calculating orientation anisotropy of a set of 3D vectors

I have a set of unit-length 3D vectors (represented as cartesian 3-tuples) with 180$^\circ$ equivalence (rotated through any plane passing through the origin i.e. $v=-v$), all situated at the origin (really their position is irrelevant). I'm interested in calculating the degree of anisotropy exhibited in the orientations of the vectors. The measurement should be 0 when the vectors orientations are uniformly distributed across the half unit-sphere, and say, 1 when all the vectors are parallel (identical).

I'm not sure exactly what statistic this is — at first I naively thought it would be the the directional variance of the set, however this clearly isn't right. The case where the vectors are split evenly into 3 orthogonal orientations should have much greater anisotropy than the uniform distribution, and even greater anisotropy if the groups were more closely oriented (not orthogonal).

I'm considering iterating through the vectors and either placing each one in a bin with other vectors of sufficiently close orientation, or creating a new bin if no previously considered vector has sufficiently similar orientation. However, this involves choosing a threshold, which I would like to avoid, and I feel like there must be a more appropriate metric.

Could anyone point me in the right direction (no pun intended)?

• The quantity you describe is the mean resultant length, which is obtained by summing all 3D vectors, dividing by n, and taking the norm. That value multiplied by 3n has approximately a chi-squared distribution with three degrees of freedom. This is an example of the Rayleigh test on a sphere. There are also further adjusted versions of this test, mentioned for example in Mardia & Jupp (2000). Apr 5 '18 at 13:20
• A more revealing statistic is discussed and illustrated in my answer at stats.stackexchange.com/a/7984/919. It assess whether, how, and in what ways the vectors differ from uniformity. To apply it in your case, double the data by including the negatives of all the vectors, too.
– whuber
Apr 5 '18 at 13:42
• @Kees hmm, I did consider doing this, but I'm not convinced that it fully captures the anisotropy (sorry I know I haven't defined this well). For example in the case I gave where the vectors are in three equally sized orthogonal groups, the resultant vector would be the zero vector which isn't very meaningful. I guess I'm expecting some sort of measure possibly related to the entropy of the distribution.
– Will
Apr 5 '18 at 13:44