
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+ y <- ifelse(y<100, y+1900, y)
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+ as.Date(paste(m,d,y, sep='/'), "%m/%d/%Y")
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+ }
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>
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> # This function takes a single subject and walks down the rate table
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> # Input: the vector of starting points, futime, and a ratetable
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> # Output: the full history of walking through said table. Let n= #unique
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> # rates that were used
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> # cell = n by #dims of the table: index of the table cell
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> # days = time spent in cell
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> # hazard= accumulated hazard = days * rate
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> # This does not do date or factor conversions -- start has to be numeric
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> #
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> ratewalk <- function(start, futime, ratetable=survexp.us) {
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+ if (!is.ratetable(ratetable)) stop("Bad rate table")
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+ ratedim <- dim(ratetable)
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+ nvar <- length(ratedim)
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+ if (length(start) != nvar) stop("Wrong length for start")
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+ if (futime <=0) stop("Invalid futime")
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+
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+ attR <- attributes(ratetable)
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+ discrete <- (attR$type ==1) #discrete categories
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+
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+ maxn <- sum(!discrete)*prod(ratedim[!discrete]) #most cells you can hit
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+ cell <- matrix(0, nrow=maxn, ncol=nvar)
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+ days <- hazard <- double(maxn)
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+
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+ eps <- 1e-8 #Avoid round off error
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+ n <- 0
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+ while (futime >0) {
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+ n <- n+1
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+ #what cell am I in?
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+ # Note that at the edges of the rate table, we use the edge: if
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+ # it only goes up the the year 2000, year 2000 is used for any
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+ # dates beyond. This effectively eliminates one boundary
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+ cell[n,discrete] <- start[discrete]
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+ edge <- futime #time to nearest edge, or finish
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+ for (j in which(!discrete)) {
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+ indx <- sum(start[j] >= attR$cutpoints[[j]]-eps)
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+ cell[n, j] <- max(1, indx)
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+ if (indx < ratedim[j])
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+ edge <- min(edge, (attR$cutpoints[[j]])[indx+1] - start[j])
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+ }
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+ days[n] <- edge #this many days in the cell
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+ # using a matrix as a subscript is so handy sometimes
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+ hazard[n] <- edge * (as.matrix(ratetable))[cell[n,,drop=F]]
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+ futime <- futime - edge #amount of time yet to account for
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+ start[!discrete] <- start[!discrete] + edge #walk forward in time
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+ }
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+ list(cell=cell[1:n,], days=days[1:n], hazard=hazard[1:n])
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+ }
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>
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> # Simple test of ratewalk: 20 years old, start on 7Sep 1960
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> # 116 days at the 1960, 20 year old male rate, through the end of the day
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> # on 12/31/1960, then 84 days at the 1961 rate.
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> # The decennial q for 1960 males is .00169.
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> zz <- ratewalk(c(20.4*365.25, 1, as.Date("1960/09/07")), 200)
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> all.equal(zz$hazard[1], -(116/365.25)*log(1-.00169))
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[1] TRUE
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> all.equal(zz$days, c(116,84))
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[1] TRUE
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>
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>
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> #
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> # Simple case 1: a single male subject, born 1/1/36 and entered on study 1/2/55
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> #
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> # Compute the 1, 5, 10 and 12 year expected survival
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>
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> temp1 <- mdy.Date(1,1,36)
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> temp2 <- mdy.Date(1,2,55)
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> exp1 <- survexp(~1, ratetable=survexp.usr,times=c(366, 1827, 3653, 4383),
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+ rmap= list(year=temp2, age=(temp2-temp1), sex=1, race='white'))
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>
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> t12 <- as.numeric(temp2-temp1) # difftimes are a PITA
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> h1 <- ratewalk(c(t12, 1, 1, temp2), 366, survexp.usr)
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> h2 <- ratewalk(c(t12, 1, 1, temp2), 1827, survexp.usr)
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> h3 <- ratewalk(c(t12, 1, 1, temp2), 3653, survexp.usr)
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> h4 <- ratewalk(c(t12, 1, 1, temp2), 4383, survexp.usr)
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>
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> aeq(-log(exp1$surv), c(sum(h1$hazard), sum(h2$hazard), sum(h3$hazard),
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+ sum(h4$hazard)))
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[1] TRUE
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>
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> # pyears should give the same result
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> dummy <- data.frame(time = 4383,
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+ year=temp2, sex = 1, age= temp2-temp1, race="white")
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> cuts <- tcut(0, c(0, 366, 1827, 3653, 4383))
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> exp1c <- pyears(time ~ cuts, data=dummy, ratetable=survexp.usr)
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> aeq(exp1$surv, exp(-cumsum(exp1c$expected)))
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[1] TRUE
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>
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>
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> # Just a little harder:
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> # Born 3/1/25 and entered the study on 6/10/55. The code creates shifted
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> # dates to align with US rate tables - entry is 59 days earlier (days from
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> # 1/1/1925 to 3/1/1925).
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> #
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> temp1 <- mdy.Date(3,1,25)
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> temp2 <- mdy.Date(6,10,55)
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> exp1 <- survexp(~1, ratetable=survexp.usr,times=c(366, 1827, 3653, 4383),
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+ rmap= list(year=temp2, age=(temp2-temp1), sex=2, race='black'))
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>
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> tyear <- temp2 - 59
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> t12 <- as.numeric(temp2-temp1)
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> h1 <- ratewalk(c(t12, 2, 2, tyear), 366, survexp.usr)
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> h2 <- ratewalk(c(t12, 2, 2, tyear), 1827, survexp.usr)
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> h3 <- ratewalk(c(t12, 2, 2, tyear), 3653, survexp.usr)
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> h4 <- ratewalk(c(t12, 2, 2, tyear), 4383, survexp.usr)
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>
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> aeq(-log(exp1$surv), c(sum(h1$hazard), sum(h2$hazard), sum(h3$hazard),
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+ sum(h4$hazard)))
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[1] TRUE
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>
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> #
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> # Simple case 2: make sure that the averages are correct, for Ederer method
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> #
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> # Compute the 1, 5, 10 and 12 year expected survival
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>
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> temp1 <- mdy.Date(1:6,6:11,1890:1895)
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> temp2 <- mdy.Date(6:1,11:6,c(55:50))
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> temp3 <- c(1,2,1,2,1,2)
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> age <- temp2 - temp1
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>
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> exp1 <- survexp(~1, rmap= list(year=temp2, age=(temp2-temp1), sex=temp3),
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+ times=c(366, 1827, 3653, 4383))
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> exp2 <- survexp(~ I(1:6),
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+ rmap= list(year=temp2, age=(temp2-temp1), sex=temp3),
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+ times=c(366, 1827, 3653, 4383))
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> exp3 <- exp2$surv
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> for (i in 1:length(temp1)){
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+ exp3[,i] <- survexp(~ 1,
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+ rmap = list(year=temp2, age=(temp2-temp1), sex=temp3),
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+ times=c(366, 1827, 3653, 4383), subset=i)$surv
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+ }
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>
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>
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> print(aeq(exp2$surv, exp3))
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[1] TRUE
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> print(all.equal(exp1$surv, apply(exp2$surv, 1, mean)))
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[1] TRUE
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>
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> # They agree, but are they right?
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> #
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> for (i in 1:length(temp1)) {
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+ offset <- as.numeric(temp1[i] - mdy.Date(1,1, 1889+i))
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+ tyear = temp2[i] - offset
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+ haz1 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 366)
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+ haz2 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 1827)
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+ haz3 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 3653)
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+ haz4 <- ratewalk(c(as.numeric(temp2-temp1)[i], temp3[i], tyear), 4383)
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+ print(aeq(-log(exp2$surv[,i]), c(sum(haz1$hazard), sum(haz2$hazard),
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+ sum(haz3$hazard), sum(haz4$hazard))))
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+ }
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[1] TRUE
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[1] TRUE
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[1] TRUE
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[1] TRUE
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[1] TRUE
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[1] TRUE
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>
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> #
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> # Check that adding more time points doesn't change things
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> #
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> exp4 <- survexp(~ I(1:6),
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+ rmap= list(year=temp2, age=(temp2-temp1), sex=temp3),
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+ times=sort(c(366, 1827, 3653, 4383, 30*(1:100))))
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> aeq(exp4$surv[match(exp2$time, exp4$time),], exp2$surv)
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[1] TRUE
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>
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> exp4 <- survexp(~1,
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+ rmap = list(year=temp2, age=(temp2-temp1), sex=temp3),
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+ times=sort(c(366, 1827, 3653, 4383, 30*(1:100))))
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> aeq(exp1$surv, exp4$surv[match(exp1$time, exp4$time, nomatch=0)])
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[1] TRUE
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>
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>
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> #
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> # Now test Hakulinen's method, assuming an analysis date of 3/1/57
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> #
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> futime <- mdy.Date(3,1,57) - temp2
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> xtime <- sort(c(futime, 30, 60, 185, 365))
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>
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> exp1 <- survexp(futime ~ 1, rmap= list(year=temp2, age=(temp2-temp1), sex=1),
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+ times=xtime, conditional=F)
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> exp2 <- survexp(~ I(1:6), times=futime,
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+ rmap= list(year=temp2, age=(temp2-temp1), sex=1))
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>
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> wt <- rep(1,6)
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> con <- double(6)
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> for (i in 1:6) {
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+ con[i] <- sum(exp2$surv[i,i:6])/sum(wt[i:6])
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+ wt <- exp2$surv[i,]
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+ }
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>
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> exp1$surv[match(futime, xtime)]
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[1] 0.9557362 0.9285840 0.9025661 0.8774220 0.8532489 0.8297416
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> aeq(exp1$surv[match(futime, xtime)], cumprod(con))
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[1] TRUE
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>
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>
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> #
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> # Now for the conditional method
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> #
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> exp1 <- survexp(futime ~ 1, rmap= list(year=temp2, age=(temp2-temp1), sex=1),
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+ times=xtime, conditional=T)
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>
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> cond <- exp2$surv
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> for (i in 6:2) cond[i,] <- (cond[i,]/cond[i-1,]) #conditional survival
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> for (i in 1:6) con[i] <- exp(mean(log(cond[i, i:6])))
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>
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> all.equal(exp1$surv[match(futime, xtime)], cumprod(con))
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[1] TRUE
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> cumprod(con)
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[1] 0.9556656 0.9284398 0.9023612 0.8771798 0.8529944 0.8294940
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>
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> #
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> # Test out expected survival, when the parent pop is another Cox model
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> #
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> test1 <- data.frame(time= c(4, 3,1,1,2,2,3),
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+ status=c(1,NA,1,0,1,1,0),
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+ x= c(0, 2,1,1,1,0,0))
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>
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> fit <- coxph(Surv(time, status) ~x, test1, method='breslow')
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>
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> dummy <- data.frame(time=c(.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5),
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+ status=c(1,0,1,0,1,0,1,1,1), x=(-4:4)/2)
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>
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> efit <- survexp(time ~ 1, rmap= list(x=x), dummy, ratetable=fit, cohort=F)
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>
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> #
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> # Now, compare to the true answer, which is known to us
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> #
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> ss <- exp(fit$coef)
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> haz <- c( 1/(3*ss+3), 2/(ss+3), 1) #truth at time 0,1,2,4+
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> chaz <- cumsum(c(0,haz))
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> chaz2 <- chaz[c(1,2,2,3,3,3,3,4,4)]
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>
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> risk <- exp(fit$coef*dummy$x)
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> efit2 <- exp(-risk*chaz2)
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>
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> all.equal(as.vector(efit), as.vector(efit2)) #ignore mismatched name attrib
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[1] TRUE
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>
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> #
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> # Now test the direct-adjusted curve (Ederer)
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> #
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> efit <- survexp( ~ 1, dummy, ratetable=fit, se=F)
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> direct <- survfit(fit, newdata=dummy, censor=FALSE)$surv
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>
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> chaz <- chaz[-1] #drop time 0
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> d2 <- exp(outer(-chaz, risk))
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> all.equal(as.vector(direct), as.vector(d2)) #this tests survfit
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[1] TRUE
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>
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> all.equal(as.vector(efit$surv), as.vector(apply(direct,1,mean))) #direct
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[1] TRUE
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>
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> # Check out the "times" arg of survexp
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> efit2 <- survexp( ~1, dummy, ratetable=fit, se=F,
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+ times=c(.5, 2, 3.5,6))
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> aeq(efit2$surv, c(1, efit$surv[c(2,2,3)]))
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[1] TRUE
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>
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> #
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> # Now test out the Hakulinen method (Bonsel's method)
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> # By construction, we have a large correlation between x and censoring
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> #
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> # In theory, hak1 and hak2 would be the same. In practice, like a KM and
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> # F-H, they differ when n is small.
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> #
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> efit <- survexp( time ~1, dummy, ratetable=fit, se=F)
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>
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> surv <- wt <- rep(1,9)
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> tt <- c(1,2,4)
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> hak1 <- hak2 <- NULL
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> for (i in 1:3) {
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+ wt[dummy$time < tt[i]] <- 0
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+ hak1 <- c(hak1, exp(-sum(haz[i]*risk*surv*wt)/sum(surv*wt)))
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+ hak2 <- c(hak2, sum(exp(-haz[i]*risk)*surv*wt)/sum(surv*wt))
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+ surv <- surv * exp(-haz[i]*risk)
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+ }
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>
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> all.equal(as.vector(efit$surv), as.vector(cumprod(hak1)))
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[1] TRUE
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>
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> #
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> # Now do the conditional estimate
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> #
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> efit <- survexp( time ~ 1, dummy, ratetable=fit, se=F,
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+ conditional=T)
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> wt <- rep(1,9)
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> cond <- NULL
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> for (i in 1:3) {
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+ wt[dummy$time < tt[i]] <- 0
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+ cond <- c(cond, exp(-sum(haz[i]*risk*wt)/sum(wt)))
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+ }
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>
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> all.equal(as.vector(efit$surv), as.vector(cumprod(cond)))
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[1] TRUE
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>
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> proc.time()
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user system elapsed
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1.083 0.077 1.150
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