xlist <-list()
for (i in 2:length(x.s)){
x.seq <- embed(length(x.s):1, i)
xlist[[i]] <- table(apply(x.seq, 1, function(z){
paste(x.s[z], collapse=":")
}))
}
xlist
--
David Winsemius
On Apr 18, 2009, at 11:46 AM, Albert Vilella wrote:
that works very well. how d
that works very well. how do I store the results into a variable instead of
doing a print?
On Fri, Apr 17, 2009 at 5:51 PM, jim holtman wrote:
> How about this:
>
> > x <- "A00096:A00096:A00096:A00096:A02178:A02178:A07776"
> > x.s <- unlist(strsplit(x, ":"))
> > for (i in 2:length(x.s)){
> +
How about this:
> x <- "A00096:A00096:A00096:A00096:A02178:A02178:A07776"
> x.s <- unlist(strsplit(x, ":"))
> for (i in 2:length(x.s)){
+ x.seq <- embed(length(x.s):1, i)
+ print(table(apply(x.seq, 1, function(z){
+ paste(x.s[z], collapse=":")
+ })))
+ }
A00096:A00096 A00096:A
Starting by the first entry:
A00096:A00096:A00096:A00096:A02178:A02178:A07776
and supposing there aren't any other subvectors identical in the set, the
algorithm will slide through the vector, first in pairs, then in trios, then
in sets of four, etc, and count the occurrences:
A00096:A00096
3
A00
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