In the documentation to the new function plotTree.boxplot
I noted:
“\code{x} can also be supplied as a formula, though in that case the factor levels need to be provided in the cladewise order of the tips in \code{tree}.”
What I didn't note is that if this is not done, then the result will be
incorrect without any error being reported. This is because the
formula
provided by the user is passed directly to the function
boxplot
without any feedback to the tree plotter.
What I realized is that one thing that I could do was first run
boxplot.formula
with plot=FALSE
which
(invisibly) returns a list containing the order of the plotted boxes. Then
I could pass that argument to plotTree
using tips
.
Finally, I could re-run boxplot
with plot=TRUE
. If
the order of the factor in formula
is a valid cladewise order of
the tips of the tree then the tree & boxplot will plot fine. If it is not,
say if we randomize the levels of our factor, then the edges of the tree
will cross when plotting.
Here is what I mean.
Here's our tree & data applied using the basic method:
library(phytools)
tree
##
## Phylogenetic tree with 26 tips and 25 internal nodes.
##
## Tip labels:
## Z, Y, X, W, V, U, ...
##
## Rooted; includes branch lengths.
x[1:30] ## first 30 elements of x
## Z Z Z Z Z
## 1.481718482 1.999288805 1.244922289 1.465959575 1.439767680
## Y Y Y Y Y
## 1.014757614 1.937999703 0.728102930 0.690196494 -0.081170095
## X X X X X
## 1.506780336 1.680023687 1.337899661 0.300120161 0.706118301
## W W W W W
## -0.615363928 1.062724934 -0.556182889 -1.070714742 -1.271925923
## V V V V V
## -1.228228110 -0.957836899 -0.617736401 0.300541684 -0.161967102
## U U U U U
## 1.722133482 1.169447096 0.624685958 2.154831186 0.009577268
plotTree.boxplot(tree,x) ## basic method
Now let's try the formula method in which our factor is in the order of our tree's tip labels:
spp<-factor(names(x),levels=reorder(tree)$tip.label)
plotTree.boxplot(tree,x~spp)
Next, we can flip the order of our factor levels. This is still a valid cladewise order so this should work fine:
spp<-factor(names(x),levels=tree$tip.label[Ntip(tree):1])
spp
## [1] Z Z Z Z Z Y Y Y Y Y X X X X X W W W W W V V V V V U U U U U T T T T T
## [36] S S S S S R R R R R Q Q Q Q Q P P P P P O O O O O N N N N N M M M M M
## [71] L L L L L K K K K K J J J J J I I I I I H H H H H G G G G G F F F F F
## [106] E E E E E D D D D D C C C C C B B B B B A A A A A
## Levels: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
plotTree.boxplot(tree,x~spp)
Ladderized - also a valid cladewise order:
spp<-factor(names(x),untangle(ladderize(tree),"read.tree")$tip.label)
spp
## [1] Z Z Z Z Z Y Y Y Y Y X X X X X W W W W W V V V V V U U U U U T T T T T
## [36] S S S S S R R R R R Q Q Q Q Q P P P P P O O O O O N N N N N M M M M M
## [71] L L L L L K K K K K J J J J J I I I I I H H H H H G G G G G F F F F F
## [106] E E E E E D D D D D C C C C C B B B B B A A A A A
## Levels: V U W S R T Q P L K J I M O N Y X Z B A C E D H G F
plotTree.boxplot(tree,x~spp)
(Note that we didn't ladderize our input tree for
plotTree.boxplot
- however the plotted tree will be ladderized.)
Finally, randomize them. This will not (generally speaking) result in a valid cladewise order:
spp<-factor(names(x),levels=sample(tree$tip.label))
spp
## [1] Z Z Z Z Z Y Y Y Y Y X X X X X W W W W W V V V V V U U U U U T T T T T
## [36] S S S S S R R R R R Q Q Q Q Q P P P P P O O O O O N N N N N M M M M M
## [71] L L L L L K K K K K J J J J J I I I I I H H H H H G G G G G F F F F F
## [106] E E E E E D D D D D C C C C C B B B B B A A A A A
## Levels: N L D C Q F J E T V A G P O B Y U K M S R H X W Z I
plotTree.boxplot(tree,x~spp)
This makes it much easier to see if we have a problem using the
formula
method.
The update can be seen here.
There was an error in this demo, which I just fixed.
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