Small modification below.
Peter Ehlers wrote:
Nick Fankhauser wrote:
I'm using functions to return a matrix of all permutations of a
specified size from a larger list for predictor selection.
For each predictor size I use a seperate function like this:
bag2 <- function(n) {
rl <- c()
for (i1 in seq(n)) {
for (i2 in seq(n)) {
if (length(unique(c(i1,i2)))==1) {next}
rl <- cbind(rl,matrix(c(i1,i2)))
}
}
rl
}
bag3 <- function(n) {
rl <- c()
for (i1 in seq(n)) {
for (i2 in seq(n)) {
for (i3 in seq(n)) {
if (length(unique(c(i1,i2,i3)))==1) {next}
rl <- cbind(rl,matrix(c(i1,i2,i3)))
}
}
}
rl
}
But I think it should be somehow possible in R to use one general
function for all sizes. Can someone help?
I don't exactly know how this kind of permutation is called, maybe this
would help to find a solution.
Nick
Looks to me like you're just doing expand.grid, but without
the 'all variables equal' cases. If that's correct and if
n > 1, then this might do it:
bag <- function(n,K){
L <- vector(mode='list', length=K)
L <- lapply(L, function(x) {x <- seq_len(n)})
d <- expand.grid(L)
keeprow <- apply(d, 1, function(x) {var(x)!=0})
## I'm blanking out on a better way to test for equal values
keeprow <- apply(d, 1, function(x) {length(unique(x)) > 1})
is probably more efficient and certainly more transparent.
-Peter Ehlers
td <- t(as.matrix(d[keeprow,]))
dimnames(td) <- c(NULL, NULL)
rtd <- td[nrow(td):1,]
rtd
}
all.equal(bag(5,2), bag2(5))
#[1] TRUE
all.equal(bag(9,3), bag3(9))
#[1] TRUE
-Peter Ehlers
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--
Peter Ehlers
University of Calgary
403.202.3921
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.