Re: [R] Finding overlaps in vector

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Sat, 22 Dec 2007 11:33:02 -0500

If you want indexes, i.e. 1, 2, 3, ... instead of the values in v you can still use split -- just split on seq_along(v) instead of v (or if v had names you might want to split along names(v)):

split(seq_along(v), ct)

and if you only want to retain groups with 2+ elements then you can just Filter then out:

twoplus <- function(x) length(x) >= 2
Filter(twoplus, split(seq_along(v), ct))

On Dec 22, 2007 5:12 AM, Johannes Graumann <johannes_graumann_at_web.de> wrote:
> But cutree does away with the indexes from the original input, which
> rect.hclust retains.
> I will have no other choice and match that input with the 'values' contained
> in the clusters ...
>
> Joh
>
>
> Gabor Grothendieck wrote:
>
> > 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
> >> >>
> >
>
> > ______________________________________________
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>
> ______________________________________________
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>



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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 22 Dec 2007 - 16:40:17 GMT

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