model
is a character or a matrix, evolution is
simulated with a Markovian model; the transition probabilities are
calculated for each branch with $P = e^{Qt}$ where $Q$ is the
rate matrix given by model
and $t$ is the branch length.
The calculation is done recursively from the root. See Paradis (2006,
p. 101) for a general introduction applied to evolution.rTraitDisc(phy, model = "ER", k = if (is.matrix(model)) ncol(model) else 2,
rate = 0.1, states = LETTERS[1:k], freq = rep(1/k, k),
ancestor = FALSE, root.value = 1, ...)
phy
. If
ancestor = TRUE
, the node labels are used if present,
otherwise, ``Node1'', ``Node2'', etc.model
:k
and rate
are ignored.} ace
: "ER"
is an equal-rates model, "ARD"
is an all-rates-different model, and "SYM"
is a symmetrical
model. Note that the argument rate
must be of the appropriate
length, i.e., 1, $k(k - 1)$, or $k(k - 1)/2$ for the three models,
respectively. The rate matrix $Q$ is then filled column-wise.}
foo(x, l)
where
x
is the trait of the ancestor and l
is the branch
length. It must return the value of the descendant as an integer.}
rTraitCont
, rTraitMult
, ace
data(bird.orders)
### the two followings are the same:
rTraitDisc(bird.orders)
rTraitDisc(bird.orders, model = matrix(c(0, 0.1, 0.1, 0), 2))
### two-state model with irreversibility:
rTraitDisc(bird.orders, model = matrix(c(0, 0, 0.1, 0), 2))
### simple two-state model:
tr <- rcoal(n <- 40, br = runif)
x <- rTraitDisc(tr, ancestor = TRUE)
plot(tr, show.tip.label = FALSE)
nodelabels(pch = 19, col = x[-(1:n)])
tiplabels(pch = 19, col = x[1:n])
### an imaginary model with stasis 0.5 time unit after a node, then
### random evolution:
foo <- function(x, l) {
if (l < 0.5) return(x)
sample(2, size = 1)
}
tr <- rcoal(20, br = runif)
x <- rTraitDisc(tr, foo, ancestor = TRUE)
plot(tr, show.tip.label = FALSE)
co <- c("blue", "yellow")
cot <- c("white", "black")
Y <- x[1:20]
A <- x[-(1:20)]
nodelabels(A, bg = co[A], col = cot[A])
tiplabels(Y, bg = co[Y], col = cot[Y])
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