Creation of binary coded image from a matrix

Hi

I am looking for a fast code to do the following:

• get a matrix with elements between 0 and 15
• create a new matrix, double the size (both rows and columns)
• for each element in the first matrix, set the binary representation of this element in the new matrix, in a 2x2 way.

first example

for a 1x1 input matrix:

A =

7

The result will be:

B =

0 1

1 1

Second example

for a 2x2 input matrix:

A =

0 2

1 3

The result will be:

B =

0 0 0 0

0 0 1 0

0 0 0 0

0 1 1 1

=

Currently, I do this with a slow for loop.

I would greatly apreciate your help.

Dror

[Walter Roberson]

for a 2x2 input matrix:

A =

0 2

1 3

The result will be:

B =

0 0 0 0

0 0 1 0

0 0 0 0

0 1 1 1

I do not understand the logic for that.

If the binary is intended to go across rows, then you have two outputs that are 0 0 0 0 which would require two elements that are the same, but no input elements are the same as each other.

If the binary is intended to go down columns, then you have two outputs that are [0; 0; 0; 1] which would require two elements that are the same, but no input elements are the same as each other.

[Voss]

Each element of the input becomes a 2x2 square in the output, e.g.

0 -> [0 0 0 0] -> [0 0; 0 0]
1 -> [0 0 0 1] -> [0 0; 0 1]
2 -> [0 0 1 0] -> [0 0; 1 0]
3 -> [0 0 1 1] -> [0 0; 1 1]
4 -> [0 1 0 0] -> [0 1; 0 0]
5 -> [0 1 0 1] -> [0 1; 0 1]
6 -> [0 1 1 0] -> [0 1; 1 0]
7 -> [0 1 1 1] -> [0 1; 1 1]
etc.

Answer 1

[Voss]

Here's one way, which will work for any non-empty matrix A.

A = [0 2; 1 3]

A = 2x2

0 2 1 3

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B = getB(A);
disp(B)

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

A larger example:

A = randi([0,15],5,6)

A = 5x6

10 12 7 7 3 9 9 11 10 9 12 2 2 12 3 0 5 14 5 4 10 14 8 5 5 9 4 8 13 2

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B = getB(A);
disp(B)

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

function B = getB(A)
    [r,c] = size(A);
    M = dec2bin(A,4)-'0';
    B = NaN(2*r,2*c);
    B(1:2:end,1:2:end) = reshape(M(:,1),r,[]);
    B(1:2:end,2:2:end) = reshape(M(:,2),r,[]);
    B(2:2:end,1:2:end) = reshape(M(:,3),r,[]);
    B(2:2:end,2:2:end) = reshape(M(:,4),r,[]);
end

[Dror Aizik]

Thank you very much.

This works perfectly and runs 50 times faster for large matrices.

[Voss]

You're welcome!

[Stephen23]

+1 very neat use of matrices and indexing.

Answer 2

[Stephen23]

A = [0,2;1,3;7,14]

A = 3x2

0 2 1 3 7 14

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Method one: BLOCKPROC (requires image toolbox, probably not very fast):

F = @(s)reshape(dec2bin(s.data,4)-'0',2,2).';
B = blockproc(A,[1,1],F)

B = 6x4

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

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Method two: RESHAPE, PERMUTE (should be reasonably fast):

S = size(A);
M = dec2bin(A(:),4)-'0';
B = reshape(permute(reshape(M,[S,2,2]),[4,1,3,2]),2*S)

B = 6x4

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

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Answer 3

[Sahas]

Hi @Dror Aizik,

From what I gather, you would like to transform an integer matrix into a binary-coded matrix with a specific pattern.

I was able to get the required matrix in a more efficient manner by using MATLAB’s “vectorization” of arrays method where I could get the required result in a much more optimized way, without using for-loops.

function B = bin_coded_matrix(A)
    % Get the size of the input matrix
    [m, n] = size(A);
    
    %Converting it to a column vector
    A_col = A(:);

    %Converting each element into 4-bit binary
    binRep = dec2bin(A_col, 4) - '0';

    %Reshaping the binary representation to 2x2 blocks as per the
    %requirement
    binMatrix = reshape(binRep.', 2, 2, []);
    binMatrix = permute(binMatrix, [2, 1, 3]);

    %Pre-allocating matrices for much faster results
    B = zeros(2*m, 2*n);

    %Placing the 2x2 binary blocks in the appropriate places in output
    for k = 1:numel(A_col)
        %Get the row and column indices for the output matrix
        [i, j] = ind2sub([m, n], k);

        %Placing the 2x2 binary block in the output matrix
        B(2*i-1:2*i, 2*j-1:2*j) = binMatrix(:, :, k);
    end
end

%Examples and timings
tic;
A3 = [7];
B3 = bin_coded_matrix(A3)
toc;

tic;
A4 = [0 2; 1 3];
B4 = bin_coded_matrix(A4)
toc;

tic; 
A6 = [0 1 2 3; 4 5 6 7; 8 9 10 11; 12 13 14 15];
B6 = bin_coded_matrix(A6)
toc;

Here is a snippet of the output matrices with different sizes and their execution times.

For learning more about vectorization, decimal to binary conversion and reshaping the matrix, refer to the MathWorks documentations below:

• Vectorization method for optimal matrix operations: https://www.mathworks.com/help/matlab/matlab_prog/vectorization.html
• Decimal to binary conversion with required bits: https://www.mathworks.com/help/matlab/ref/dec2bin.html
• Reshaping the matrix for the required output without using for loops: https://www.mathworks.com/help/matlab/ref/reshape.html https://www.mathworks.com/help/matlab/ref/permute.html

[Image Analyst]

"without using for-loops"???

" for k = 1:numel(A_col)"

Not that for loops are slow. It may have been true decades ago but it's a myth now. I've done tens of millions of iterations in less than a second.