Getting Robinson-Foulds distances for a set of trees

The phangorn function RF.dist - for computing the Robinson-Foulds distance between two trees - is very fast. However if we want to compare a set of trees stored in an object of class "multiPhylo" there is some opportunity to recycle internal calculations & thus improve on just calling RF.dist n × (n - 1) / 2 times for n trees. Specifically, here is some code that scales to multiple trees a little bit better than n × (n - 1) / 2 RF.dist by calling the ape function prop.part (which finds all bipartitions in the tree) only once for each tree:

multiRF<-function(trees){
 if(class(trees)!="multiPhylo")
  stop("trees should be an object of class \"multiPhylo\"")
 N<-length(trees)
 RF<-matrix(0,N,N)
 if(any(sapply(unclass(trees),is.rooted))){
  cat("Some trees are rooted. Unrooting all trees.\n")
  trees<-lapply(unclass(trees),unroot)
 }
 foo<-function(pp) lapply(pp,function(x,pp)
  sort(attr(pp,"labels")[x]),pp=pp)
 xx<-lapply(unclass(trees),function(x) foo(prop.part(x)))
 for(i in 1:(N-1)) for(j in (i+1):N)
  RF[i,j]<-RF[j,i]<-2*sum(!xx[[i]]%in%xx[[j]])
 RF
}

It works OK compared to calling RF.dist a whole bunch of times independently. First, here is an alternate version of the function which uses RF.dist:

multiRF2<-function(trees){
 N<-length(trees)
 RF<-matrix(0,N,N)
 for(i in 1:(N-1)) for(j in (i+1):N)
  RF[i,j]<-RF[j,i]<-RF.dist(trees[[i]],trees[[j]])
 RF
}

Now, a very simple comparison with random trees:

> N<-100
> n<-100
> trees<-rmtree(N,n,rooted=FALSE)
> system.time(X1<-multiRF(trees))
   user  system elapsed
   9.16    0.03    9.19
> system.time(X2<-multiRF2(trees))
   user  system elapsed
  19.98    0.00   19.99
> all(X1==X2)
[1] TRUE