tests/expected.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$hazard[1], -(116/365.25)*log(1-.00169))
all.equal(zz$days, c(116,84))

        
#
# 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'))
#older style call
exp1b <- survexp(~ratetable(year=temp2, age=(temp2-temp1), sex=1, race='white'),
		    ratetable=survexp.usr, times=c(366, 1827, 3653, 4383))

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$hazard)))
aeq(exp1$surv, exp1b$surv)

# 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)))


# 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)))

#
# 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(~ratetable(year=temp2, age=(temp2-temp1), sex=temp3),
		       times=c(366, 1827, 3653, 4383))
exp2 <- survexp(~ratetable(year=temp2, age=(temp2-temp1), sex=temp3) + I(1:6),
			times=c(366, 1827, 3653, 4383))
exp3 <- exp2$surv
for (i in 1:length(temp1)){
    exp3[,i] <- survexp(~ratetable(year=temp2, age=(temp2-temp1), sex=temp3),
                        times=c(366, 1827, 3653, 4383), subset=i)$surv
    }

             
print(aeq(exp2$surv, exp3))
print(all.equal(exp1$surv, apply(exp2$surv, 1, mean)))

# 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))))
    }

#
# Check that adding more time points doesn't change things
#
exp4 <- survexp(~ratetable(year=temp2, age=(temp2-temp1), sex=temp3) + I(1:6),
		times=sort(c(366, 1827, 3653, 4383, 30*(1:100))))
aeq(exp4$surv[match(exp2$time, exp4$time),], exp2$surv)

exp4 <- survexp(~ratetable(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)])


#
# 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 ~ ratetable(year=temp2, age=(temp2-temp1), sex=1),
		times=xtime, conditional=F)
exp2 <- survexp(~ratetable(year=temp2, age=(temp2-temp1), sex=1) + I(1:6),
			times=futime)

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$surv[match(futime, xtime)]
aeq(exp1$surv[match(futime, xtime)], cumprod(con))


#
# Now for the conditional method
#
exp1 <- survexp(futime ~ ratetable(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))
cumprod(con)

#
# 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 ~ ratetable(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

#
# Now test the direct-adjusted curve (Ederer)
#
efit <- survexp( ~ ratetable(x=x), 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

all.equal(as.vector(efit$surv), as.vector(apply(direct,1,mean)))  #direct

# Check out the "times" arg of survexp
efit2 <- survexp( ~ ratetable(x=x), dummy, ratetable=fit, se=F,
                  times=c(.5, 2, 3.5,6))
aeq(efit2$surv, c(1, efit$surv[c(2,2,3)]))

#
# 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 ~ ratetable(x=x), 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)))

#
#  Now do the conditional estimate
#
efit <- survexp( time ~ ratetable(x=x), 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)))

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survival documentation built on Aug. 24, 2021, 5:06 p.m.