Binarize a metrix with some threshold

Hi Matlab experts,

How can I binarize a matrix according to a certain threshold? for exapmle the highest 400 values or the highest 20% of the values?

Thanks!

Tamir

Answer 1

[Bhaskar R]

A = randi([0, 1000], 5)  % dummy marix

A =

   392   277   317   766   646
   656    46   951   795   710
   171    97    34   187   755
   706   824   439   490   276
    31   695   381   446   680

>> A>400    % command to compare with 400 threshold

ans =

  5×5 logical array

   0   0   0   1   1
   1   0   1   1   1
   0   0   0   0   1
   1   1   1   1   0
   0   1   0   1   1

>> A>(400*0.2)

ans =

  5×5 logical array

   1   1   1   1   1
   1   0   1   1   1
   1   1   0   1   1
   1   1   1   1   1
   0   1   1   1   1

[Tamir Eisenstein]

Dear Bhaskar, thank you, but I think you misunderstood my two examples:

1) Let's say I have a matrix with 1000 components - I want the highest 400 values to be assigned a value of 1, and the rest to become 0.

2) Let's say I have a matrix with 1000 components - I want the highest 20% values to be assigned a value of 1, and the rest to become 0.

Answer 2

[Arthur Roué]

You can use the 2nd output of sort function.

The code bellow should do what you want.

% Random matrix
M = randi([0, 1000], 5);

M =
   699   480   806    28   500
   198   905   577   490   471
    30   610   183   168    59
   744   618   240   979   682
   500   860   887   713    42
   
% Sort
[~,I1] = sort(M(:));
[~,I2] = sort(flipud(I1)); % flip for ascending order

% Reshape
Index = (1:numel(M))';
Index = Index(I2);
Index = reshape(Index, size(M));

Index =
     8    16     5    25    13
    19     2    12    15    17
    24    11    20    21    22
     6    10    18     1     9
    14     4     3     7    23
    
% For the 5 greatest values
Index <= 5

ans =

  5×5 logical array

   0   0   1   0   0
   0   1   0   0   0
   0   0   0   0   0
   0   0   0   1   0
   0   1   1   0   0

[Image Analyst]

Or you can use the other input of sort() to avoid flipping:

[sortedI1, sortOrder] = sort(I1, 'ascend'); % No need to flip for ascending order

[Arthur Roué]

Indeed, you could to that, thanks for the complement.