Description Usage Arguments Details Value Examples

This function provides a means to specify the prior for the location of shifts across the phylogeny. Certain combinations are not allowed. For example, a maximum shift number of Inf on one branch cannot be combined with a maximum shift number of 1 on another. Thus, bmax must be either a vector of 0's and Inf's or a vector of 0's and 1's. Also, if bmax == 1, then all probabilities must be equal, as bayou cannot sample unequal probabilities without replacement.

1 2 3 |

`sb` |
A vector giving the branch numbers (for a post-ordered tree) |

`ntips` |
The number of tips in the phylogeny |

`bmax` |
A single integer or a vector of integers equal to the number of branches in the phylogeny indicating the maximum number of shifts allowable in the phylogeny. Can take values 0, 1 and Inf. |

`prob` |
A single value or a vector of values equal to the number of branches in the phylogeny indicating the probability that a randomly selected shift will lie on this branch. Can take any positive value, values need not sum to 1 (they will be scaled to sum to 1) |

`log` |
A logical indicating whether the log probability should be returned. Default is 'TRUE' |

`k` |
The number of shifts to randomly draw from the distribution |

`dsb`

calculates the probability of a particular arrangement of shifts for a given set of assumptions.

The log density of the particular number and arrangement of shifts.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | ```
n=10
tree <- sim.bdtree(n=n)
tree <- reorder(tree, "postorder")
nbranch <- 2*n-2
sb <- c(1,2, 2, 3)
# Allow any number of shifts on each branch, with
# probability proportional to branch length
dsb(sb, ntips=n, bmax=Inf, prob=tree$edge.length)
# Disallow shifts on the first branch, returns -Inf
# because sb[1] = 1
dsb(sb, ntips=n, bmax=c(0, rep(1, nbranch-1)),
prob=tree$edge.length)
# Set maximum number of shifts to 1, returns -Inf
# because two shifts are on branch 2
dsb(sb, ntips=n, bmax=1, prob=1)
#Generate a random set of k branches
rsb(5, ntips=n, bmax=Inf, prob=tree$edge.length)
``` |

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