Comparing Matrices in a Struct

Asked by hamzah khan on 27 Oct 2019

Latest activity Edited by per isakson on 10 Nov 2019

Accepted Answer by per isakson

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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 by per isakson on 28 Oct 2019

Edited by per isakson on 8 Nov 2019

Accepted Answer

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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per isakson on 8 Nov 2019

I added version 3 to the answer. It handles size a bit more explicit and use smaller sizes of the value of model_data.

Running version 3 returns

>> full(Out(1,1).model_data)
ans =
      0.19092      0.58951      0.25181          NaN          NaN
      0.42825      0.22619      0.29044          NaN          NaN
      0.48202      0.38462      0.61709          NaN          NaN
      0.12061      0.58299      0.26528          NaN          NaN
>> full(Out(1,2).model_data)
ans =
      0.82438      0.73025      0.58407          NaN          NaN
      0.98266      0.34388      0.10777          NaN          NaN
          NaN          NaN          NaN          NaN          NaN
          NaN          NaN          NaN          NaN          NaN
>> full(Out(1,3).model_data)
ans =
      0.90631      0.26073      0.42526      0.17877      0.59852
      0.87965      0.59436      0.31272      0.42289      0.47092
      0.81776     0.022513      0.16148     0.094229      0.69595
          NaN          NaN          NaN          NaN          NaN
>> 

hamzah khan on 10 Nov 2019

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 on 10 Nov 2019

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)

Proposal: Post a new question in which you describe in some detail the algorithm you try to implement.

You are now following this question

hamzah khan (view profile)