selecting a triangle in an array of delaunay points

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luc 2014-12-16

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https://ww2.mathworks.cn/matlabcentral/answers/166840-selecting-a-triangle-in-an-array-of-delaunay-points

回答： luc 2014-12-18

采纳的回答： luc

在 MATLAB Online 中打开

The code explains it all.

Try to run it, it might give you a good result, but that would be random.

This code is supposed to display a point, and color the nearest triangle green. It does not do this, and gives "nan" errors on seemingly random iterations of the loop.

I would appreciate some help :)

clc;
clear all;
close all;
P = [ 2.5    8.0
      6.5    8.0
      2.5    5.0
      6.5    5.0
      1.0    6.5
      8.0    6.5];
  dt=delaunayTriangulation(P);
  triplot(dt);
  hold on
  %%example functions 
  dt.Points
  dt.ConnectivityList
  %abs vals x
  a=min(P(:,1))
  b=max(P(:,1))
  %abs vals y
  c=min(P(:,2))
  d=max(P(:,2))
    for n=1:5
        random_x=a + (b-a).*rand(1);
        random_y=c + (d-c).*rand(1);
         triangleId=pointLocation(dt, random_x,random_y) %select the triangle ID of the triangle closest to the point generated by the random generator
         scatter(random_x,random_y)                       %display the points
         %%now... the part that does not work
         %%selecting the right triangle and turning it a different color
         tri = dt(triangleId, [1:end 1]);
         patch(P(tri,1), P(tri,2), 'r', 'LineWidth',1, 'FaceColor','g')
    end

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John D'Errico 2014-12-17

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My point is, pointLocation will return NaN in that case. SURPRISE!

How often will that event happen? As it turns out, that should happen roughly 21.429% of the time that a point falls outside of your triangulated region, but inside the rectangle that you use to generate the random points.

(1 - (8-2.5)*3/(3*(8-1)))*100
ans =
     21.429

luc 2014-12-17

Thanks! This solved my problem. I think an addition to the source code of this function would be to still pick the nearest triangle to the point, accompanied by an error message saying that the point itself is not inside a triangle.

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