Hi all,
To complete the previous suggestions from David and Santiago, the
function makeNodeLabel with its option method = "md5sum" creates node
labels depending only the composition of the tip labels descending from
each node. Then using match(), it is possible to find nodes representing
identical clades in distinct trees.
A small example with n = 4:
## t1 and t2 are the same tree:
R> t1 <- read.tree(text = "((a,b),(c,d));")
R> t2 <- read.tree(text = "((c,d),(a,b));")
## the default of makeNodeLabel(); "Node2" or "Node3" does not
## label the same clade in both trees:
R> write.tree(makeNodeLabel(t1))
[1] "((a,b)Node2,(c,d)Node3)Node1;"
R> write.tree(makeNodeLabel(t2))
[1] "((c,d)Node2,(a,b)Node3)Node1;"
## much longer labels, but they identify the clades:
R> write.tree(makeNodeLabel(t1, method = "md5sum"))
[1]
"((a,b)dd8c6a395b5dd36c56d23275028f526c,(c,d)2611be5ce337526d271d4925a1833b0c)47ece2e49e5c0333677fc34e044d8257;"
R> write.tree(makeNodeLabel(t2, method = "md5sum"))
[1]
"((c,d)2611be5ce337526d271d4925a1833b0c,(a,b)dd8c6a395b5dd36c56d23275028f526c)47ece2e49e5c0333677fc34e044d8257;"
Best,
Emmanuel
Le 14/01/2016 17:11, David Bapst a écrit :
Lev,
This is easy, if the newick strings are structured the same so that
the resulting edge matrix and tip labels are identical. Here's a
worked example:
```
library(ape)
# with edge lengths
newick1<-"(Homo:30,(Echinus:18,(Cephalodiscus:12,(Rhabdopleura:8,(Dictyonema:1,(Rhabdinopora:1,Dicellograptus:3):1):1):4):4):10);"
# with node names
newick2<-"(Homo,(Echinus,(Cephalodiscus,(Rhabdopleura,(Dictyonema,(Rhabdinopora,Dicellograptus)Graptoloida)Eugraptolithina)Graptolithina)Pterobranchia)Ambulacraria)Deuterostomia;"
# convert newick strings to phylo objects
tree1<-read.tree(text=newick1)
plot(tree1)
tree2<-read.tree(text=newick2)
plot(tree2)
nodelabels(tree2$node.label)
# combine
if(identical(tree1$edge,tree2$edge) &
identical(tree1$tip.label,tree2$tip.label)){
tree3<-tree1
tree3$node.label<-tree2$node.label
}
plot(tree3)
nodelabels(tree3$node.label)
```
If $edge and $tip.label are not identical, then this is a somewhat
more complex issue.
Cheers,
-Dave
On Wed, Jan 13, 2016 at 4:39 PM, Yampolsky, Lev <yampo...@mail.etsu.edu> wrote:
Dear Colleagues,
Thank you very much for your help, it’s all clear now. (I got way more
than I asked for, for example, my trees include only extant taxa). And
yes, branching.times() and getBtimes() return exact same numbers of
course, I just didn’t understand ow indexing works.
Next question, a very simple one again.
I have two newick trees, identical topology. In one I have branch lengths,
in the other I have node names. I would like a tree with both. How do I do
this?
Thanks in advance!
--
Lev Yampolsky
Professor
Department of Biological Sciences
East Tennessee State University
Box 70703
Johnson City TN 37614-1710
Cell 423-676-7489
Office/lab 423-439-4359
Fax 423-439-5958
On 1/13/16, 4:20 AM, "Emmanuel Paradis" <emmanuel.para...@ird.fr> wrote:
To pile a little bit more after David's message, ape has the function
ltt.plot.coords which does the same thing than phytools::ltt but with
different options, including 'tol' which specifies the tolerance for
considering a tree as ultrametric or not.
Best,
Emmanuel
Le 13/01/2016 03:29, Liam J. Revell a écrit :
Hi Lev.
You could also look at ltt in phytools. The object returned has the time
of all the events on the tree - including the end of lineages that go
extinct before the present. To access these, you can do:
obj<-ltt(tree,plot=FALSE)
obj$times
obj$ltt
& to see how these are associated with events, you can do:
plot(obj,show.tree=TRUE)
All the best, Liam
Liam J. Revell, Associate Professor of Biology
University of Massachusetts Boston
web: http://faculty.umb.edu/liam.revell/
email: liam.rev...@umb.edu
blog: http://blog.phytools.org
On 1/12/2016 10:22 PM, Dan Rabosky wrote:
Dear Lev-
I don't think branching.times can compute these for non-ultrametric
trees.
You can do this with package BAMMtools, but you need a "hidden"
internal function. You can access it as
"BAMMtools:::NU.branching.times"
It returns branching times relative to the most recently-occurring tip
in the tree. It's a R-based recursion that is a little slower that the
ape function, so it's not recommended as a replacement for
branching.times if you have an ultrametric tree.
I'm not actively maintaining laser, but getBtimes returns the output
of branching.times after sorting the times and stripping out the node
names (this was useful for something many years ago!). If you plot
sort(getBtimes(x)) and sort(branching.times(x)) they should be
identical.
~Dan Rabosky
On Jan 12, 2016, at 7:37 PM, Yampolsky, Lev <yampo...@mail.etsu.edu>
wrote:
Dear Colleagues,
Does anyone know what is the difference between ape�s
branching.times() and laser�s getBtimes()?
And why they may be giving rather different results, particularly for
internal branches? (From an ultrametric tree created by
chronotree <- chronos(tree, lambda = 1, model = "correlated", quiet =
FALSE, calibration = makeChronosCalib(tree), control =
chronos.control())
Thank you very much in advance for your help!
PS. A related but less important question: I am curious how does
branching.times() calculate branching times from a non-ultrametric
tree?
--
Lev Yampolsky
Professor
Department of Biological Sciences
East Tennessee State University
Box 70703
Johnson City TN 37614-1710
Cell 423-676-7489
Office/lab 423-439-4359
Fax 423-439-5958
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_____________________
Dan Rabosky
Assistant Professor & Curator of Herpetology
Museum of Zoology &
Department of Ecology and Evolutionary Biology
University of Michigan
Ann Arbor, MI 48109-1079 USA
drabo...@umich.edu
http://www-personal.umich.edu/~drabosky
http://www.lsa.umich.edu/ummz/
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