# [R] Finding overlaps in vector

Gabor Grothendieck ggrothendieck at gmail.com
Fri Dec 21 17:09:48 CET 2007

```This may not be as direct as Jim's in terms of specifying granularity but
will uses conventional hierarchical clustering to create the clusters and also
draws a nice dendrogram for you.   I have split the dendrogram at a
height of 0.5
to define the clusters but you can change that to whatever granularity you like:

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