remove negative cells in a matrix by averaging

Hello,

I have a matrix 700*700*1 in which there are some negative numbers. these negative numbers have no physical interpretation for my work and are caused by an error that has to be fixed. I need to convert these negative numbers into positive ones by putting them equal to the average of their adjacent cells (8 adjacent cells). I will really appreciate it if someone can help me in this regard.

Best regards

[Azzi Abdelmalek]

For example the cell A(10,20): What is its adjacent cells?

[Ali Faridkhou]

Dear Azzi those would be like this

A(9,19) A(9,20) A(9,21) A(10,19) A(10,21) A(11,19) A(11,20) A(11,21)

Thank you

[Cedric]

Is it possible that the average over neighbors is negative due to some clustering of negative values? What is the density of negative values, or roughly how many negative values do you have in this array?

If your matrix is A, you can visualize negative values by evaluating

 spy(A < 0)

which also indicates the number of non-zero elements, or you can count them with

 sum(A(:) < 0)

Answer 1

[Youssef Khmou]

hi, you can try :

 % Given your matrix A :

1)

    A=randn(10);
    A(A<0)=A(A<0)*-1

2) Or use your idea :

 for x=2:size(A,1)-1
  for y=2:size(A,2)-1

   if A(x,y)<0
    A(x,y)=A(x-1,y-1)+A(x-1,y)+A(x-1,y+1)+A(x,y-1)+A(x,y+1)+...
   A(x+1,y-1)+A(x+1,y)+A(x+1,y+1);
    A(x,y)=A(x,y)/8;
 end
 end
 end

But the first solution is Very efficient, because this second one does not treat the boundaries

Answer 2

[Cedric]

See my comment below your question and note that none of the solutions below solves the problem of managing cases where the average over neighbors is negative.

1. If you have a large amount of negative values:

..you can build a global array of averages over neighbors using CONV2, and use its value to replace negative values in A:

 k = [1, 1, 1; 1, 0, 1; 1, 1, 1] ;
 M = conv2(A, k, 'same') ./ conv2(ones(size(A)), k, 'same') ;
 A(A < 0) = M(A < 0) ;

2. If you have only a few negative values:

.. you can loop over them and compute the average "manually":

 [r,c] = find(A < 0) ;
 for k = 1 : numel(r)
    rId  = r(k) + [1, 1, 1, 0, 0, -1, -1, -1] ;
    cId  = c(k) + [1, 0, -1, 1, -1, 1, 0, -1] ;
    isOk = rId > 0 & rId <= size(A,1) & cId > 0 & cId <= size(A,2) ;
    ind  = sub2ind(size(A), rId(isOk), cId(isOk)) ;
    A(r(k),c(k)) = mean(A(ind)) ;
 end

Note 1: you could partially solve the issue of negative means by setting all remaining negative elements of A to 0 after executing either solution above.

 A(A<0) = 0 ;

PS: I've not fully tested these solutions, so there a still a bit of work on your side.