多维数据集的边坐标

Question

我有一个与多维数据集8 vortex对应的矩阵，

CubeVortex = [3 3 0; 
              0 3 0; 
              0 3 3; 
              3 3 3; 
              0 0 3;
              3 0 3; 
              3 0 0;
              0 0 0];

现在我想得到所有边缘的坐标，分为3，例如，

如您所见，将有12x2 = 24坐标。

写它们会有点困难。

有没有办法从CubeVortex计算出来？

Answer 1

一种方法：

Cube = [
    3 3 0; 
    0 3 0; 
    0 3 3; 
    3 3 3; 
    0 0 3;
    3 0 3; 
    3 0 0;
    0 0 0];

% find edges by looking for all combinations of points on cube that
% differ by only one coordinate

sections_per_edge = 3;
weights = ((1:sections_per_edge-1) / sections_per_edge).';

edges = []; % indices into Cube
points = []; 
n = size(Cube, 1);
for i = 1:n-1
    pointA = Cube(i, :);
    for j = i+1:n
        pointB = Cube(j, :);
        if nnz(pointA - pointB) == 1
            edges = [edges; i, j];
            % find points along edge as weighted average of point A and B
            points = [points; weights * pointA + (1 - weights) * pointB];
        end
    end
end

% plot corners
plot3(Cube(:,1), Cube(:,2), Cube(:,3), '.r', 'markersize', 20)
hold on

% plot points along edges
plot3(points(:,1), points(:,2), points(:,3), '.b', 'markersize', 20)

% draw edges
line([Cube(edges(:,1), 1), Cube(edges(:,2), 1)].', ...
     [Cube(edges(:,1), 2), Cube(edges(:,2), 2)].', ...
     [Cube(edges(:,1), 3), Cube(edges(:,2), 3)].', 'color', 'k')

axis([-1,4,-1,4])

结果：

将sections_per_edge增加到10，即可获得

Answer 2

一种方法可能就是这个 -

n = 3; %// number of IDs
m = 3; %// number of columns
combs = dec2base(0:(n+1)^m-1,n+1,m)-'0' %// form repeated combinations
out = c1(sum(ismember(combs,[1 2]),2)==1,:) %// combinations for intermediate points

您可以为N-point案例制作此通用名称，并使用此作为更高效的案例

N = 3;
[x,y,z] = ndgrid(0:N,0:N,0:N)
combs = [z(:) y(:) x(:)]
out = combs(sum(combs~=0 & combs~=N,2)==1,:)

因此，对于你的3分（0到3）的情况，你会有 -

out =
     0     0     1
     0     0     2
     0     1     0
     0     1     3
     0     2     0
     0     2     3
     0     3     1
     0     3     2
     1     0     0
     1     0     3
     1     3     0
     1     3     3
     2     0     0
     2     0     3
     2     3     0
     2     3     3
     3     0     1
     3     0     2
     3     1     0
     3     1     3
     3     2     0
     3     2     3
     3     3     1
     3     3     2

Answer 3

这应该这样做：

NewVortex=[];          
for i=1:3
    NewVortex=[CubeVortex*i/3;NewVortex];
end
NewVortex