Suppose we have a 3D array r(x,y,z) filled with values 1 and 2. 1. How to visualise it into a 3D model? 2. How to visualise only "2" values, i.e. make "1" invisible? Thank you!

 采纳的回答

Ok, so try this one. It includes a fast way to generate a random array with only 1 or 2 entries.
%generate a 3d-matrix of size 10x5x5 with values 1,2.
R = randi(2,10,5,5);
%define coordinate grid
x = 1:10; y = 1:5; z = 1:5;
[X,Y,Z] = meshgrid(x,y,z);
%detect points with label=1
I1 = R==1;
%plot points with label=1
scatter3(X(I1),Y(I1),Z(I1),'filled');

更多回答(1 个)

Try this out. Given a set of 10 points (3dim) and a set of labels (1,2) it returns the plot of the points according to the labels.
XYZ = rand(10,3);
R = [1,1,2,1,2,2,2,1,2,1];
R1 = (R==1);
R2 = (R==2);
figure;
scatter3(XYZ(R1,1),XYZ(R1,2),XYZ(R1,3),'r','filled');
hold on;
scatter3(XYZ(R2,1),XYZ(R2,2),XYZ(R2,3),'b','filled');
hold off;
You may also want to map colors of the colormap directly to labels, in automated way.
figure;
scatter3(XYZ(:,1),XYZ(:,2),XYZ(:,3),30,R,'filled');

8 个评论

r=[1 2 1; 1 1 1; 2 2 2]
R1=(r==1);
R2=(r==2);
figure;
scatter3(r(R1,1),r(R1,2),r(R1,3));
If I test your code using this trivial 3D matrix, I get an error "
Index exceeds matrix dimensions.
Error in test1 (line 6)
scatter3(r(R1,1),r(R1,2),r(R1,3)); "
Can you please correct my script?
Is your "r" the matrix of coordinates or labels? They are different.
r is the set of values
And where you stored the 3d point coordinates? Why is your "r" a matrix instead of a vector?
Initially I have a big grid 100x50x50, filled with values 1 and 2. r(x,y,z). r is the coordinates, not values. But each r(x,y,z) can be either 1 or 2.
So for each 3d point, you have 3 labels??
Can you give me an examplt of an input yuo have to display?

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