How to create a specific matrix from a regular matrix

I wanna create a specific matrix B from a regular matrix A as following in Matlab. Is there anyone who can help me to use a better method to generate it, not for loop? The matrix size could be various. In example, it's 3 by 3 matrix. It could be 3 by n matrix. Thanks

A=[ a b c; 
    d e f; 
    g h i]

B=[a 0 0;  
   d b 0;  
   g e c; 
   0 h f; 
   0 0 i]

[Azzi Abdelmalek]

What is the relation between A and B?

[Image Analyst]

B is like A that is rotated 45 degrees down.

[Pierre]

all diagonal lines(right top to left down) from matrix A

Answer 1

[Azzi Abdelmalek]

A=[ 1 2 3;4 5 6;7 8 9] 
[n,m]=size(A)
B=[A;zeros(n-1,m)]
out=cell2mat(arrayfun(@(x) circshift(B(:,x),[x-1 0]),1:m,'un',0))

[Pierre]

Thanks for your answer. The solution you provided can just solve 3 by 3 matrix, not for 3 by n matrix dynamically, such as: 3 by 5 matrix. Please help me for that. Thanks

[Azzi Abdelmalek]

There is no big changes

[n,m]=size(A)
B=[A;zeros(m-1,m)]
out=cell2mat(arrayfun(@(x) circshift(B(:,x),[x-1 0]),1:m,'un',0))

[Azzi Abdelmalek]

Speed test for 3x1000 array in a for loop (100)

    'Azzi'      'Cedric'    'Andrei'    'Jos'   
    [9.5960]    [0.1596]    [3.5598]    [2.4450]

[Cedric]

Huh, I wouldn't have guessed that SPDIAGS was that fast, thank you for the test! Did you try with a square A matrix, or, to take the other extreme, a 1000x3 matrix?

[Azzi Abdelmalek]

I did the test with a 1000x3 matrix

[Cedric]

You wrote 3x1000 actually, which should produce a significantly sparse 1002x1000 output, whereas the 1000x3 should produce a significantly dense 1002x3 output.

Answer 2

[Cedric]

Here is a one-liner. Assume

 >> A = randi(20, 3, 4)             % Example.
 A =
    19    16     2    16
     6     8     2    19
    16    12    11     3

then

 >> B = full(spdiags(A.', 1-(1:size(A,1)), sum(size(A))-1, size(A,2)))
 B =
    19     0     0     0
     6    16     0     0
    16     8     2     0
     0    12     2    16
     0     0    11    19
     0     0     0     3

is what you are looking for. Note that it is not the fastest solution, but it is appropriate for A matrices with nCols >> nRows. In such case, remove the call to FULL and keep the solution sparse.

Answer 3

[Image Analyst]

Very simple:

B = [A(1,1), 0, 0; 
A(2,1), A(1,2), 0; 
A(3,1), A(2,2), A(1,3); 
0, A(3,2), A(2,3); 
0, 0, A(3,3)];

[Image Analyst]

For the 3 row by n column case, maybe something with kron, hankel, or toeplitz but I'll leave it to someone smarter than me to figure that out.

[Pierre]

This is not what I expected. This matrix size could be various dynamically, for example: 3 by 7, 3 by 21. Therefore, I need few code to create it. Thanks

[Image Analyst]

What would you expect this to give as a result:

A=[  1  2  3  4  5  6;
     7  8  9 10 11 12;
    13 14 15 16 17 18]

Answer 4

[Jos (10584)]

Just for fun and to show the power of logical indexing:

A = reshape(1:12,3,[]).' % input

[m,n] = size(A) ;
tmp = cumsum(cumsum(eye(n+m-1,n)))
B = double(tmp & (tmp < m+1))
B(logical(B)) = A

Answer 5

[Andrei Bobrov]

[m,n] = size(A);
B = ones(m+n-1,n);
p = tril(B) & rot90( tril(B),2);
B = B - 1;
B(p) = A;