Add a number to a region of cells bounded by nonzero numbers in a matrix

If I have for example, A matrix of any size, with data surrounded by zeros.

Notice that within the region of interest, the area bounded by data there are some zeros within. However, the surrounding zeros are not useful data (zero strain from the background, zero strain within the sample). However I want to + 0.1 to the entire region of numerical (nonzero) data, which is my sample.

I have tried to + 0.1 by using the nonzero function for matrixes, but have noticed the zeros within the region of interest gets left behind so I was hoping to try and define the region boundaries by taking the min and max horizontal value, and vertical value, and then + 0.1 on this region. It would be even better if I could identify the pockets of zeros surrounded by nonzero numbers, and add + 0.1 to this.

A = [0 0 0 0 0 0 0 0 0 0; 
    0 0 0 0 3 9 2 1 3 0;
    0 0 2 4 1 0 0 0 0 0;
    0 0 3 3 2 4 9 2 0 0;
    0 5 1 3 0 0 4 2 0 0;
    0 0 2 1 3 0 0 4 8 0;
    0 2 1 2 3 3 1 3 7 0;
    0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0]

 B = A;
 isNonzero = A~=0;
 B(isNonzero) = B(isNonzero)+0.1;
 % This adds +0.1 to the non zero values but does not consider the zeros
 % encapsulated by the other numbers
 

I would hope to get something like this:

 B = [0 0 0 0 0 0 0 0 0 0; 
    0 0 0 0 3.1 9.1 2.1 1.1 3.1 0;
    0 0 2.1 4.1 1.1 0 0 0 0 0;
    0 0 3.1 3.1 2.1 4.1 9.1 2.1 0 0;
    0 5.1 1.1 3.1 0.1 0.1 4.1 2.1 0 0;
    0 0 2.1 1.1 3.1 0.1 0.1 4.1 8.1 0;
    0 2.1 1.1 2.1 3.1 3.1 1.1 3.1 7.1 0;
    0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0]

[Mathieu NOE]

hello

maybe the start of a solution :

matlab - Determine matrix boundary without loops - Stack Overflow

and do the correction of the data inside the boundary

Answer 1

[Jan]

A = [0 0 0 0 0 0 0 0 0 0; 
    0 0 0 0 3 9 2 1 3 0;
    0 0 2 4 1 0 0 0 0 0;
    0 0 3 3 2 4 9 2 0 0;
    0 5 1 3 0 0 4 2 0 0;
    0 0 2 1 3 0 0 4 8 0;
    0 2 1 2 3 3 1 3 7 0;
    0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0];

M = (A ~= 0);
M = cumsum(M, 1, 'forward') & cumsum(M, 1, 'reverse') & ...
    cumsum(M, 2, 'forward') & cumsum(M, 2, 'reverse');
A(M) = A(M) + 0.1

A = 9×10

0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.1000 9.1000 2.1000 1.1000 3.1000 0 0 0 2.1000 4.1000 1.1000 0 0 0 0 0 0 0 3.1000 3.1000 2.1000 4.1000 9.1000 2.1000 0 0 0 5.1000 1.1000 3.1000 0.1000 0.1000 4.1000 2.1000 0 0 0 0 2.1000 1.1000 3.1000 0.1000 0.1000 4.1000 8.1000 0 0 2.1000 1.1000 2.1000 3.1000 3.1000 1.1000 3.1000 7.1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

[Nicole Wan]

Thanks for your help it does exactly what I wanted it to do!

Answer 2

[Voss]

Here's one way:

A = [0 0 0 0 0 0 0 0 0 0; 
     0 0 0 0 3 9 2 1 3 0;
     0 0 2 4 1 0 0 0 0 0;
     0 0 3 3 2 4 9 2 0 0;
     0 5 1 3 0 0 4 2 0 0;
     0 0 2 1 3 0 0 4 8 0;
     0 2 1 2 3 3 1 3 7 0;
     0 0 0 0 0 0 0 0 0 0;
     0 0 0 0 0 0 0 0 0 0];
isNonzero = A~=0;
idx = cummax(isNonzero,1) ...
    & cummax(isNonzero,1,'reverse') ...
    & cummax(isNonzero,2) ...
    & cummax(isNonzero,2,'reverse')

idx = 9×10 logical array
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   1   1   0
   0   0   1   1   1   0   0   0   0   0
   0   0   1   1   1   1   1   1   0   0
   0   1   1   1   1   1   1   1   0   0
   0   0   1   1   1   1   1   1   1   0
   0   1   1   1   1   1   1   1   1   0
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0

A(idx) = A(idx)+0.1

A = 9×10

0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.1000 9.1000 2.1000 1.1000 3.1000 0 0 0 2.1000 4.1000 1.1000 0 0 0 0 0 0 0 3.1000 3.1000 2.1000 4.1000 9.1000 2.1000 0 0 0 5.1000 1.1000 3.1000 0.1000 0.1000 4.1000 2.1000 0 0 0 0 2.1000 1.1000 3.1000 0.1000 0.1000 4.1000 8.1000 0 0 2.1000 1.1000 2.1000 3.1000 3.1000 1.1000 3.1000 7.1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

[Jan]

How nice: Two brains, one idea.

[Voss]

Haha, exactly.

[Nicole Wan]

Thank you!!

[Voss]

You're welcome!

Answer 3

[Image Analyst]

Looks like no one on this thread is familiar with the functions in the Image Processing Toolbox that make this operation trivial, and a lot simpler than the other code.

A = [0 0 0 0 0 0 0 0 0 0; 
    0 0 0 0 3 9 2 1 3 0;
    0 0 2 4 1 0 0 0 0 0;
    0 0 3 3 2 4 9 2 0 0;
    0 5 1 3 0 0 4 2 0 0;
    0 0 2 1 3 0 0 4 8 0;
    0 2 1 2 3 3 1 3 7 0;
    0 0 0 0 0 0 0 0 0 0;
    0 0 0 0 0 0 0 0 0 0];
% Find non-zeros and fill "pockets/holes" of zeros.
mask = imfill(A ~= 0, 'holes')

mask = 9×10 logical array
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   1   1   1   1   1   0
   0   0   1   1   1   0   0   0   0   0
   0   0   1   1   1   1   1   1   0   0
   0   1   1   1   1   1   1   1   0   0
   0   0   1   1   1   1   1   1   1   0
   0   1   1   1   1   1   1   1   1   0
   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0

% Add 0.1 everywhere that the mask is true:
A(mask) = A(mask) + 0.1

A = 9×10

0 0 0 0 0 0 0 0 0 0 0 0 0 0 3.1000 9.1000 2.1000 1.1000 3.1000 0 0 0 2.1000 4.1000 1.1000 0 0 0 0 0 0 0 3.1000 3.1000 2.1000 4.1000 9.1000 2.1000 0 0 0 5.1000 1.1000 3.1000 0.1000 0.1000 4.1000 2.1000 0 0 0 0 2.1000 1.1000 3.1000 0.1000 0.1000 4.1000 8.1000 0 0 2.1000 1.1000 2.1000 3.1000 3.1000 1.1000 3.1000 7.1000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0