Finding overlaps in vector

If we don't need any plotting we don't really need rect.hclust at
all.  Split the output of cutree, instead.  Continuing from the
prior code:

for(el in split(unname(vv), names(vv))) print(el)

[1] 0.00 0.45
[1] 1
[1] 2
[1] 3.00 3.25 3.33 3.75 4.10
[1] 5
[1] 6.00 6.45
[1] 7.0 7.1
[1] 8

On Dec 21, 2007 3:24 PM, Johannes Graumann <johannes_graumann at web.de>
wrote:

Hm, hm, rect.hclust doesn't accept "plot=FALSE" and cutree doesn't retain
the indexes of membership ... anyway short of ripping out the guts of
rect.hclust to achieve the same result without an active graphics device?

Joh

# cluster and plot
hc <- hclust(dist(v), method = "single")
plot(hc, lab = v)
cl <- rect.hclust(hc, h = .5, border = "red")

# each component of list cl is one cluster.  Print them out.
for(idx in cl) print(unname(v[idx]))

[1] 8
[1] 7.0 7.1
[1] 6.00 6.45
[1] 5
[1] 3.00 3.25 3.33 3.75 4.10
[1] 2
[1] 1
[1] 0.00 0.45

# a different representation of the clusters
vv <- v
names(vv) <- ct <- cutree(hc, h = .5)
vv

   1    1    2    3    4    4    4    4    4    5    6    6    7    7  
    8
0.00 0.45 1.00 2.00 3.00 3.25 3.33 3.75 4.10 5.00 6.00 6.45 7.00 7.10
8.00


On Dec 21, 2007 4:56 AM, Johannes Graumann <johannes_graumann at web.de>
wrote:

<posted & mailed>

Dear all,

I'm trying to solve the problem, of how to find clusters of values in
a vector that are closer than a given value. Illustrated this might
look as follows:

vector <- c(0,0.45,1,2,3,3.25,3.33,3.75,4.1,5,6,6.45,7,7.1,8)

When using '0.5' as the proximity requirement, the following groups
would result:
0,0.45
3,3.25,3.33,3.75,4.1
6,6.45
7,7.1

Jim Holtman proposed a very elegant solution in
http://tolstoy.newcastle.edu.au/R/e2/help/07/07/21286.html, which I
have modified and perused since he wrote it to me. The beauty of this
approach is that it will not only work for constant proximity
requirements as above, but also for overlap-windows defined in terms
of ppm around each value. Now I have an additional need and have found
no way (short of iteratively step through all the groups returned) to
figure out how to do that with Jim's approach: how to figure out that
6,6.45 and 7,7.1 are separate clusters?

Thanks for any hints, Joh