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creating smaller matrix from a large matrix

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I want to create a 16*16 square matrix from 4*4 square matric based on a condition that 4*4 (i.e. 256) elements add together to form a new element for smaller matrix. Example is shown below.
  2 Commenti
Torsten
Torsten il 4 Ott 2023
If this is an IQ test, I didn't pass it...
Manoj Kumar V
Manoj Kumar V il 4 Ott 2023
All the 16 elements in the blue boxes were added to get a number 72. And all the 16 elements in yellow boxes were added to get a number 92. Likewise, I want to add all 4*4 (16 elements) to form a new smaller 4*4 matrix in matlab. It is about contracting the dimensions of a matrix.

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Risposta accettata

Voss
Voss il 4 Ott 2023
M = randi(10,16,16); % 16-by-16 matrix of integers between 1 and 10
disp(M);
7 4 5 5 2 3 4 10 10 5 1 8 8 4 1 4 3 5 6 7 10 6 6 8 7 5 3 2 10 5 6 4 7 4 9 6 1 9 8 7 2 8 3 8 2 7 4 5 3 10 6 7 9 7 6 1 9 3 7 5 3 2 7 5 8 8 7 7 2 7 4 10 4 3 5 5 6 8 1 5 4 9 2 4 9 3 9 10 10 7 6 3 7 8 3 10 9 7 1 2 10 5 4 6 5 6 9 2 9 10 4 10 10 3 7 2 6 4 10 3 5 6 2 6 2 5 5 1 7 6 7 8 7 8 2 2 5 8 7 2 5 4 10 2 10 9 10 9 6 5 10 9 5 2 8 6 3 8 8 1 10 7 8 2 6 9 9 3 5 2 10 3 4 8 1 4 1 10 2 8 8 8 8 2 4 2 7 5 9 10 10 2 1 9 1 3 3 7 3 1 8 5 4 3 4 1 1 6 7 10 3 6 2 5 3 8 3 1 1 4 1 1 1 7 3 4 3 2 9 2 8 9 4 2 10 8 6 8 8 1 5 6 7 4 3 1 4 8 3 6 10 3 6 5 6 10
siz = [4 4]; % size of smaller matrices to divide M into
args = arrayfun(@(n,s)n*ones(1,s/n),siz,size(M),'UniformOutput',false);
m = cellfun(@(b)sum(b,'all'),mat2cell(M,args{:}))
m = 4×4
94 97 86 77 90 102 84 94 114 102 81 89 74 76 75 72
% check
m(1,1) == sum(M(1:4,1:4),'all')
ans = logical
1
m(2,1) == sum(M(5:8,1:4),'all')
ans = logical
1
m(3,4) == sum(M(9:12,13:16),'all')
ans = logical
1
  1 Commento
Voss
Voss il 4 Ott 2023
Or using some loops:
M = randi(10,16,16); % 16-by-16 matrix of integers between 1 and 10
disp(M)
9 2 2 5 4 1 5 9 5 5 1 10 3 3 3 3 1 7 6 10 4 2 4 3 7 3 1 3 6 3 7 9 10 2 9 8 8 8 1 8 1 7 2 7 3 6 3 1 3 8 6 2 4 8 1 10 5 10 7 4 8 7 5 10 8 9 5 4 10 10 8 2 7 3 6 9 2 2 4 8 3 2 9 3 7 10 7 4 3 6 8 8 2 4 2 7 8 10 4 5 9 8 5 4 2 5 1 4 5 4 9 3 9 10 3 7 1 5 4 5 1 1 7 1 9 5 7 7 9 1 10 8 3 7 6 6 8 1 9 1 9 1 8 5 4 5 3 6 6 9 4 8 1 10 2 9 7 10 7 8 7 2 9 9 5 1 10 1 7 3 9 2 4 1 4 2 10 5 6 9 5 7 3 8 3 9 8 8 3 8 7 3 5 4 9 2 4 5 6 1 7 1 9 1 3 5 8 9 7 8 8 4 4 5 4 4 8 3 7 7 5 4 10 10 5 7 5 4 1 7 7 10 7 9 10 4 10 9 8 6 3 6 7 4 8 5 1 2 6 8 6 2 1 5 7 3
siz = [4 4]; % size of smaller matrices to divide M into
n = size(M)./siz;
m = zeros(n);
for ii = 1:n(1)
for jj = 1:n(2)
m(ii,jj) = sum(M((ii-1)*siz(1)+(1:siz(1)),(jj-1)*siz(2)+(1:siz(2))),'all');
end
end
disp(m)
90 80 78 80 99 99 72 80 103 89 90 87 88 74 95 103
% check
m(1,1) == sum(M(1:4,1:4),'all')
ans = logical
1
m(2,1) == sum(M(5:8,1:4),'all')
ans = logical
1
m(3,4) == sum(M(9:12,13:16),'all')
ans = logical
1

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Più risposte (2)

Torsten
Torsten il 4 Ott 2023
Modificato: Torsten il 4 Ott 2023
You mean
rng("default")
A = rand(16);
for i = 1:4
for j = 1:4
B = A((i-1)*4+1:i*4,(j-1)*4+1:j*4);
A_compressed(i,j) = sum(B(:));
end
end
A_compressed
A_compressed = 4×4
9.5135 8.3613 7.5381 7.0071 7.4615 8.0673 5.5216 8.6224 10.1127 6.9479 8.0587 7.2866 7.9292 8.0528 7.6726 7.9785
?
  2 Commenti
Manoj Kumar V
Manoj Kumar V il 4 Ott 2023
Thank You Torsten. There seems to be numbers with decimal points for the answer, which shall not be expected. Voss has explained with the example.
Torsten
Torsten il 4 Ott 2023
Modificato: Torsten il 4 Ott 2023
Well, you didn't mention the condition that the matrix elements should be integers - so I generated doubles for the 16x16 matrix. But no problem ...

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Adam Danz
Adam Danz il 4 Ott 2023
Modificato: Adam Danz il 4 Ott 2023
Here are two more options. Neither are very readable but I knew there was a grouping solution and a vectorized solution and I was in the mood for a challenge.
Grouping solution
The variable group is a matrix the same size as A that groups the values of A into 4x4 sub-matrixes.
rng("default")
A = randi(10,16);
group = repelem(reshape(1:16,4,4)',4,4);
sumVec = groupsummary(A(:),group(:),'sum'); % alternative: =splitapply(@sum,A(:),group(:));
m = reshape(sumVec,4,4)'
m = 4×4
104 92 85 80 82 88 62 94 107 78 87 83 87 87 86 89
Vectorized solution
This reshapes matrix A into a 4-dimensional array and then permutes the array so that sum can be applied acros the first two dimensions. Then the results are reshaped back into a matrix.
m = reshape(sum(permute(reshape(A',4,4,4,4),[1 3 2 4]),[1,2]),4,4)'
m = 4×4
104 92 85 80 82 88 62 94 107 78 87 83 87 87 86 89

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