Dear Petr Pikal and Petr Savicky thank you for your replies..
If the y vector contains different elements my algorithm gives this result;
y <- c(1,2,3,4,5,6,7,8,9,10)
x <- c(1,0,0,1,1,0,0,1,1,0)
n <- length(x)
t <- matrix(cbind(y,x), ncol=2)
z = x+y
for(j in 1:length(x)) {
out <- vector("list", )
for(i in 1:10) {
t.s <- t[sample(n,n,replace=T),]
y.s <- t.s[,1]
x.s <- t.s[,2]
z.s <- y.s+x.s
out[[i]] <- list(ff <- (z.s), finding=any (y.s==y[j]))
kk <- sapply(out, function(x) {x$finding})
ff <- out[! kk]
}
}
> kk
[1] TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE
> ff
[[1]]
[[1]][[1]]
[1] 5 7 3 2 2 6 7 2 6 6
[[1]]$finding
[1] FALSE
[[2]]
[[2]][[1]]
[1] 7 10 6 2 2 2 6 6 9 3
[[2]]$finding
[1] FALSE
Here, the two situations are FALSE, that is 5th and 10th bootstrap
re-samples do not contain one (or more) element(s) of original vector ("y").
How can I get the similar result when the y vector includes the only
response variable (1 or 0) ? That is
y <- c(1,1,1,0,0,1,0,1,0,0)
Many thanks.
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