SplitStatus {Quartet}R Documentation

Matching partitions

Description

Calculates how many of the partitions present in tree 1 are also present in tree 2 (s), how many of the partitions in tree 1 are absent in tree 2 (d1), and how many of the partitions in tree 2 are absent in tree 1 (d2). The Robinson-Foulds (symmetric partition) distance is the sum of the latter two quantities, i.e. d1 + d2.

Usage

SplitStatus(trees, cf = trees[[1]])

SharedSplitStatus(trees, cf)

Arguments

trees

A list of trees of class phylo, with identically labelled tips.

cf

Comparison tree of class phylo. If unspecified, each tree is compared to the first tree in trees.

Value

Returns a two dimensional array. Rows correspond to the input trees, and are named if names were present. Columns report:

N: The total number of partitions present in the two trees, i.e. P1 + P2.

P1: The number of partitions present in tree 1.

P2: The number of partitions present in tree 2.

s: The number of partitions present in both trees.

d1: The number of partitions present in tree 1, but contradicted by tree 2.

d2: The number of partitions present in tree 2, but contradicted by tree 1.

r1: The number of partitions present in tree 1, and neither present nor contradicted in tree 2.

r2: The number of partitions present in tree 2, and neither present nor contradicted in tree 1.

Functions

Author(s)

Martin R. Smith (martin.smith@durham.ac.uk)

References

See Also

Other element-by-element comparisons: CompareQuartetsMulti(), CompareQuartets(), CompareSplits(), PairSharedQuartetStatus(), QuartetState(), SharedQuartetStatus()

Examples

data("sq_trees")

# Calculate the status of each quartet
splitStatuses <- SplitStatus(sq_trees)

# Calculate the raw symmetric difference (i.e. Robinson–Foulds distance)
RawSymmetricDifference(splitStatuses)

# Normalize the Robinson Foulds distance by dividing by the number of 
# splits present in the two trees:
RawSymmetricDifference(splitStatuses) / splitStatuses[, "N"]

# Normalize the Robinson Foulds distance by dividing by the total number of 
# splits that it is possible to resolve for `n` tips:
nTip <- length(sq_trees[[1]]$tip.label)
nPartitions <- 2 * (nTip - 3L) # Does not include the nTip partitions that 
                               # comprise but a single tip
RawSymmetricDifference(splitStatuses) / nPartitions



[Package Quartet version 1.2.6 Index]