Hi everyone,
Consider a large MxN matrix of randomly placed ones and zeros (call it A). I need to create a matrix (call it B) that shifts the location of the ones in the original matrix to the right depending on the number of preceding ones in the row. For example, consider an arbitrary row:
A(i,:) = [0 1 0 0 0 1 0 0 0 1 1 1 1 0 0 0];
This should become:
B(i,:) = [0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0];
Because the number of ones in each row is random, B will need to have as many columns as necessary to accommodate the row with the most ones. Trailing zeros can be added to the end of any row to equalize the number of columns. Note that consecutive ones stay grouped together.
I have implemented this using a loop (see MWE below). But because M and N are both large, my current code is relatively slow.
M=10000;
N=1000;
out0 = binornd(1,0.3,M,N);
out_cum0 = zeros(M,N);
out_cum1 = zeros(M,N);
for i=2:N
out_cum0(:,i)=out_cum0(:,i-1)+out0(:,i-1);
if i>1
out_cum1( and(out0(:,i-1)==1,out0(:,i)==1 ), i ) = out_cum1( and(out0(:,i-1)==1,out0(:,i)==1 ), i-1 );
out_cum1(~and(out0(:,i-1)==1,out0(:,i)==1 ), i ) = out_cum0(~and(out0(:,i-1)==1,out0(:,i)==1 ), i );
end
end
out1 = zeros(M,2*N);
loc0 = find(out0==1);
[row,col] = ind2sub(size(out0),loc0);
loc1 = sub2ind(size(out1),row,col+out_cum1(loc0));
out1(loc1) = 1;
I suspect there's a faster way to do this, particularly the loop, but I'm not sure what it is. Any suggestions? Thanks in advance.
