[R] indexing??
helin_susam
helin.susam at gmail.com
Tue Feb 28 20:42:32 CET 2012
Dear Petr Savicky,
Actually, this is based on jackknife after bootstrap algorithm. In summary,
I have a data set, and I want to compute some values by using this
algorithm.
Firstly, using bootstrap, I create some bootstrap re-samples. This step O.K.
Then, for each data point within these re-samples, I want to get a subset
which do not contain that data point ( this point would be any point of the
original data set), in general, if B is the number of bootstrap-resamples,
there are B/e resamples obtained for each data point. And finally, I want
to calculate some values for each of this re samples.
Explanation of my algorithm;
#My data set: (x and y)
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),] # Here is the bootstrap step
y.s <- t.s[,1]
x.s <- t.s[,2]
z.s <- y.s+x.s
nn <- sum (z.s) # For example, I want to calculate this value
out[[i]] <- list(ff <- (nn), finding=any (y.s==y[j])) # I get the mentioned
subset in here
kk <- sapply(out, function(x) {x$finding})
ff <- out[! kk]
}
}
I obtained the following results of an experiment;
> kk
[1] FALSE TRUE TRUE FALSE TRUE TRUE FALSE TRUE TRUE TRUE
> ff
[[1]]
[[1]][[1]]
[1] 47
[[1]]$finding
[1] FALSE
[[2]]
[[2]][[1]]
[1] 46
[[2]]$finding
[1] FALSE
[[3]]
[[3]][[1]]
[1] 52
[[3]]$finding
[1] FALSE
It is easy to do when "y" contains different elements. "out[[i]] <- list(ff
<- (nn), finding=any (y.s==y[j]))"
But, when y contains the same element, doing this process can be confusing
confusing..
Because, (y <- c(1,1,1,0,0,1,0,1,0,0)) for y[j] when j= 1 there are some
other 1 in the y. Is there something special about the y to an j ?
Thanks
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