I believe the original poster was looking for runs of consecutive
values. Here's a generalization of Tony's solution:
findlong = function(seq){
rr = rle(seq)
lens = rr$length
lens[rr$value == FALSE] = 0
ll = which.max(lens)
start = cumsum(c(1,rr$length))[ll]
list(start=start,length=rr$lengths[ll])
}
sq <- c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1)
Then
findlong(diff(sq) == 1) # starts at position 1, run of 3
$start
[1] 1
$length
[1] 3
findlong(diff(sq) == -1) # starts at position 8, run of 5
$start
[1] 8
$length
[1] 5
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
[EMAIL PROTECTED]
On Tue, 13 May 2008, Tony Plate wrote:
If the increases or decreases could be any size, rle(sign(diff(x))) could do
it:
x <- c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1)
r <- rle(sign(diff(x)))
r
Run Length Encoding
lengths: int [1:5] 3 2 2 5 4
values : num [1:5] 1 0 1 -1 0
i1 <- which(r$lengths==max(r$lengths[r$values==1]) & r$values==1)[1]
i2 <- which(r$lengths==max(r$lengths[r$values==-1]) & r$values==-1)[1]
i1
[1] 1
i2
[1] 4
rbind(up=c(start=cumsum(c(1, r$lengths))[i1], len=r$lengths[i1]),
down=c(start=cumsum(c(1, r$lengths))[i2], len=r$lengths[i2]))
start len
up 1 3
down 8 5
Ingmar Visser wrote:
rle(diff(sq)) could be helpful here,
best, Ingmar
On May 13, 2008, at 11:19 PM, Marko Milicic wrote:
Hi all R helpers,
I'm trying to comeup with nice and elegant way of "detecting" consecutive
increases/decreases in the sequence of numbers. I'm trying with
combination
of which() and diff() functions but unsuccesifuly.
For example:
sq <- c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1);
I'd like to find way to calculate
a) maximum consecutive increase = 3 (from 1 to 4)
b) maximum consecutive decrease = 5 (from 6 to 1)
All ideas are highly welcomed!
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