在矩阵的每一行中找到前N个非零元素

Question

I have a matrix in MATLAB with zeroes and I would like to get another matrix with the first N non-zero elements in each row. Let's say for example N = 3 , and the matrix is

A = [ 0 0 2 0 6 7 9;
      3 2 4 7 0 0 6;
      0 1 0 3 4 8 6;
      1 2 0 0 0 1 3]

I'd like the result to be:

B = [2 6 7;
     3 2 4;
     1 3 4;
     1 2 1]

I have a huge matrix so I would like to do it without a loop, could you please help me? Thanks a lot!

Answer 1

Since MATLAB stores a matrix according to column-major order , I first transpose A , bubble up the non-zeros, and pick the first N lines, and transpose back:

N = 3;
A = [ 0 0 2 0 6 7 9;
      3 2 4 7 0 0 6;
      0 1 0 3 4 8 6;
      1 2 0 0 0 1 3];

Transpose and preallocate output B

At = A';
B = zeros(size(At));

At =
     0     3     0     1
     0     2     1     2
     2     4     0     0
     0     7     3     0
     6     0     4     0
     7     0     8     1
     9     6     6     3

Index zeros

idx = At == 0;

idx =
     1     0     1     0
     1     0     0     0
     0     0     1     1
     1     0     0     1
     0     1     0     1
     0     1     0     0
     0     0     0     0

Bubble up the non-zeros

B(~sort(idx)) = At(~idx);

B =
     2     3     1     1
     6     2     3     2
     7     4     4     1
     9     7     8     3
     0     6     6     0
     0     0     0     0
     0     0     0     0

Select first N rows and transpose back

B(1:N,:)'

You can do the bubbling in row-major order, but you would need to retrieve the row and column subscripts with find, and do some sorting and picking there. It becomes more tedious and less readable.

Answer 2

Using accumarray with no loops:

N = 3;
[ii,jj] = find(A); [ii,inds]=sort(ii); jj = jj(inds);
lininds = ii+size(A,1)*(jj-1);
C = accumarray(ii,lininds,[],@(x) {A(x(1:N)')}); %' cell array output
B = vertcat(C{:})
B =
     2     6     7
     3     2     4
     1     3     4
     1     2     1

Answer 3

Usually I don't go with a for loop solution, but this is fairly intuitive:

N = 3;
[ii,jj] = find(A);
B = zeros(size(A,1),N);
for iRow = 1:size(A,1),
    nzcols = jj(ii==iRow);
    B(iRow,:) = A(iRow,nzcols(1:N));
end

Since you are guaranteed to have more than N nonzeros per row of A , that should get the job done.

Answer 4

N = 3;
for ii=1:size(A,1);
    B(ii,:) = A( ii,find(A(ii,:),N) );
end

Answer 5

One-liner solution:

B = cell2mat(cellfun(@(c) c(1:N), arrayfun(@(k) nonzeros(A(k,:)), 1:size(A,1), 'uni', false), 'uni', false)).'

Not terribly elegant or efficient, but so much fun!

Answer 6

Actually , you can do it like the code blow:

---

N=3
 for n=1:size(A,1)
   [a b]=find(A(n,:)>0,N);
   B(n,:)=A(n,transpose(b));
end

---

Then I think this B matrix will be what you want.