On 15-02-2012, at 15:27, Martin Morgan wrote: > On 02/14/2012 11:45 PM, Petr Savicky wrote: >> On Wed, Feb 15, 2012 at 02:17:35PM +1000, 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 > > match(exmpl, patrn) returns indexes that differ by 1 if the sequence patrn > occurs > > n = length(patrn) > r = rle(diff(match(exmpl, patrn)) == 1) > > we're looking for a run of TRUE's of length 3, and can find their ends (of > the runs of diffs) as cumsum(r$length) > > cumsum(r$length)[r$values & r$length == (n - 1)] - (n - 2) > > Seems like there could be edge cases that I'm missing...
Clever. However it is quite slow. Berend ______________________________________________ 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.
