yet another solution (I think)
> 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)
> indx <- embed(rev(seq_along(exmpl)), length(patrn))
> matches <- apply(indx, 1, function(.indx){
+ all(exmpl[.indx] == patrn)
+ })
> indx[matches, 1L]
[1] 23 13 6
>
On Wed, Feb 15, 2012 at 6:32 PM, Redding, Matthew
<[email protected]> wrote:
> 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: [email protected]
>>[mailto:[email protected]] On Behalf Of Berend Hasselman
>>Sent: Thursday, 16 February 2012 1:35 AM
>>To: Martin Morgan
>>Cc: [email protected]
>>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
>>
>>______________________________________________
>>[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
>>and provide commented, minimal, self-contained, reproducible code.
>>
>>
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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