Replace matrix entries with entries from another matrix

There are two matrices A and B as under:

A = [0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0];
B = [1 1 1;2 2 2;3 3 3];

I want to create a third matrix C in which the columns of B replace columns of A incrementally i.e columns of B move inside A. For example:

C1 = [0 0 0 0 1;0 0 0 0 2;0 0 0 0 3];
C2 = [0 0 0 1 1;0 0 0 2 2;0 0 0 3 3];
C3 = [0 0 1 1 1;0 0 2 2 2;0 0 3 3 3];
C4 = [0 1 1 1 0;0 2 2 2 0;0 3 3 3 0];

[dpb]

I'm not at all sure what your sample C arrays are to represent as a "third matrix C", but note

>> fliplr(tril(ones(4,5)))
ans =
     0     0     0     0     1
     0     0     0     1     1
     0     0     1     1     1
     0     1     1     1     1
>> 

to build the pattern...

[Waseem Akhtar]

It's not a pattern, rather, the columns entries of B matrix should replace that of A matrix in a loop. For example, in the first increment the C matrix should be:

C1 =

0 0 0 0 1

0 0 0 0 2

0 0 0 0 3

In next increment, it should be:

C2 =

0 0 0 1 1

0 0 0 2 2

0 0 0 3 3

and likewise for next increments. You may better understand it by running the matrices defined in my query in the Matlab.

THanks

Answer 1

[Simon Chan]

I do it in a cell array:

clear; clc;
A = [0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0];
B = [1 1 1;2 2 2;3 3 3];
combine = [A,B];
idx = -1:-1:-4;
M=arrayfun(@(x) circshift(combine,[0,x]),idx,'UniformOutput',false);
C = cellfun(@(x) x(:,1:5),M,'UniformOutput',false)

If you really want the name to be C1, C2, C3 and C4, then

C1 = C{1};
C2 = C{2};
C3 = C{3};
C4 = C{4};

[Waseem Akhtar]

Thanks Simon, this do creats the desired matrix C. But, is it possible to create/update C incrementally in a loop using a counter rather than using cells.

[Simon Chan]

Try the following:

However, the matrix C will be overwritten every time in the loop.

So additional manipulation on matrix C must be performed inside the loop.

clear; clc;
A = [0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0];
B = [1 1 1;2 2 2;3 3 3];
for idx = -1:-1:-4
	M = circshift([A,B],[0,idx]);
	C = M(:,1:5)
end

Result like this:

C =

     0     0     0     0     1
     0     0     0     0     2
     0     0     0     0     3


C =

     0     0     0     1     1
     0     0     0     2     2
     0     0     0     3     3


C =

     0     0     1     1     1
     0     0     2     2     2
     0     0     3     3     3


C =

     0     1     1     1     0
     0     2     2     2     0
     0     3     3     3     0

[Waseem Akhtar]

That's it.

Thanks alot!