On Feb 7, 2011, at 6:43 PM, B77S wrote:
So, after thinking about this a bit, I realized that the previous
solution
wasn't exactly what I needed. I really needed replacement=F and to
be able
to choose any sample size (n.sample) less than or equal to the site
(row)
with the lowest total abundance.
The reason I suggested , replace =FALSE, is that I thought those
were population parameters. Furthermore, even if we think of them as
samples, it seems unlikely that they are the entire universe for
inference, since knowing such a universe would make statistics
superfluous. My advice is to consult a statistician before you set
replace=FALSE.
Anyway, I think this works. Forgive me if I have misunderstood
something
regarding the previous solutions output. I do not pretend to be
intelligent.
Better watch out.... amusing utterances such as that are fortune-fodder.
Cheers!
############### start function ###############
RAND_L <- function(L.matrix, n.sample){
mainout <- vector("list")
for(i in 1:nrow(L.matrix)){
## decomposes species (1:ncol(L.matrix)) into a list of counts per
each
out<- vector("list")
for(j in 1:length(L.matrix[i,])){
out[[j]] <- rep(names(L.matrix[i,])[j], L.matrix[i,j])
}
## puts previous loop products (counts) in a row
out2 <- vector()
for(k in 1:length(out)){
out2 <- append(out2, as.character(unlist(out[k])))
}
out3<- sample(out2, n.sample, replace=F)
mainout[[i]] <- out3
mainout[[i]] <- factor(mainout[[i]], levels= colnames(L.matrix))
}
finalout <- t(sapply(mainout, table))
rownames(finalout)<-rownames(L.matrix)
return(finalout)
}
################### end function ##################
RAND_L(abund2, 100)
spA spB spC spD spa spF spG
site1 11 12 18 8 0 24 27
site2 24 24 0 0 27 25 0
site3 0 0 6 38 0 0 56
site4 27 20 0 0 16 37 0
--
View this message in context:
http://r.789695.n4.nabble.com/Subsampling-out-of-site-abundance-matrix-tp3263148p3265402.html
David Winsemius, MD
West Hartford, CT
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