How to Fill matrix rows with another row?

Hello everyone,

I want fill the zero containing rows of this matrix with the row on top of it. For example, row (2:9) should be a copy of row one. Row (11:18), copy of row ten so on and so forth. This algorithm should be implemented till the end of (3240x9) matrix which is in the attachment.

1.843 6.97 -0.1 -0.001420 1.556600 -7.39140 2.1620 -3.3124 2.084

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

1.835 7.0751 -5.86 0.179 -0.0030 4.378 -3.37 1.476 -2.81

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

Answer 1

[KSSV]

Let A be your matrix.

[m,n] = size(A) ; 
for i = 1:m
    if ~any(A(i,:))
        A(i,:) = A(i-1,:) ; 
    end
end

Answer 2

[Fawad Farooq Ashraf]

Well you could find the indices of all the nonzero rows for a matrix A as

idx = find(~all(A==0,2));

and then you can replicate rows from idx(1)+1:idx(2)-1 equal to A(idx(1),:) using repmat function in matlab.

There could be a better logic for this but right now the one that's coming to my mind is something like this,

for i = 1:length(idx)-1
   A(idx(i)+1:idx(i+1)-1,:) = repmat(A(idx(i),:),length(idx(i)+1:idx(i+1)-1),1); 
end
A(idx(end)+1:size(A,1),:) = repmat(A(idx(end),:),length(idx(end)+1:size(A,1)),1);

You can develop your own logic but this kind of works as well.