Minimum width across any part of a connected component in matlab

In order to label connected components in a matrix composed of double format values (not binary values), I am using a Matlab Central File Exchange program: LABEL

So, let's say I have a double format matrix of the form:

A =     [1 3 3 3 3 6 6; 
         1 3 3 0 6 6 6;
         1 3 3 0 4 4 6;
         1 2 2 0 4 4 6;
         1 2 2 0 5 5 6;
         1 1 1 0 5 5 6;
         1 1 1 0 5 5 6]

If I run the above mentioned LABEL function with 4-connectivity, I get the results in the form of labels, area of each connected components and number of connected components.

Labels:

1	2	2	2	2	5	5
1	2	2	4	5	5	5
1	2	2	4	6	6	5
1	3	3	4	6	6	5
1	3	3	4	7	7	5
1	1	1	4	7	7	5
1	1	1	4	7	7	5

Area of each connected component:

11	8	8	8	8	10	10
11	8	8	6	10	10	10
11	8	8	6	4	4	10
11	4	4	6	4	4	10
11	4	4	6	6	6	10
11	11	11	6	6	6	10
11	11	11	6	6	6	10

and 7 number of components.

If we look at the A matrix that I have written above, we can see the connected component with value 1 is only 1 cell wide at some places (see first column of A matrix). Similarly for components with elements 0, 6, 3.

Is there any way that I can replace the cells which are only 1 cell wide (in any component) with NaN values. Basically, from all the components, I would like to remove the parts which are only 1 cell wide, and replace those parts (cells) with NaN values.

My attempts: I tried using the ratio of horizontal width to vertical width of a

 BoundingBox

and

Eccentricity

from

regionprops

function, but I could not achieve any satisfactory results.

Can anybody please help me out with this? Thanks!

P.S. : The "1" cell wide value can be any threshold that we can decide in our code. If possible, I would like to be able to remove any parts which are only "threshold" cells wide (threshold can be 1 or 2 or any positive integer).

[Ankit Sahay]

So I came across this answer by Image Analyst about Solidity, where he mentions that

"a thin "L" shape (or T or E, etc.) would have a very low solidity. The more bays, nooks, and crannies it has, the lower the solidity."

Maybe this might be of some use. I will give an update if this solves my problem.

Answer 1

[Cris LaPierre]

Here's a quick and simple way to do it

for c = 0:max(A,[],'all')
    tmp = sum(A==c,2);
    if any(tmp==1)
        A(A==c)=missing;
    end
end

A = 7×7    
   NaN     3     3     3     3   NaN   NaN
   NaN     3     3   NaN   NaN   NaN   NaN
   NaN     3     3   NaN     4     4   NaN
   NaN     2     2   NaN     4     4   NaN
   NaN     2     2   NaN     5     5   NaN
   NaN   NaN   NaN   NaN     5     5   NaN
   NaN   NaN   NaN   NaN     5     5   NaN

[Ankit Sahay]

This looks good, but I had a slightly different objective in mind. I would like to remove only those parts (of any component) that are 1 cell wide, not the whole component.

For example: In the original matrix A in my question, if we focus on the cells marked 1 (i.e. left and bottom-left parts of the matrix), I would like to have the following result:

NaN 3 3 3 3 6 6

NaN 3 3 0 6 6 6

NaN 3 3 0 4 4 6

NaN 2 2 0 4 4 6

NaN 2 2 0 5 5 6

1 1 1 0 5 5 6

1 1 1 0 5 5 6

Thank you for your prompt answer. It does solve a part of my problem (does remove the cells marked 0!). I will try to see if I can work further on your answer.

[Ankit Sahay]

If I use a slightly modified version of your answer, I am able to solve a part of my problem, but then I get incorrect results with cells marked 0 (4th column), and cells marked 6 (right side of original A matrix).

for c = 0:max(A,[],'all')
    tmp = sum(A==c,2);
    tmp1 = sum(A==c,1);
    if any(tmp==1) && any(tmp1==1)
        A(A==c)=missing;
    end
end
A
A =
   NaN     3     3     3     3   NaN   NaN
   NaN     3     3     0   NaN   NaN   NaN
   NaN     3     3     0     4     4   NaN
   NaN     2     2     0     4     4   NaN
   NaN     2     2     0     5     5   NaN
     1     1     1     0     5     5   NaN
     1     1     1     0     5     5   NaN

[Cris LaPierre]

for c = 0:max(A,[],'all')
    tmp = sum(A==c,2);
    A(A==c & tmp==1)=missing;
end

A = 7×7    
   NaN     3     3     3     3     6     6
   NaN     3     3   NaN     6     6     6
   NaN     3     3   NaN     4     4   NaN
   NaN     2     2   NaN     4     4   NaN
   NaN     2     2   NaN     5     5   NaN
     1     1     1   NaN     5     5   NaN
     1     1     1   NaN     5     5   NaN

[Ankit Sahay]

This is perfect. A very small modification I need to do is to repeat the same computation along dimension 1 also. Thank you so much!

A =
     1     3     3     3     3     6     6
     1     3     3     0     6     6     6
     1     3     3     0     4     4     6
     1     2     2     0     4     4     6
     1     2     2     0     5     5     6
     1     1     1     0     5     5     6
     1     1     1     0     5     5     6
     
     
A_new = A;
for c = 0:max(A,[],'all')
    tmp = sum(A==c,2);
    A_new(A==c & tmp==1)=missing;
end

for c = 0:max(A,[],'all')
    tmp = sum(A==c,1);
    A_new(A==c & tmp==1)=missing;
end


A_new =
   NaN     3     3   NaN   NaN     6     6
   NaN     3     3   NaN   NaN     6     6
   NaN     3     3   NaN     4     4   NaN
   NaN     2     2   NaN     4     4   NaN
   NaN     2     2   NaN     5     5   NaN
     1     1     1   NaN     5     5   NaN
     1     1     1   NaN     5     5   NaN

[Ankit Sahay]

Hi Cris! Would it be possible for you to explain how your answer works?

for c = 0:max(A,[],'all')
    tmp = sum(A==c,2);
    A(A==c & tmp==1)=missing;
end

I have a few minor issues in my computations, and I would like to solve them myself. I tried debugging the code to understand each line of your answer, but I couldn't. Or you can direct to me some similar answer which I can refer to for understanding your code. I tried going through this answer, but it didn't help much.

Thanks!

[Cris LaPierre]

Try removing the semi-colons and then run the code. You could also try splitting the code into parts to see what each bit is doing.

c=1
A==c
tmp = sum(A==c,2)
A(A==c)=missing
A(tmp==1)=missing
A(A==c & tmp==1)=missing

Change the value of c to see how it handles each value.

[Ankit Sahay]

Okay. Thanks!