Comparing Matrices in a Struct

Question

Chris Dan 2019-10-27

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/487659-comparing-matrices-in-a-struct

编辑： per isakson 2019-11-10

采纳的回答： per isakson

Hello Guys, I am new to MATLAB, I have some matrices in a struct more like an array of arrays.

I want to compare their sizes and see if they are equal or not.

and how to make them equal to the biggest matrice, maybe by adding zeros to the smaller matrices.

I am attaching a picture

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Answer 1

per isakson 2019-10-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/487659-comparing-matrices-in-a-struct#answer_398624

编辑：per isakson 2019-11-8

在 MATLAB Online 中打开

Try this

S(1).model_data = rand( 4, 10 );
S(2).model_data = rand( 2, 10 );
S(3).model_data = rand( 3, 11 );
Out = cssm_( S );
function    Out = cssm_( S )
    
    sz(1) = max( arrayfun( @(s) size( s.model_data, 1 ), S ) );
    sz(2) = max( arrayfun( @(s) size( s.model_data, 2 ), S ) );
    
    Out = struct( 'model_data', repmat( {nan(sz)}, 1,numel(S) ) );
    
    for jj = 1 : numel(S)
        sz = size(S(jj).model_data);
        Out(jj).model_data(1:sz(1),1:sz(2)) = S(jj).model_data;
    end
end

In response to comment

This will handle sparse, test it. ( cssm_ assumes that all values of data_model are either full or sparse.)

S(1).model_data = sparse( rand( 4, 10 ) );
S(2).model_data = sparse( rand( 2, 10 ) );
S(3).model_data = sparse( rand( 3, 11 ) );
Out = cssm_( S );
function    Out = cssm_( S )
    
    sz(1) = max( arrayfun( @(s) size( s.model_data, 1 ), S ) );
    sz(2) = max( arrayfun( @(s) size( s.model_data, 2 ), S ) );
    
    if issparse( S(1).model_data )
        Out = struct( 'model_data', repmat( {sparse(nan(sz))}, 1,numel(S) ) );
    else
        Out = struct( 'model_data', repmat( {nan(sz)}, 1,numel(S) ) );
    end
    
    for jj = 1 : numel(S)
        sz = size(S(jj).model_data);
        Out(jj).model_data(1:sz(1),1:sz(2)) = S(jj).model_data;
    end
end

Version 3 in response to a later comment

S(1,1).model_data = sparse( rand( 4, 3 ) );
S(1,2).model_data = sparse( rand( 2, 3 ) );
S(1,3).model_data = sparse( rand( 3, 5 ) );
Out = cssm_( S );
function    Out = cssm_( S )
    
    sz(1) = max( arrayfun( @(s) size( s.model_data, 1 ), S ) );
    sz(2) = max( arrayfun( @(s) size( s.model_data, 2 ), S ) );
    
    if issparse( S(1).model_data )
        Out = struct( 'model_data', repmat( {sparse(nan(sz))}, size(S) ) );
    else
        Out = struct( 'model_data', repmat( {nan(sz)}, size(S) ) );
    end
    
    for jj = 1 : numel(S)
        sz = size(S(jj).model_data);
        Out(jj).model_data(1:sz(1),1:sz(2)) = S(jj).model_data;
    end
end

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Chris Dan 2019-11-10

编辑：per isakson 2019-11-10

在 MATLAB Online 中打开

per isakson

Hey, Thanks again for you help!! :) :)

I just have one more question, it is a different problem but is related matrix addition.

I am copying my code, the thing is I have the logic developed somehow, but it is not giving me correct answers.

B = [ 1 -1 1;
     -1  1 1;
      1 -1 1];
C = [ 1 -1 2;
     -1  1 2;
      4  5 5];
D = [ 1 -1 1;
     -1  1 4;
      5  6 7];
T = zeros(7,7)  
d=zeros(3,3,3);
d(:,:,1) = B;       
d(:,:,2) = C;
d(:,:,3) = D;
for k = 1:1:3   
    for i = 1:1:3
        for j = 1:1:3
            T((i+k)-1,(j+k)-1) = T((i+k)-1,(j+k)-1) + d(i,j,k)
        end
    end
end

The answer which I am getting is this : which is wrong

What should I get is

T = [ 1 -1 1 0 0 0 0
      -1 1 1 0 0 0 0
      1 -1 2 -1 2 0 0
       0 0 -1 1 2 0 0
       0 0 4 5 6 -1 1
       0 0 0 0 -1 1 4
       0  0 0 0 5 6 7]

If it is possible, can you help me and tell me what mistake I am making, that would be very nice :)

per isakson 2019-11-10

编辑：per isakson 2019-11-10

It's far from obvious to me what algorithm you try to implement and I cannot easily deduce it from the example.

Regarding your code I directly observe that

all T_actual(6:7,:) and all T_actual(:,6:7) are equal to zero
the maximum values of the counters, k, i, j, are 3
the maximum value of (i+k)-1 and (j+k)-1, respectively is 5, which explains why all T_actual(6:7,:) and all T_actual(:,6:7) are equal to zero
all T_actual(5,1:5)==T_expected(7,3:7)