Please help me in generating a matrix of ones for a given matrix

Suppose if I already have a matrix 'X' having only one '1' in each row for example matrix given below

X = [0     0     0     0     1     0     0     0     0     0;
     0     1     0     0     0     0     0     0     0     0;
     0     0     1     0     0     0     0     0     0     0;
     1     0     0     0     0     0     0     0     0     0;
     0     0     0     0     0     1     0     0     0     0]

i need a matrix a few (specified number) '1's after existing '1' in each row for example output matrix is

Y = [0     0     0     0     1     1     1     1     0     0;
     0     1     1     1     1     1     0     0     0     0;
     0     0     1     1     1     0     0     0     0     0;
     1     1     0     0     0     0     0     0     0     0;
     0     0     0     0     0     1     1     1     1     0]

code for matrix 'X'is

P = 5; % The number of competitive projects
T = 10; % The number of time periods
D = [4; 5; 3; 2; 4];
excludedcount = D-1;
X = zeros(P,T);
for row = 1:P
   rv = [1, zeros(1, T - excludedcount(row) - 1)];
   X(row, 1 : (T - excludedcount(row))) = rv(randperm(numel(rv)));
end

[Stephen23]

Question continued here:

http://www.mathworks.com/matlabcentral/answers/293884-please-help-me-in-inserting-ones-in-a-given-matrix-in-special-rows

See also my two-line vectorized solution here:

http://www.mathworks.com/matlabcentral/answers/293884#answer_227811

Answer 1

[dpb]

"Dead-ahead" solution is simply

D = [4; 5; 3; 2; 4];
for row = 1:P
  ic=find(X(row,:);             % the '1' column index
  X(row,ic:ic+D-1)=ones(1,D);   % insert the ones vector at that location
end

With some thought this could undoubtedly be transformed to use arrayfun and the multiple outputs from find on the array w/o the explicit loop; whether that would be any faster is questionable I'd venture...

[MANISH KUMAR]

if any row contains zeros only then it shows error in this line

X(row,ic:ic+D(row)-1)=ones(1,D(row));

For example matrix is

Y =

     0     1     0     0     0     0     0     0     0     0
     0     0     0     1     0     0     0     0     0     0
     0     0     0     0     0     0     1     0     0     0
     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0

in this matrix last two rows contains zeros only. So this code doesn't work.