Randomly generate clustered points given a center coordinate in 3D

Question

I would like to be able to generate a cluster of points in 3D space that would create a majority of the points within a specified sphere radius (in this case, 4) from the starting point (in this case, 0,0,0). I'd also like it to generate the occasional outlier too.

It would be similar to this:

3D Point Clustering

But I'd like to be able to tweak some of the settings. I will be generating this in C# and I'm familiar with the Random class, and have some familiarity with spheres. I'm not sure how I can piece it all together.

I have this which can generate a point in a sphere, but I want it to cluster around that center point:

// elsewhere in code
private static readonly Random Random = new Random();

private static double NextDouble(double minValue, double maxValue)
{
    double next = Random.NextDouble();

    return minValue + (next * (maxValue - minValue));
}

// generates random points in sphere
const int r = 6;

double x;
double y;
double z;

do
{
    x = NextDouble(-r, 1 + r);
    y = NextDouble(-r, 1 + r);
    z = NextDouble(-r, 1 + r);

} while (Math.Pow(x, 2) + Math.Pow(y, 2) + Math.Pow(z, 2) > Math.Pow(r, 2));

score 1 · Answer 1 · answered Mar 25 '15 at 19:46

1

I think it is easiest to generate random points in spherical coordinates and transform them to cartesian coordinates.

You could generate the radius from the distribution of your choice. Get the two angles from a uniform distribution; one ranging from -pi to pi, the other from 0 to pi.

answered Mar 25 '15 at 19:46

Martijn Pot

86
3

I feel like I'm close with my current code, I think I'm just missing some concept to have it cluster. – test Mar 25 '15 at 20:12
If you feel you're almost there, you can simply add the coordinates of what you want as your center to the x,y,z coordinates that you generated. – Martijn Pot Mar 25 '15 at 20:15

score 1 · Answer 2 · answered Mar 25 '15 at 20:40

1

You can check this article about Spherical Coordinate System. Martin is correct, you will need

a random number from 0 to r set as r
a random number from 0 to pi set as theta
a random number from 0 to 2*pi set as phi

then to transform them to the Cartesian Coordinate System you have to do the following

x = r sin(theta) cos(phi)
y = r sin(theta) sin(phi)
x = r cos(theta)

What you are doing right now will lead to a point cloud of a cubic shape.

If you need further help with the code please do tell.

answered Mar 25 '15 at 20:40

Panagiotis Kazakos

41
3

This is wrong. Generating `theta` uniformly from 0 to \pi won't work – Severin Pappadeux Mar 25 '15 at 21:20
Why is that? What do you suggest? – Panagiotis Kazakos Mar 25 '15 at 21:22
please take a look at my answer. Your proposal for `theta` sampling is wrong, it is not uniform in angles – Severin Pappadeux Mar 25 '15 at 21:33
ok I see, you are basing your answer on the Uniformity of the distribution, while all I am trying to do is point to him that he is creating a cubic form from which he excludes the points that are not inside the sphere. – Panagiotis Kazakos Mar 25 '15 at 21:49

Severin Pappadeux · Answer 3 · 2015-03-25T21:32:22.917

Ok, let's start with distribution, which obviously only depends on r So Probability Density Function would be

PDF(x, y, z) = f(x, y, z) dx dy dz

We know f(x,y,z) depends only on r=sqrt(x^2+y^2+z^2) Lets move to spherical coordinates

PDF(r,theta,phi) = f(r) r^2 sin(theta) dr dtheta dphi

Where r^2 sin(theta) are coming from Jacobian It could be easily factorized into

P(r)     = f(r) r^2 dr
P(theta) = sin(theta) dtheta
P(phi)   = dphi

ok, only phi is distribution uniformly so

phi = 2*pi*random()

for theta it is uniform cos(theta), so

cos(theta) = 2*random()-1

and for r you have to know what f(r) is, but distribution would be affected by r^2 term

P(r) = f(r) r^2 dr

I might suggest using gaussian for f(r), but you might want another choice. And of course

sin(theta) = sqrt(1-cos(theta)^2)
x = r * sin(theta) * cos(phi)
y = r * sin(theta) * sin(phi)
z = r * cos(theta)

Randomly generate clustered points given a center coordinate in 3D

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