Hi Frank,

It seems that you can use the (apparently not used a lot) option from makeNodeLabel(, method = "md5sum") which creates node labels with the MD5SUM algorithm using the tip labels descending from each node (considering the tree as rooted). The result is, for each node, a label that is unique for a given set of tip labels (even among different trees).


Taking your 1st tree:

R> tr <- read.tree(text = "(A:0.1, (B:0.2, (C:0.3, D:0.4)100:0.5)95:0.55);")
R> BP <- tr$node.label
R> trbis <- makeNodeLabel(tr, "md5sum")
R> names(BP) <- trbis$node.label

BP is a vector which you can index with labels built in the same way from another tree.

Cheers,

Emmanuel

PS: if the labelled topologies are identical for all trees as in your example below, then the node labels will be ordered in the same way in all trees and the above procedure is not needed.

Le 16/12/2016 à 17:57, Frank T Burbrink a écrit :
Hi Everyone,

Thank you Jacob, Keith and Emmanuel for your responses. What I am trying to do 
is a little different than what the solutions posted here would do. I am not 
important a distribution of trees, but rather a single tree with the support 
values already included. So imagine something like comparing support for three 
trees (the first two are bootstraps is from bootstraps, the third is Pp):

  1.  (A:0.1, (B:0.2, (C:0.3, D:0.4)100:0.5)95:0.55);
  2.  (A:0.1, (B:0.2, (C:0.3, D:0.4)60:0.5)20:0.55);
  3.  (A:0.1, (B:0.2, (C:0.3, D:0.4)1:0.5)0.8:0.55);

This would show that support for clade BCD would be 95, 20, 0.8 in each of the 
three trees and support for CD would 100, 60 and 1.0 respectively.

Is there a method that could read the trees with support, find the shared 
clades, and generate a table by node for all three support values?


Thanks!

Frank

Frank T. Burbrink, Ph.D.
Associate Curator
Department of Herpetology
American Museum of Natural History
Central Park West at 79th Street
New York, NY 10024-5192

fburbr...@amnh.org<mailto:fburbr...@amnh.org>

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