How can I assign a unique label to all points lying inside a single sub-region of divided 3D space?

1 vue (au cours des 30 derniers jours)
Suppose I have a 3D space. I have divided it using grids with X = 0:2:6, Y = 0:1:6 and Z = 0:3:9. So that now I have 54 small 3D regions. Few points are scattered in the entire 3D region. I have to find how many points are lying inside individual small region. Also I have to assign a label for all points in each region, e.g. All points in region 1 will have label 1, all in region 2 will be label 2, so on. I have added the picture for the reference.
Is there any direct function in matlab to find a solution for this problem ? Using if else conditions would be complex if the number of grids and dimensions increases. Is there any idea or hint ?
  2 commentaires
Walter Roberson
Walter Roberson le 25 Juin 2018
Please recheck Z=0:2:10 . Your image is Z=0:3:9 . With 0:2:10 you would have 90 bins.
Varun Pai
Varun Pai le 25 Juin 2018
Thank u for the observation. It is Z =0:3:9 actually.

Connectez-vous pour commenter.

Réponses (2)

KSSV
KSSV le 25 Juin 2018
Go for logical indexing......If (X,y,z) are you points. To get the points lying between (x1,y1,z1) and (x2,y2,z2) ;
idx = (x>x1 & x<=x2) & (y>y1 & y<=y2) & (z>z1 & z<=z2) ;
Initilize your label matrix and fill the value you want for the above idx.
  2 commentaires
Varun Pai
Varun Pai le 25 Juin 2018
Logical indexing would be difficult if the dimension and grid size gets increased. The more number of grids, the more conditions we may have to write

Connectez-vous pour commenter.


Walter Roberson
Walter Roberson le 25 Juin 2018
Let the coordinates of the points be stored in x, y, z
nx = 3; ny = 6; nz = 3;
xb = floor(x/2)+1;
yb = floor(y)+1;
zb = floor(z/3)+1;
counts = accumarray([xb(:), yb(:), zb(:)], 1, [nx, ny, nz]);
labels = sub2ind([nx, ny, nz], xb, yb, zb);

Catégories

En savoir plus sur Scatter Plots dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by