El 02/06/10 15:14, Joris Meys escribió:
> Hi,
>
> with a little hack you can use the function cophenetic.phylo from ape.
> You just set all branch lengths to 1 :
>
> require(ape)
>
> >tree <- rtree(5,rooted=F)
> >n <- length(tree$edge.length)
> >tree$edge.length <- rep(1,n)
> >cophenetic.phylo(tree)
>
> t3 t1 t2 t4 t5
> t3 0 3 3 3 3
> t1 3 0 2 4 4
> t2 3 2 0 4 4
> t4 3 4 4 0 2
> t5 3 4 4 2 0
>
> Cheers
>
>
> On Wed, Jun 2, 2010 at 2:47 PM, Arnau Mir Torres <arnau....@uib.es
> <mailto:arnau....@uib.es>> wrote:
>
> Hello.
>
> I'd like to compute the associate vector of distances between
> leaves in a binary non-rooted tree. The definition of a distance
> between two leaves in a binary non-rooted tree is the number of
> edges in the path joining the two leaves.
> I've tried the ape package but I'm unable to find this vector.
> For example, using rtree(5,rooted=F) I've obtained the following tree:
>
> $edge
> [,1] [,2]
> [1,] 6 7
> [2,] 7 1
> [3,] 7 8
> [4,] 8 2
> [5,] 8 3
> [6,] 6 4
> [7,] 6 5
>
> $tip.label
> [1] "t4" "t3" "t2" "t1" "t5"
>
> $edge.length
> [1] 0.9126727 0.2765674 0.4996832 0.7904400 0.8508797 0.8174133
> 0.9027958
>
> $Nnode
> [1] 3
>
>
> My question is: how to compute the vector of distances between the
> 5 leaves. This vector is in this case:
>
> v=(d(t1,t2),d(t1,t3),d(t1,t4),d(t1,t5),d(t2,t3),d(t2,t4),d(t2,t5),d(t3,t4),d(t3,t5),d(t4,t5))=(4,4,3,2,2,3,4,3,4,3).
>
>
> Thanks in advance,
>
> Arnau.
> ------------------------------------------------------------
> Arnau Mir Torres
> Edifici A. Turmeda
> Campus UIB
> Ctra. Valldemossa, km. 7,5
> 07122 Palma de Mca.
> tel: (+34) 971172987
> fax: (+34) 971173003
> email: arnau....@uib.es <mailto:arnau....@uib.es>
> URL: http://dmi.uib.es/~arnau <http://dmi.uib.es/%7Earnau>
> ------------------------------------------------------------
>
>
>
>
>
>
>
>
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>
> --
> Joris Meys
> Statistical Consultant
>
> Ghent University
> Faculty of Bioscience Engineering
> Department of Applied mathematics, biometrics and process control
>
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Thank you very much. It has been very useful.
I'd like to go one step further.
Would it be possible to obtain a "distance matrix" but with no
permutation of the leaves? In the above example, the "distance matrix"
must be transformed into:
t1 t2 t3 t4 t5
t1 0 2 3 4 4
t2 2 0 3 4 4
t3 3 3 0 3 3
t4 4 4 3 0 2
t5 4 4 3 2 0
Thanks,
Arnau.
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