this is ugly, but...
l <-length(patrn)
l2 <-length(exmpl)
out <- vector("list")
for(i in 1:(l2-l+1))
{
exmpl[i:(i+l-1)]
patrn==exmpl[i:(i+l-1)]
if(all(patrn==exmpl[i:(i+l-1)]))
{ out[[i]] <- i } else { out[[i]] <- "NA"}
}
out <- do.call(c, out)
as.numeric(out[which(out!="NA")])
## Cheers and HTH
Redding, Matthew-2 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
>
> Suggestions very much appreciated!
>
> Kind regards,
>
>
>
>
> Matt Redding, Ph.D.
> Principal Scientist
> Geochemist/Soil Chemist
> Queensland Primary Industries & Fisheries
> DEEDI
> PO Box 102, Toowoomba, 4350, Qld
> ph: 0746 881372
> fax: 0746 881192
>
>
> ********************************DISCLAIMER**************...{{dropped:15}}
>
> ______________________________________________
> R-help@ mailing list
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> 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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