Replace NaN with nearest numerical value to the left in the same row

Greetings, I have a matrix with numerical and NaN values. I am searching for a vectorized solution to replace all of the NaN values with the nearest numerical value to the left in the same row.

I am guaranteed the first column will always be completely populated with numerical values.

For example, if I start with this matrix:

       8   NaN   NaN
       3   NaN     7
       4     9   NaN

I should end up with this:

     8     8     8
     3     3     7
     4     9     9

I am using the loop below to accomplish this: [~,cols] = size(array);

for col = 2:cols
    array(isnan(array(:,col)),col) = array(isnan(array(:,col)),col-1);
end

Does anyone know how I can do this without a loop?

Thanks, Miles

Answer 1

[Andrei Bobrov]

lo = ~isnan(z);
m = size(z,1);
out = z(bsxfun(@plus,(cumsum(lo,2)-1)*m,(1:m)'));
out(lo) = z(lo)

Answer 2

[Image Analyst]

This works:

m = [8   NaN   NaN
  3   NaN     7
  4     9   NaN]

nanMap = isnan(m)
m(nanMap) = -inf
mOut = imdilate(m, [1, 1])
mOut = imdilate(mOut, [1, 1])

Answer 3

[Roger Stafford]

Let A be your original matrix. The following is vectorized but I doubt if it is at all superior to, or even the equal of, your simple for-loop.

   B = A.’; % To use ‘find’ we have to convert transpose of A to a column vector
   t = [~isnan(B(:));true];
   f = find(diff([true;t])~=0); % Now we can use ‘find’
   t = cumsum(t+accumarray(f(2:2:end),f(2:2:end)-f(1:2:end-1),size(t)));
   A = reshape(B(t(1:end-1)),size(B)).’;

For the above to work properly it is necessary that the left side of each row of A be numerical as promised.