hypervolume#

moocore.hypervolume(data, /, ref, maximise=False)[source]#

Hypervolume indicator.

Compute the hypervolume metric with respect to a given reference point assuming minimization of all objectives. For 2D and 3D, the algorithm used [1][2] has \(O(n \log n)\) complexity. For 4D or higher, it uses a recursive algorithm that has the 3D algorithm as a base case [1], which has \(O(n^{d-2} \log n)\) time and linear space complexity in the worst-case, but experimental results show that the pruning techniques used may reduce the time complexity even further. Andreia P. Guerreiro improved the integration of the 3D case with the recursive algorithm, which leads to significant reduction of computation time. She has also enhanced the numerical stability of the algorithm by avoiding floating-point comparisons of partial hypervolumes.

Parameters:

  • data (ArrayLike) – Numpy array of numerical values, where each row gives the coordinates of a point. If the array is created from the read_datasets() function, remove the last column.

  • ref (ArrayLike) – Reference point as a 1D vector. Must be same length as a single point in the data.

  • maximise (bool | list[bool], default: False) – Whether the objectives must be maximised instead of minimised. Either a single boolean value that applies to all objectives or a list of booleans, with one value per objective. Also accepts a 1D numpy array with value 0/1 for each objective

Returns:

float – A single numerical value, the hypervolume indicator

See also

Hypervolume

object-oriented interface.

References

Examples

>>> dat = np.array([[5, 5], [4, 6], [2, 7], [7, 4]])
>>> moocore.hypervolume(dat, ref=[10, 10])
38.0

Default is minimization, we can easily assume maximization.

>>> dat = np.array([[5, 5], [4, 6], [2, 7], [7, 4]])
>>> moocore.hypervolume(dat, ref=0, maximise=True)
39.0

Merge all the sets of a dataset by removing the set number column:

>>> dat = moocore.get_dataset("input1.dat")[:, :-1]
>>> len(dat)
100

Dominated points are ignored, so this:

>>> moocore.hypervolume(dat, ref=[10, 10])
93.55331425585321

gives the same hypervolume as this:

>>> dat = moocore.filter_dominated(dat)
>>> len(dat)
6
>>> moocore.hypervolume(dat, ref=[10, 10])
93.55331425585321