Description Usage Arguments Details Value Author(s) See Also Examples
root reroots a phylogenetic tree with respect to the specified
outgroup or at the node specified in node.
unroot unroots a phylogenetic tree, or returns it unchanged if
it is already unrooted.
is.rooted tests whether a tree is rooted.
1 2 3 |
phy |
an object of class |
outgroup |
a vector of mode numeric or character specifying the new outgroup. |
node |
alternatively, a node number where to root the tree. |
resolve.root |
a logical specifying whether to resolve the new root as a bifurcating node. |
interactive |
if |
The argument outgroup can be either character or numeric. In
the first case, it gives the labels of the tips of the new outgroup;
in the second case the numbers of these labels in the vector
phy$tip.label are given.
If outgroup is of length one (i.e., a single value), then the
tree is rerooted using the node below this tip as the new root.
If outgroup is of length two or more, the most recent common
ancestor (MRCA) is used as the new root. Note that the tree is really
unrooted before being rerooted, so that if outgroup is already
the outgroup, then the returned tree is not the same than the original
one (see examples). If outgroup is not monophyletic, the
operation fails and an error message is issued.
If resolve.root = TRUE, root adds a zero-length branch
below the MRCA of the ingroup.
A tree is considered rooted if either only two branches connect to the
root, or if there is a root.edge element. In all other cases,
is.rooted returns FALSE.
an object of class "phylo" for root and unroot; a
single logical value for is.rooted.
Emmanuel Paradis
bind.tree, drop.tip,
nodelabels, identify.phylo
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | data(bird.orders)
plot(root(bird.orders, 1))
plot(root(bird.orders, 1:5))
tr <- root(bird.orders, 1)
is.rooted(bird.orders) # yes!
is.rooted(tr) # no!
### This is because the tree has been unrooted first before rerooting.
### You can delete the outgroup...
is.rooted(drop.tip(tr, "Struthioniformes"))
### ... or resolve the basal trichotomy in two ways:
is.rooted(multi2di(tr))
is.rooted(root(bird.orders, 1, r = TRUE))
### To keep the basal trichotomy but forcing the tree as rooted:
tr$root.edge <- 0
is.rooted(tr)
|
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