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!

$\endgroup$
  • 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 Sep 23 at 11:10
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.

| cite | improve this answer | |
$\endgroup$
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.

| cite | improve this answer | |
$\endgroup$
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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.