On Wed, Feb 15, 2012 at 10:26:44AM +0100, Berend Hasselman wrote: > > On 15-02-2012, at 05:17, Redding, Matthew wrote: > > > Hi All, > > > > > > I've been trawling through the documentation and listserv archives on this > > topic -- but > > as yet have not found a solution. I'm sure this is pretty simple with R, > > but I cannot work out how without > > resorting to ugly nested loops. > > > > As far as I can tell, grep, match, and %in% are not the correct tools. > > > > Question: > > given these vectors -- > > patrn <- c(1,2,3,4) > > exmpl <- c(3,3,4,2,3,1,2,3,4,8,8,23,1,2,3,4,4,34,4,3,2,1,1,2,3,4) > > > > how do I get the desired answer by finding the occurence of the pattern and > > returning the starting indices: > > 6, 13, 23 > > > > patrn.rev <- rev(patrn) > w <- embed(exmpl,length(patrn)) > w.pos <- apply(w,1,function(r) all(r == patrn.rev)) > which(w.pos)
Hi. If the speed is an issue and exmpl is long, the following modification may be faster. patrn.rev <- rev(patrn) w <- embed(exmpl,length(patrn)) which(rowSums(w == rep(patrn.rev, each=nrow(w))) == ncol(w)) [1] 6 13 23 For length(patrn) = 11 and length(exmpl) = 10000, i obtained a speed up by a factor of 10. Hope this helps. How large are the vectors "patrn" and "exmpl" in your application? Petr Savicky. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
