Create a matrix of this type?

Hello,

I want to make a matrix this type

     1     0     0     0     0     0
     2     0     0     0     0     0
     3     4     0     0     0     0
     5     6     0     0     0     0
     7     8     9     0     0     0
    10    11    12     0     0     0
    13    14    15    16     0     0
    17    18    19    20     0     0
    % Alternate rows have same number of element
    %Each element of the matrix is 1 larger than previous one

How to achieve it ?

Answer 1

[Andrei Bobrov]

m = 8;
n = 6;
P = kron(triu(ones(n,m/2)),[1,1]);
P(P>0) = 1:nnz(P);
out = P';

[P K]

Thanks.@Andrei Bobrov. I had to create one more matrix. It would be like

A=[1 0 0 0 0 0;0 2 0 0 0 0;3 0 4 0 0 0;0 5 0 6 0 0;7 0 8 0 9 0;0 10 0 11 0 12];

I can do it with 'for loop'. But can you show some efficient way as you had done in the earlier case.

Thanks in advance.

[Andrei Bobrov]

m = 6;
n = 6;

lo = triu(~rem((1:n)' + (1:m),2));
out = int64(lo);
out(lo) = 1:nnz(lo);
out = out';

with kron

m = 6;
n = 6;

out = triu(kron(ones(ceil(n/2),ceil(m/2)),[1,0;0,1]));
out = out(1:n,1:m);
out(out~=0) = 1:nnz(out);
out = out';

[P K]

Sir Andrei Bobrov, it took me "SO LONG" to do it and you have posted two ways to achieve it. Thank you.

Answer 2

[madhan ravi]

n=6; % number of elements in a row
B=mat2cell((1:20).',repelem(1:4,2));
B=cellfun( @transpose,B,'un',0);
R=cellfun( @(x) [x zeros(1,6-numel(x))],B,'un',0);
vertcat(R{:})

[P K]

Thanks Madhan ravi. It works.Andrei Bobrov answer is more general.