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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