[R] I think not so hard question
Liaw, Andy
andy_liaw at merck.com
Thu Jan 12 20:41:29 CET 2006
Mark,
It's a bit unclear whether you're looking for a run of exactly 3, or at
least 3. If it's exactly 3, what Prof. Koenker suggested (using rle())
should help you a lot. If you want runs of at least 3, it should work
similarly as well.
> set.seed(1)
> (x <- sample(c(-1, 1), 100, replace=TRUE))
[1] -1 -1 1 1 -1 1 1 1 1 -1 -1 -1 1 -1 1 -1 1 1 -1 1 1 -1 1
-1 -1 -1
[27] -1 -1 1 -1 -1 1 -1 -1 1 1 1 -1 1 -1 1 1 1 1 1 1 -1 -1 1
1 -1 1
[53] -1 -1 -1 -1 -1 1 1 -1 1 -1 -1 -1 1 -1 -1 1 -1 1 -1 1 -1 -1 -1
1 1 -1
[79] 1 1 -1 1 -1 -1 1 -1 1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 1 1
> r <- rle(x)$length
> p <- which(r == 3)
> idx <- cumsum(r)[p] - 2
> sapply(idx, function(i) x[i:(i+2)])
[,1] [,2] [,3] [,4]
[1,] -1 1 -1 -1
[2,] -1 1 -1 -1
[3,] -1 1 -1 -1
> idx
[1] 10 35 62 73
Andy
From: Mark Leeds
>
> I'm sorry to bother this list so much
> But I haven't programmed in
> A while and I'm struggling.
>
> I have a vector in R of 1's and -1's
> And I want to use a streak of size Y
> To predict that the same value will
> Be next.
>
> So, suppose Y = 3. Then, if there is a streak of three
> ones in a row, then I will predict that the next value is
> a 1. But, if there is a streak of 3 -1's in a row,
> then I will predict that a -1 is next. Otherwise,
> I don't predict anything.
>
> I am really new to R and kind of struggling
> And I was wondering if someone could show
> how to do this ?
>
> In other words, given a vector of -1's
> And 1's, I am unable ( I've been trying
> For 2 days ) to create a new vector that
> Has the predictions in it at the appropriate
> places ? Thanks.
>
>
>
>
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