hv_contributions {moocore}

Hypervolume contribution of a set of points

Description

Computes the hypervolume contribution of each point given a set of points with respect to a given reference point assuming minimization of all objectives. Dominated points have zero contribution. Duplicated points have zero contribution even if not dominated, because removing one of them does not change the hypervolume dominated by the remaining set.

Usage

hv_contributions(x, reference, maximise = FALSE)

Arguments

x

matrix()|data.frame()
Matrix or data frame of numerical values, where each row gives the coordinates of a point.

reference

numeric()
Reference point as a vector of numerical values.

maximise

logical()
Whether the objectives must be maximised instead of minimised. Either a single logical value that applies to all objectives or a vector of logical values, with one value per objective.

Value

numeric()
A numerical vector

Author(s)

Manuel López-Ibáñez

References

Carlos M. Fonseca, Luís Paquete, Manuel López-Ibáñez (2006). “An improved dimension-sweep algorithm for the hypervolume indicator.” In Proceedings of the 2006 Congress on Evolutionary Computation (CEC 2006), 1157–1163. IEEE Press, Piscataway, NJ. doi: 10.1109/CEC.2006.1688440.

Nicola Beume, Carlos M. Fonseca, Manuel López-Ibáñez, Luís Paquete, Jan Vahrenhold (2009). “On the complexity of computing the hypervolume indicator.” IEEE Transactions on Evolutionary Computation, 13(5), 1075–1082. doi: 10.1109/TEVC.2009.2015575.

See Also

hypervolume()

Examples

data(SPEA2minstoptimeRichmond)
# The second objective must be maximized
# We calculate the hypervolume contribution of each point of the union of all sets.
hv_contributions(SPEA2minstoptimeRichmond[, 1:2], reference = c(250, 0),
            maximise = c(FALSE, TRUE))

# Duplicated points show zero contribution above, even if not
# dominated. However, filter_dominated removes all duplicates except
# one. Hence, there are more points below with nonzero contribution.
hv_contributions(filter_dominated(SPEA2minstoptimeRichmond[, 1:2], maximise = c(FALSE, TRUE)),
                 reference = c(250, 0), maximise = c(FALSE, TRUE))