Trying to Vectorise Cell Allocation to Remove For Loops

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Paolo Olson
Paolo Olson 2021-3-3
编辑： Walter Roberson 2023-10-14
I am trying to speed up my code and was wondering how I can vectorise my code to remove the for loops. The code essentially discretises space into cubes and this section finds the 8 coordinates which define each cube.
Thanks for all your help! :)
x=1:6
y=1:7
z=1:7
coord_matrix=cell(size(x,2)-1,size(y,2)-1,size(z,2)-1);
for ik=1:length(z)-1
for ij=1:length(y)-1
for ii=1:length(x)-1
coord_matrix(ii,ij,ik)={[x(ii),y(ij),z(ik);...
x(ii+1),y(ij),z(ik);...
x(ii),y(ij+1),z(ik);...
x(ii),y(ij),z(ik+1);...
x(ii+1),y(ij),z(ik+1);...
x(ii),y(ij+1),z(ik+1);...
x(ii+1),y(ij+1),z(ik);...
x(ii+1),y(ij+1),z(ik+1)]};
end
end
end
  1 个评论
Walter Roberson
Walter Roberson 2023-10-14
编辑：Walter Roberson 2023-10-14
I did a vectorized 2D analog of this a couple of days ago; see https://www.mathworks.com/matlabcentral/answers/2032169-create-mesh-from-matrix#answer_1331984
You probably do not need to keep storing the coordinates over and over again: you can probably just store the indices. And besides, if you create the index array into a numeric array, you can use it for vectorized lookups of the coordinates if you really do need the array of coordinates.

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回答（2 个）

Ravi
Ravi 2023-10-14
Hi Paolo Olson,
I understand you are trying to eliminate the nested for loops and want to optimize your code by switching to vectorization approach for finding the cube indices.
You can try using the “meshgrid” function,which is used to create a grid of indices.
  1. Create a grid of indices using meshgrid with ii, ij, and ik values where the ii values range from 1 to length(x) – 1, ij values range from 1 to length(y) – 1, and ik values range from 1 to length(z) – 1.
  2. Now you have the indices, and you need to apply the indexing on operation on arrays x, y, and z with the meshgrid indices. For this purpose, you can use the “reshape” function. For example, you can obtain the first coordinate using the following,
reshape(x(ii), [], 1), reshape(y(ij), [], 1), reshape(z(ik), [], 1)
For an in-depth understanding on “meshgrid” and “reshape” functions, kindly look at the below mentioned resources,
Hope the above-mentioned approach solves the issue you are facing.
Walter Roberson
Walter Roberson 2023-10-14
Your coordinates are the same size for each cuboid, and the size is predicatable. Use a numeric array instead of a cell array.
x=1:6
y=1:7
z=1:7
coord_matrix = zeros(size(x,2)-1,size(y,2)-1,size(z,2)-1, 8, 3);
for ik=1:length(z)-1
for ij=1:length(y)-1
for ii=1:length(x)-1
coord_matrix(ii,ij,ik,:,:)= [x(ii),y(ij),z(ik);...
x(ii+1),y(ij),z(ik);...
x(ii),y(ij+1),z(ik);...
x(ii),y(ij),z(ik+1);...
x(ii+1),y(ij),z(ik+1);...
x(ii),y(ij+1),z(ik+1);...
x(ii+1),y(ij+1),z(ik);...
x(ii+1),y(ij+1),z(ik+1)];
end
end
end

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