Linear indexing does this simply and efficiently. The trick is to work down the columns, which requires transposing:
>> A = [1,3,0,2,5;0,2,0,1,0;2,0,0,0,0;3,1,1,0,0]
A =
1 3 0 2 5
0 2 0 1 0
2 0 0 0 0
3 1 1 0 0
>> N = 3;
>> Z = A.';
>> S = size(Z);
>> [~,R] = sort(Z==0,1);
>> [~,C] = ndgrid(1:N,1:S(2));
>> X = sub2ind(S,R(1:N,:),C);
>> B = Z(X).'
B =
1 3 2
2 1 0
2 0 0
3 1 1
Probably the most efficient approach would be to use a simple loop, e.g. (not particularly optimized):
R = size(A,1);
B = zeros(R,N);
for k = 1:R
tmp = nonzeros(A(k,:));
idx = 1:min(N,numel(tmp));
B(k,idx) = tmp(idx);
end
Some timings (1e3 iterations):
Elapsed time is 5.358 seconds. % madhan ravi's with loop and CELLFUN
Elapsed time is 0.606 seconds. % my answer with SUB2IND
Elapsed time is 0.265 seconds. % my answer with loop and indexing
