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How to caluclate centroid of a voronoi cell of a voronoi diagram.

34 vues (au cours des 30 derniers jours)
Pallov Anand
Pallov Anand le 3 Jan 2023
Commenté : Pallov Anand le 4 Jan 2023
Suppose I have generated a voronoi diagram by following code:
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
voronoi(X,Y)
Now, once the voronoi diagram is generated, how to find the centroid of each voronoi cell. I know there are lot of algorithms. One of them is Lloyd's Algorithm, the code of which is given here
but after running the script , I am getting this error:
'poly2cw' requires Mapping Toolbox.
Error in lloydsAlgorithm>VoronoiBounded (line 178)
[X2, Y2] = poly2cw(V(C{ij},1),V(C{ij},2));
Error in lloydsAlgorithm (line 89)
[v,c]=VoronoiBounded(Px,Py, crs);
Can anyone help me in this.
  1 commentaire
Bruno Luong
Bruno Luong le 3 Jan 2023
Modifié(e) : Bruno Luong le 3 Jan 2023
The Vorinoi cells that contain a seed on the hull are unbounded, there is no centroid for such cells.
MATLAB voronoi just cuts them with an empiric bounding box. So of you use those outer vertexes the result is randomly cut.

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Réponse acceptée

Matt J
Matt J le 3 Jan 2023
Modifié(e) : Matt J le 3 Jan 2023
The voronoi cells are always convex, so assuming it is bounded, you can just take the mean of all the vertices of each cell.
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
[V,C]=voronoin([X;Y]');
centroids = cell2mat( cellfun(@(c) mean(V(c,:),1)' , C','uni', 0) )
centroids = 2×11
Inf 1.9180 1.1707 1.2800 1.1995 1.6029 Inf Inf Inf Inf Inf Inf 1.3544 1.4421 1.8802 0.5374 1.6873 Inf Inf Inf Inf Inf
  4 commentaires
Matt J
Matt J le 4 Jan 2023
Because you didn't download the files at the link I gave you.
Pallov Anand
Pallov Anand le 4 Jan 2023
Oh yes, my bad. I just downloaded voronoiPolyhedrons.m and was going through the overview given here
Now, I downloaded and its working now. Thanks a lot mate.

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Plus de réponses (1)

Constantino Carlos Reyes-Aldasoro
This is not a trivial problem especially because the voronoi algorithm does not give you a series of closed polygons, i.e. change the axis of your current problem
X=[1 2 1.1 1.3 1.4 1.5 1.3 1.2 1.8 2.1 2.2];
Y=[1.5 1.3 1.5 1.8 1.4 1.6 2.5 2.3 2.4 1.1 1.8];
voronoi(X,Y)
axis([-2 4 -3 4])
You will see that the voronoi is finding lines that divide the points that you have provided, but it is not essentially generating polygons. Only some of these would be closed and then a centroid makes sense, but not for all of them
  1 commentaire
Pallov Anand
Pallov Anand le 3 Jan 2023
Modifié(e) : Pallov Anand le 3 Jan 2023
Ya, right. Lets take one region with vertices:
(1.7038, 1.3731) ; (1.8661, 1.6435); (1.7780, 1.9520); (1.6864, 1.9864); (1.29, 1.59); (1.2929, 1.5786)
For this region, I think centroid can be calculated and it should lie inside the region.

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