Thank you very much David - R is so rich, the easy way can be hard to find.
Just to close this out for others, the final solution I used was:
Peak2Return <- function(v) {
S <- cummax(v)
L <- which((v ==S) & (diff(c(0,v)<0))
R <- sapply(v[L], function(x,S) {which(x < S)[1]; }, S)
now you have L for the left index, and R for the corresponding right
index. If there is no right index due to the curve, the R value is NA.
On 11/24/2011 7:35 AM, David Winsemius wrote:
On Nov 24, 2011, at 4:52 AM, Scott Tetrick wrote:
So I have a problem that I'm trying to get through, and I just can't
seem to get it to run very fast in R.
What I'm trying to do is to find in a vector a local peak, then the
next time that value is crossed later. I don't care about peaks that
may be lower than this first one - they can be ignored. I've tried
some sapply methods along the way, but they all are slower. The best
solution I have is a loop, and I just know there are smart R folks
that could help me eliminate it.
It looks as though you are reinventing hte function:
?cummax
Peak2Return <- function(v) {
Q <- (1:m)[diff(v)<0] ; find all the peaks
L <- Q[c(TRUE,v[Q[-1]] > v[Q[-length(Q)]])]
;
eliminate lower peaks
R <- sapply(L,function (x,v) { ((x+1):length(v))[v[x] <
v[(x+1):m]][1]; }, v)
;
find the next crossing
out <- data.frame(peak=L,Return=R)
out
}
Thanks in advance!
David Winsemius, MD
West Hartford, CT
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