Divide the 3D feature space into grid and then get their labels

Dear all,

I have 3D locations in three vectors, x, y, and z. I want to divide the space into fix numbe of grids lets say 2 x 2 x 2. Lets take example in 2D space. I have two vectors x, and y, as below;

x = [1 1 2 2 3 10];
y = [1 2 1 1 2 5];
plot(x,y, '+')

now I divide in to 2 x 2. So the interval for x would be 0-to-5 and then 5.1-to-10, and likewise for y(0-to-2.5, and 2.6-to-5). I have drawn the points and divided the space manually as shown below.

now the there are four cells. I have vector r that should be r = [3,3,3,3,3,2] where all points are in cell 3 and only point 5,10 is on cell 2.

Answer 1

[Turlough Hughes]

Example in 2D

% Data
x = rand(1,100)*10;
y = rand(1,100)*5;
r = zeros(size(x));     % group index

r(x>0 & x<=5  & y>0 & y<=2.5) = 1;
r(x>5 & x<=10 & y>0 & y<=2.5) = 2;
r(x>0 & x<=5  & y>2.5 & y<=5) = 3;
r(x>5 & x<=10 & y>2.5 & y<=5) = 4;

figure(), gscatter(x,y,r)
hold on, axis padded, legend off
xline(5), yline(2.5)

Here I'm using conditions > as well as <= so that you can select one edge, eg 5 instead of switching between 5 and 5.1 etc, though if you need to modify from this then do so as required.

Example in 3D

% Additional data for z
z = rand(1,100)*5;

% E for Edges
xE = [0 5 10];
yE = [0 2.5 5];
zE = [0 2.5 5]; % You can see why just one value for the edge is useful.

% Group the data according to the edges (r gives the groups).
idx = 1;
for ii = 1:numel(xE)-1
    for jj = 1:numel(yE)-1
        for kk = 1:numel(zE)-1
            r(x>xE(ii) & x<=xE(ii+1)  & y>yE(jj) & y<=yE(jj+1) & z>zE(kk) & z<=zE(kk+1)) = idx;
            idx = idx + 1;
        end
    end
end
cmap = jet(8);
figure(), scatter3(x,y,z,80,cmap(r,:),'filled')

Not the best visualisation but you can see that it does the job.

[Aravin]

Thank you Huges. Wonderfull. As I stated, the grid size which we have kept 2 x 2 or 2 x 2 x 2 is just an example. Could you please help in making n x n or n x n x n ? As I have to experiment with different values of n.

Your support is highly appreciated on this.

[Turlough Hughes]

Just use linspace to define your edges, for example:

% Data
x = rand(1,1000)*10;
y = rand(1,1000)*5;
z = rand(1,1000)*5;
r = zeros(size(x));     % group index

% E for Edges
n = 3;
xE = linspace(min(x)-1e-6, max(x), n+1);
yE = linspace(min(y)-1e-6, max(y), n+1);
zE = linspace(min(z)-1e-6, max(z), n+1);

% Group the data according to the edges
idx = 1;
for ii = 1:numel(xE)-1
    for jj = 1:numel(yE)-1
        for kk = 1:numel(zE)-1
            r(x>xE(ii) & x<=xE(ii+1)  & y>yE(jj) & y<=yE(jj+1) & z>zE(kk) & z<=zE(kk+1)) = idx;
            idx = idx + 1;
        end
    end
end
cmap = colorcube(n^3);
figure(), scatter3(x,y,z,200,cmap(r,:),'filled')

[Aravin]

Thank you so much. Please illusarate in 2D too. sorry for being naive.

[Turlough Hughes]

% Data
x = rand(1,500)*10;
y = rand(1,500)*5;
r = zeros(size(x));     % group index
% E for Edges
n = 6;
xE = linspace(min(x)-1e-6, max(x), n+1);
yE = linspace(min(y)-1e-6, max(y), n+1);
% Group the data according to the edges
idx = 1;
for ii = 1:numel(xE)-1
    for jj = 1:numel(yE)-1
        r(x>xE(ii) & x<=xE(ii+1)  & y>yE(jj) & y<=yE(jj+1)) = idx;
        idx = idx + 1;
    end
end
cmap = colorcube(n^2);
figure(), scatter(x,y,100,cmap(r,:),'filled')
xline(xE)
yline(yE)

[Aravin]

What version of matlab you are using? I don't have xline in 2018.

[Turlough Hughes]

The xline() and yline() functions were introduced in 2018b.