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...
Martin
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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