I guess I am coming a bit late to this but here is a one-line zoo solution. The lapply creates a zoo object from each foo component with the constant value 1 and the contents as the times of that object. Merging the objects together using cbind gives columns of 1s and NAs and removing the NAs leaves a zoo object at only the intersection of times, which we then extract.
library(zoo) time(na.omit(do.call(cbind, lapply(foo, zoo, x = 1)))) # 2, 3 On 4/8/06, Andy Bunn <[EMAIL PROTECTED]> wrote: > > The original post is ambiguous: do you want to find the intersection or do > > you want to find whether a prespecified set is in the > > intersection? Patrick > > provided you an answer to the latter while you provided an answer to the > > former. Actually, I thought using table as you did (mod the need for no > > replicates) was clever. A more direct but I think considerably slower > > approach would be to use intersect() in a loop: > > > > inall<-intersect(foo[[1]],foo[[,2]]) > > for(i in seq(3, to=length(foo))inall<-intersect(inall,foo[[i]]) > > > > I suspect you already thought of this and rejected it. Other than > > transparency, I think the only advantage it has is that it will work for > > something other than lists of numerics, e.g. it will work for lists of > > factors, which the table() solution would not. > > > Indeed, the first post was ambiguous. I was after the type of solution that I > posted: the intersection rather than a prespecified set. Given that, I > suppose my use of table is satisfactory. > > Thanks Patrick and Bert. > > -Andy > > ______________________________________________ > [email protected] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
