matching a sequence in a vector?
Thankyou all for your great, and creative solutions. There is definately more than one way to skin a cat. A colleague alerted me to another solution: seq.strt <- which( sapply( 1:(length(exmpl)-length(patrn)+1), function(i) isTRUE( all.equal( patrn, exmpl[ i + 0:(length(patrn)-1) ] ) ) ) ) Apparently this came from a post in the help archive that my searches missed. I think the solutions you have put up are more readable. Kind regards Matt
-----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Berend Hasselman Sent: Thursday, 16 February 2012 1:35 AM To: Martin Morgan Cc: r-help at r-project.org Subject: Re: [R] matching a sequence in a vector? 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
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