3D Matrix to 2D Matrix

I have four 3D matrix like Y= ones(2,2,4) and i want to make a new matrix like YT=zeros(8,8).

I want to take values from 3D "Y" matrix and insert them into 2D "YT" matrix in a following fashion

YT(1,1)=Y(1,1,1)
YT(1,2)=Y(1,2,1)
YT(2,1)=Y(2,1,1)
YT(2,2)=Y(2,2,1)

YT(3,3)=Y(1,1,2)
YT(3,4)=Y(1,2,2)
YT(4,3)=Y(2,1,2)
YT(4,4)=Y(2,2,2)

YT(5,5)=Y(1,1,3)
YT(5,6)=Y(1,2,3)
YT(6,5)=Y(2,1,3)
YT(6,6)=Y(2,2,3)

YT(7,7)=Y(1,1,4)
YT(7,8)=Y(1,2,4)
YT(8,7)=Y(2,1,4)
YT(8,8)=Y(2,2,4)

Can some one suggest me what would be the best way to do this thing automatically? I only assume the values of Y to 1,1,1,1 but in my real case, values of Y are different complex numbers.

Answer 1

[Star Strider]

This works:

y = 1:16;                                               % Create Data
Y = reshape(y, 2, 2, 4);                                % Create Data
RY = reshape(Y(:), 2, []);                              % Reshape ‘Y’
RC = mat2cell(RY, 2, ones(1,size(RY,2)/2)*2);           % Create Submatrices
YT = blkdiag(RC{:})                                     % Create ‘YT’

[Adan91h]

Dear Strider,

Thank you very much for your prompt response.

I want to send values automatically to these positions. Your solution works well if we write following command manually

YT=blkdiag(RC{1},RC{2},RC{3},RC{4});

but i want to create a loop with which i can take values from "Y" or "RY" (in your code) and send them automatically to required positions of YT.

I hope this clarify my problem.

BR, Adnan

[Star Strider]

The cell array creates a commas-separated list of the kind that blkdiag wants. I changed it my original Answer replacing the ‘YT’ assignment with:

YT = blkdiag(RC{:})                                     % Create ‘YT’

That does exactly what you want.

(I should have remembered that before. I wasn’t completely awake when I wrote it originally.)

[Adan91h]

Thanks once again :). Now its working perfectly. You are great!!!

BR, Adnan

[Star Strider]

My pleasure! Thank you for your compliment!

(I apologize for the earlier confusion. I’m UTC-7 here, and wasn’t fully awake then.)

Answer 2

[John BG]

Hi Adnan

1.

generating test matrices

Y=randi([-10 10],2,2,4)
YT=zeros(8,8)

YT([1 2],[1 2])=Y(:,:,1)
YT([3 4],[1 2])=Y(:,:,2)
YT([5 6],[1 2])=Y(:,:,3)

2. moving values

[sz1 sz2 sz3]=size(Y)

for k=1:1:sz3
    YT([2*k-1 2*k],[2*k-1 2*k])=Y(:,:,k)
end

3. the values of Y and YT can be complex, there's no need to build 2 matrices containing real and imaginary parts.

if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?

To any other reader, please if you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>