# Comparing Matrices in a Struct

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hamzah khan on 27 Oct 2019
Edited: per isakson on 10 Nov 2019
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

per isakson on 28 Oct 2019
Edited: per isakson on 8 Nov 2019
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

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
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.