Here is a solution to identify the unique topologies in your dataset and
then average some quantity over them. Let us know if it works:
http://blog.phytools.org/2016/05/identifying-unique-tree-topologies-in.html
All the best, Liam
Liam J. Revell, Associate Professor of Biology
University of Massachusetts Boston
web: http://faculty.umb.edu/liam.revell/
email: [email protected]
blog: http://blog.phytools.org
On 5/30/2016 1:52 PM, John Denton wrote:
Hi Liam,
Thanks for this. Yes, it looked back at the script and it is actually a data
frame. I realized my question wasn't clear. Th second column has the actual
trees as each entry (not just A, B, C, etc), which I suppose makes it more
complicated--isn't an RF.dist/multiRF call needed?
Best,
~John
John S. S. Denton, Ph.D.
Department of Vertebrate Paleontology
American Museum of Natural History
www.johnssdenton.com
________________________________________
From: Liam J. Revell [[email protected]]
Sent: Monday, May 30, 2016 12:53 PM
To: John Denton; [email protected]
Subject: Re: [R-sig-phylo] identify duplicate topologies and add corresponding
values
Hi John.
This would have to be a data frame (not a matrix) as you could not have
different data types in your matrix. Try the function aggregate, e.g.:
obj<-data.frame(number=c(1,1.5,0.2,0.8,1.0,3.2),
tree=factor(c("A","A","B","C","A","B")))
obj ## data frame
aggregate(obj$number,by=list(obj$tree),sum)
- Liam
Liam J. Revell, Associate Professor of Biology
University of Massachusetts Boston
web: http://faculty.umb.edu/liam.revell/
email: [email protected]
blog: http://blog.phytools.org
On 5/30/2016 12:13 PM, John Denton wrote:
Hi all,
I have a two-column matrix, containing trees in one column and a numeric value
associated with each tree in the other column. I would like to go through the
list to identify any duplicated topologies, and for each set of duplicates sum
the corresponding numeric values and prune away the duplicate trees.
For example,
1.0 A
1.5 A
0.2 B
0.8. C
1.0. A
3.2. B
Would become
3.5. A
3.4. B
0.8. C
Any help would be great!
Thanks,
~John
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