Does the following do what you want:
rseq <- function(n=1, length.=2){
s1 <- sample(x=length., size=n, replace=TRUE)
s2 <- sample(x=length., size=n, replace=TRUE)
ranseq <- array(0, dim=c(n, length.))
for(i in 1:n)
ranseq[i, s1[i]:s2[i]] <- 1
ranseq
}
set.seed(1)
rseq(9, 5)
> set.seed(1)
> rseq(9, 5)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 0 0 0
[2,] 0 1 0 0 0
[3,] 1 1 1 0 0
[4,] 0 0 0 1 1
[5,] 0 1 0 0 0
[6,] 0 0 0 1 1
[7,] 0 0 1 1 1
[8,] 0 0 0 1 0
[9,] 0 0 0 1 1
>
hope this helps. spencer graves
Jim Lemon wrote:
On Tuesday 29 June 2004 01:48 pm, Steve S wrote:
Dear All,
I wonder if there is a probability distribution where you can specify when
a certain event start and finish within a fixed period? For example I might
specify the number of period to be 5, and a random vector from this
distribution might give me:
0 1 1 1 0
where 1 is always adjacent to each other?
This can never happen: 0 0 1 0 1 for example.
Well, I'll have a go. Let's call it the start-finish distribution. We have a
p (period) and d (duration). As there must be an "off" observation (otherwise
we don't know the duration), It's fairly easy to enumerate the outcomes for a
given period:
d start(s) finish(f) count
1 1:n-1 2:n n-1
2 1:n-2 3:n n-2
...
n-1 1 n-1 1
Assuming that all outcomes are equally likely, the total number of outcomes
is:
n(n-1)/2
thus the probability of a given d occurring is:
P[d|n] = 2(n-d)/n(n-1)
The probabilities of s and f over all d are inverse over the values k in 1:n
P[s=k|n] = (n-k-1)/(n-1)
P[f=k|n] = (k-1)/(n-1)
giving, I think, a monotonic function for s and f.
My apology for this strange question!
My apology if this is no use at all.
Jim
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