How to fill a 3D zeros matrix array

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Hugo Hernández Hernández
编辑： James Tursa 2020-12-9
I have created a 3D zeros Matrix
R3_T = zeros(3,3,1442);
And want to fill it using the following computation:
R3_Thot = [cos(Tho_t), sin(Tho_t), z_0; -sin(Tho_t), cos(Tho_t), z_0; z_0, z_0, z_1];
Where:
Tho_t is an array of 1x1442
z_0 is an array of 1x1442 (all zeros)
z_1 is an array of 1x1442(all ones)
Actually I created a Matrix but was filled in the wrong way due to the fact that the elements were not in the correct order, so I got something like this:
While my R3_Thot matrix specifies another order.
Thanks.
Hugo
  1 个评论
Hugo Hernández Hernández
Hugo Hernández Hernández 2020-12-9
编辑：Hugo Hernández Hernández 2020-12-9
I am trying with the following structure:
for i=1:4326
for t=1:1442
R3_T(:,:,t) = [1,2,3;4,5,6;7,8,9];
end
end
But still in a trouble with the arrays of:
R3_Thot = [cos(Tho_t), sin(Tho_t), z_0; -sin(Tho_t), cos(Tho_t), z_0; z_0, z_0, z_1];
Because each element is an array of 1x1442 and does not fit with the for loop above, there should be a way to insert each element with its different 1442 values inside the created matrix. The above for loop is an example which I got this matrices:
Now I want to fill them with those R3_Thot data, but still do not know how.

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James Tursa
James Tursa 2020-12-9
编辑：James Tursa 2020-12-9
One way:
c = arrayfun(@(a)[cos(a), sin(a), 0; -sin(a), cos(a), 0; 0, 0, 1],Tho_t,'Uni',false);
result = cat(3,c{:});
Your for-loop would have worked also if you had done this by itself without the outer loop:
for t=1:numel(Tho_t)
R3_T(:,:,t) = [cos(Tho_t(t)), sin(Tho_t(t)), 0; -sin(Tho_t(t)), cos(Tho_t(t)), 0; 0, 0, 1];
end
  1 个评论
Hugo Hernández Hernández
Hugo Hernández Hernández 2020-12-9
Thank you very much James!, the first solution worked as expected but for my surprise the second one also worked, I wasn't sure if that could compute the solution that I wanted so I was trying different ways and concatenations in order to get the assignment of my data.
Thanks!

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