Creating a block matrix of matrices?
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I have a problem where I'm trying to create a matrix of the form
[A B 0 0; 0 A B 0; 0 0 A B; 0 0 0 A];
However, this is in block matrix notation. That means all of the elements are matrices of appropriate size so that this concatenation works. I saw the blkdiag function, but it doesn't look like it's going to work for this, because the elements overlap in certain columns.
For example, if A = [1 1] and B = [2 2] this matrix would look like:
[1 1 2 2 0 0 0 0; 0 0 1 1 2 2 0 0; 0 0 0 0 1 1 2 2; 0 0 0 0 0 0 1 1]
Does anyone have any suggestions how to do this in a parsimonious way?
2 件のコメント
Image Analyst
2016 年 5 月 6 日
Your second row is [0 A B 0] yet your example does not show that. Your example shows [0 0 A B 0 0], so which is it? Does it shift over by one zero, or the number of elements in A, or by the number of elements in B?
We can't give the answer you want until we know for sure which it is.
採用された回答
Hang Qian
2016 年 5 月 6 日
Hello,
If the FOR loop is not your choice, you may consider the following:
>> A = [1 1];
>> B = [2 2];
>> C = blkdiag(A,A,A,A);
>> C(1:end-1,3:end) = C(1:end-1,3:end) + blkdiag(B,B,B)
If you’d like something slightly more general,
>> nrow=2;ncol=3;
>> A=rand(nrow,ncol);B=rand(nrow,ncol);
>> C = blkdiag(A,A,A,A);
>> C(1:end-nrow,ncol+1:end) = C(1:end-nrow,ncol+1:end) + blkdiag(B,B,B)
Regards,
- Hang Qian
その他の回答 (1 件)
Mark Britten-Jones
2019 年 3 月 14 日
This is by far the easiest way to do this. Create the blocks. Create a 2-D cell array and place the blocks into the appropriate cells. And then convert to a matrix by cell2mat. I have used this where I have used loops over the cell blocks to create quite complicated matrices and you do not have to worry about the indexes at the matrix level. Bonus! See the cell2mat documenation for rules regarding permissable block sizes. (They do not all have to be of the same size.)
A = [1 1];
B = [2 2];
Z = [ 0 0]
Ccell = {A, B, Z, Z; Z, A, B, Z; Z, Z, A, B; Z, Z, Z, A};
C = cell2mat(Ccell);
1 件のコメント
Jon
2019 年 4 月 4 日
Actually in the example you give it is not necessary to use the cell arrays at all. You can directly assign:
C =[A B Z Z;Z A B Z;Z Z A B;Z Z Z A]
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