sharing of numbers in the matrix

If

A=[0.1 0 0.2 0 0 0 0 0;0 0 0 0.3 0.4 0 0 0;0 0.5 0 0 0 0 0 0.6;0 0 0 0 0 0.7 0.8 0]

I want to have

B=[0.1 0 0.2 0.3 0.4 0 0 0;0.1 0 0.2 0.3 0.4 0 0 0;0 0.5 0 0 0 0 0.7 0.8 0.6;0 0.5 0 0 0 0 0.7 0.8 0.6]

could anyone tell me how to get it?

[per isakson]

The rows of B don't have the same length.

[Jos (10584)]

What is "sharing of numbers"?

[Prabha Kumaresan]

A =

    0.1000         0    0.2000         0         0         0         0         0
         0         0         0    0.3000    0.4000         0         0         0
         0    0.5000         0         0         0         0         0    0.6000
         0         0         0         0         0    0.7000    0.8000         0

B =

    0.1000         0    0.2000    0.3000    0.4000         0         0         0
    0.1000         0    0.2000    0.3000    0.4000         0         0         0
         0    0.5000         0         0         0    0.7000    0.8000    0.6000
         0    0.5000         0         0         0    0.7000    0.8000    0.6000

I made the mistake in B.Now could you tell me how to get b from A.

[Jos (10584)]

What is the exact connection between A and B? Why, for instance, is B(2,2) 0 and not 0.5?

[Prabha Kumaresan]

A is a matrix of subcarrier allocation to users,where rows denote users and columns denote subcarriers.B(2,2) is 0 which means subcarrier is not being used by user1 and user2.

[Jos (10584)]

I am completely lost ... Stephen and Andrei, the floor is yours

Answer 1

[Stephen23]

Solution to your original question:

>> reshape(repmat(max(reshape(A,2,[])),2,1),size(A))
ans =
   0.10000   0.00000   0.20000   0.30000   0.40000   0.00000   0.00000   0.00000
   0.10000   0.00000   0.20000   0.30000   0.40000   0.00000   0.00000   0.00000
   0.00000   0.50000   0.00000   0.00000   0.00000   0.70000   0.80000   0.60000
   0.00000   0.50000   0.00000   0.00000   0.00000   0.70000   0.80000   0.60000

Solution to your later comment:

>> repmat(reshape(max(reshape(A,2,[])),2,[]),2,1)
ans =
   0.10000   0.00000   0.20000   0.30000   0.40000   0.00000   0.00000   0.00000
   0.00000   0.50000   0.00000   0.00000   0.00000   0.70000   0.80000   0.60000
   0.10000   0.00000   0.20000   0.30000   0.40000   0.00000   0.00000   0.00000
   0.00000   0.50000   0.00000   0.00000   0.00000   0.70000   0.80000   0.60000

[Andrei Bobrov]

+1

Answer 2

[Andrei Bobrov]

out = repelem(squeeze(max(permute(reshape(A,2,2,[]),[1,3,2]))),1,2)';

or

out = A;
[ii,jj] = find(A);
t = rem(ii,2);
out(sub2ind(size(out),ii - ~t + t,jj)) = A(A>0);

[Prabha Kumaresan]

Thanks

[Prabha Kumaresan]

If i need to have B such that B =

    0.1000    0.5000    0.2000         0         0         0         0    0.6000
         0         0         0    0.3000    0.4000    0.7000    0.8000         0
    0.1000    0.5000    0.2000         0         0         0         0    0.6000
         0         0         0    0.3000    0.4000    0.7000    0.8000         0
how it can be done.

[Andrei Bobrov]

See Stephen's answer (first part).

B = reshape(repmat(max(reshape(A,2,2,[]),[],2),1,2),size(A));

or

[ii,jj] = find(A);
B = A;
B(sub2ind(size(A),rem(ii + 1,4)+1,jj)) = A(A>0);