在Matlab中获取平面上一组3D点的边界矩形的最佳方法

Question

我需要将一组3D点的边界矩形的四个边缘存储为3xN矩阵（tt）。 N> = 4。这些点位于一个平面上。

代码示例：

% Simulate some points
deltaXY = 20;
[xx,yy] = meshgrid(-100:deltaXY:100,-100:deltaXY:100);
XYZ = [xx(:)'; yy(:)'; zeros(1,numel(xx))];

% Add some imperfection to data removing the top rigth point
maxXids = find(XYZ(1,:) == max(XYZ(1,:)));
maxYids = find(XYZ(2,:) == max(XYZ(2,:)));
id = intersect(maxXids,maxYids);
XYZ = removerows(XYZ',id)';

% Lets rotate a bit
XYZ = roty(5)*rotx(7)*rotz(0)*XYZ;

% Plot points
figure;
grid on;
rotate3d on;
axis vis3d;
hold on;
plot3(XYZ(1,:),XYZ(2,:),XYZ(3,:),'.r');

% Find bounding rectangle
% ??? :(
%Currently I'm using this code:
tt = XYZ;
%Get the max and min indexes
minX = find(tt(1,:) == min(tt(1,:)));
minY = find(tt(2,:) == min(tt(2,:)));
maxX = find(tt(1,:) == max(tt(1,:)));
maxY = find(tt(2,:) == max(tt(2,:)));
%Intersect to find the common index
id1 = intersect(minX,minY);
id2 = intersect(maxX,minY);
id3 = intersect(maxX,maxY);
id4 = intersect(minX,maxY);
%Get the points
p1 = tt(:,id1(1));
p2 = tt(:,id2(1));
p3 = tt(:,id3(1));
p4 = tt(:,id4(1));

示例点图：

问题是交叉一些时间可以为null，例如：如果点不形成矩形。导致此错误：

  

    

指数超出矩阵维度。

  

Answer 1

第一个解决方案：使用逻辑索引来摆脱find来电

p1=tt(:,tt(1,:)==min(tt(1,:))&tt(2,:)==min(tt(2,:)));
p2=tt(:,tt(1,:)==max(tt(1,:))&tt(2,:)==min(tt(2,:)));
p3=tt(:,tt(1,:)==max(tt(1,:))&tt(2,:)==max(tt(2,:)));
p4=tt(:,tt(1,:)==min(tt(1,:))&tt(2,:)==max(tt(2,:)));

第二个解决方案：使用convhull获取角落：

k=convhull(tt(1,:),tt(2,:));
Corners=[tt(:,k(1:end-1))];

Answer 2

好的找到了解决方案：

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