4
$\begingroup$

I have circles containing points (x,y). I would like to measure the scattering of the points within the circle.
For example, in the following picture, circle A will have a higher value since the points are much more scattered across the circle.

Notice that the circles have varying value - so we can have a circles with different radiuses.
For example in the following picture, although the points are the same - circle C will have a higher value because the points are scattered across the whole circle.

Do you know a measurement which I can use for such purpose?
Thanks!

$\endgroup$
1
  • 2
    $\begingroup$ Wouldn't mean-square distance from the points' center of mass, divided by the radius of the circle, do the trick? $\endgroup$
    – gmvh
    Commented Sep 23, 2020 at 11:10

3 Answers 3

4
$\begingroup$

There exists a variety of measures of uniformity of a point set. See, for example, On assessing spatial uniformity of particle distributions... for an overview, and a critical comparison when applied to real-world data.

There are two distinct classes of uniformity measures: Quadrat-based measures divide the region into a number of small grids, called quadrats, and count the number of points falling into each grid. Distance-based methods focus on the distances between points, such as those between nearest neighbors or between randomly selected locations.

$\endgroup$
3
$\begingroup$

Judging from your pictures, it should be sufficient to consider the root-mean-square distance $\rho$ of the points $\vec{x}_k$ from their center of mass $\vec{\mu}$, divided by the radius $R$ of the circle: \begin{align} \vec{\mu} &= \frac{1}{N}\sum_{k=1}^N\vec{x}_k, \\ \rho &= \sqrt{\left(\frac{1}{N}\sum_{k=1}^N\left\lVert\vec{x}_k-\vec{\mu}\right\rVert^2\right)}, \\ S &=\frac{\rho}{R}, \end{align} where $S$ is your measure of scattering within the circle.

$\endgroup$
0
$\begingroup$

Star discrepancy

The star discrepancy is usually used when thinking about random numbers and low discrepancy sequences, and seems to fit the bill for your task.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .