prodlim (version 1.1.3)

jackknife: Jackknife pseudo-values based on the product-limit method

Description

Jackknife pseudo-values based on the Kaplan-Meier and Aalen Johannsen estimates.

Usage

jackknife(object, times, ...)

Arguments

object

times

...

Value

Examples

## Survival
dd <- prodlim:::SimSurv(100)
ff <- prodlim(Hist(time,status)~1,data=dd)
jj <- jackknife.survival(ff,times=c(10,30,100))

dd <- prodlim:::SimSurv(1000)
library(pseudo)
system.time(jack <- with(dd,pseudosurv(time,status,c(10,30,100))))
system.time({ff <- prodlim(Hist(time,status)~1,data=dd);jack2 <- jackknife(ff,times=c(10,30,100))})
all(round(jack2[,1],9)==round(jack[[3]],9))

## CompRisk
ddd <- data.frame(time=1:100,event=rbinom(100,2,.5),X=rbinom(100,1,.5))
fff <- prodlim(Hist(time,event)~1,data=ddd)
fk <- leaveOneOut.competing.risks(fff,times=c(10,30,100,500))
jk <- jackknife.competing.risks(fff,times=c(10,30,100))

library(prodlim)
library(pseudo)
ddd <- data(sampleCompriskFrame)
fff <- prodlim(Hist(time,event)~1,data=ddd)
fff7 <- prodlim(Hist(time,event)~1,data=ddd[-7,])
fff1 <- prodlim(Hist(time,event)~1,data=ddd[-1,])
ttt <- ddd$time
p1 <- predict(fff1,times=ttt)
p7 <- predict(fff7,times=ttt)
summary(fff1,times=1:10,cause=1)
sf <- leaveOneOut.competing.risks(fff,times=ttt)
cbind(loo=sf[7,],true=p7)
cbind(loo=sf[1,],true=p1)
jack <- with(ddd,pseudoci(time,event,ttt))[,-c(1:2,seq(4,22,2))]
jack2 <- jackknife.competing.risks(fff,times=ttt)
all(sapply(1:10,function(x){
a <- round(jack[,x],4)
b <- round(jack2[,x],4)
# all(a[!is.na(a)]==b[!is.na(b)])
all(a[!is.na(a)]==b[!is.na(a)])
}))


N <- 2000
ddd <- data.frame(time=1:N,event=rbinom(N,2,.5),X=rbinom(N,1,.5))
ttt <- c(3,5,10)
# ttt <- ddd$time
fff <- prodlim(Hist(time,event)~1,data=ddd)
system.time(jack <- with(ddd,pseudoci(time,event,ttt))[,-c(1:2,seq(4,N+2,2))])
system.time({jack2 <- jackknife.competing.risks(fff,times=ttt)})
all(round(jack2[,1],9)==round(jack[,1],9))

all(sapply(1:N,function(x){
a <- round(jack[x,],4)
b <- round(jack2[x,],4)
# all(a[!is.na(a)]==b[!is.na(b)])
all(a[!is.na(a)]==b[!is.na(a)])

}<keyword>~kwd1</keyword>
<keyword>~kwd2</keyword>