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
Hi.
If the pattern is not too long, try
m <- length(patrn)
n <- length(exmpl)
ind <- seq.int(length=n-m+1)
occur <- rep(TRUE, times=n-m+1)
for (i in seq.int(length=m)) {
occur <- occur & (patrn[i] == exmpl[ind + i - 1])
}
which(occur)
[1] 6 13 23
Hope this helps.
Petr Savicky.
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