How do I make matrix of ones and zeros alternating depending on size and elements of an array?

I have an array, A, and I want to make a matrix, B, that has the size dependent on A, and has 1s and 0s alternating, depending on the value of elements in A

E.g. A=[1;2;3;4;5]

B should be of size: ((length(A)+1)/2,sum(A)]

In this case B would be a matrix of 3x15

Then the values for B needs to be be thus:

row 1:[ones(A(1)),zeros(A(2)),zeros(A(3)),zeros(A(4)),zeros(A(5))]

row2: [zeros(A(1)),zeros(A(2)),ones(A(3)),zeros(A(4)),zeros(A(5))]

row3: [zeros(A(1)),zeros(A(2)),zeros(A(3)),zeros(A(4)),ones(A(5))]

B should look like:

[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;

0 0 0 1 1 1 0 0 0 0 0 0 0 0 0;

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]

I've been trying to formulate something with a nested for look for each dimension of B, but have not been succesful!

Would really appreciate any help thanks!

Answer 1

[Fabio Freschi]

% initialize
iRow = [];
jCol = [];

% preallocate
B = zeros((length(A)+1)/2,sum(A));

% indices for columns
idx = [0; cumsum(A)]+1;

for i = 1:length(A)
    if mod(i,2) ~= 0
        iRow = [iRow; repmat((i+1)/2,A(i),1)];
        jCol = [jCol; (idx(i):idx(i+1)-1).'];
    end
end

B(sub2ind(size(B),iRow,jCol)) = 1;

[Fabio Freschi]

if you want B to be sparse (not a bad idea)

Bsp = sparse(iRow,jCol,1,(length(A)+1)/2,sum(A));

Answer 2

[Guillaume]

Here's a fairly simple way:

B = zeros((numel(A)+1)/2, sum(A));
for row = 1:2:numel(A)
    B(ceil(row/2), :) = repelem([0, 1, 0], [sum(A(1:row-1)), A(row), sum(A(row+1:end))]);
end