How to filter a matrix Row by Row to get the highest value?

Dear all,

I have this matrix

A= [ 31,62.3,  31,96.3,  31,52.8,  57,91.4  ;...
     71,64.4,  31,93.5,  31,36.2,  57,83.1 ] ;

I want to scan this matrix row by row such that values at the odd columns will not be repeated + it will get the highest value at the even column attached to it

So I want the following answer:

Answer = [ 31,96.3    57,91.4      0,0     ;...  
           71,64.4    31,93.5      57,83.1 ]

The oreder of pairs in each row is not important. The following answe is also OK

Answer = [ 57,91.4     31,96.3        0,0    ;...  
           31,93.5      57,83.1      71,64.4 ]

Many thanks in advance.

[Guillaume]

A matrix cannot have two elements on one row and 3 on another. All the rows must have the same number of elements.

You need to review your requirements.

[Ghost]

Thank you for your comment.

I corrected my question and added the zeros.

[Image Analyst]

It's such a quirky thing to do that you may have to end up just writing a custom program from low level functions like for loops and max(). I don't think there is a built in function to do exactly this thing. But there are a lot of ways it could be done from low level functions.

Why do you need to do this unusual thing anyway - what's the use case?

By the way, "raw" means "uncooked food", and a line of an array is "row", not "raw".

[Ghost]

Thank you Image Analyst. I corrected the spelling.

I just needed this for one idea that I am trying to solve. I don't think it has many applications.

Answer 1

[madhan ravi]

See if this satisfies your needs:

C = zeros(size(A));
for k = 1:size(A,1)
    B = reshape(A(k,:),2,[]).';
    G = findgroups(B(:,1));
    BB = reshape(B(any(B(:,2)==splitapply(@(x)max(x),...
        B(:,2),G).',2),:).',1,[]);
    C(k,1:numel(BB)) = BB;
end
C(:,all(C==0,1)) = []

Note: Not tested under all circumstances.

[Ghost]

Thank you very much Madhan.

Answer 2

[Guillaume]

A version with no loop, but a few accumarray so may not be faster:

%build demo data
A = [31,62.3,  31,96.3,  31,52.8,  57,91.4  ;...
     71,64.4,  31,93.5,  31,36.2,  57,83.1 ];
A = [A; A]; A(4, 5) = 32

%algorithm
workingA = [repelem((1:size(A, 1))', size(A, 2)/2), reshape(A.', 2, []).']; %reshape in 2 columns of pairs and prepend with a column indicating which row the data came from
[urows, ~, id] = unique(workingA(:, [ 1 2]), 'rows');                       %identify unique 1st pair element per row
offsets = accumarray(workingA(:, 1), id, [], @max);                         %need to reset id for each row, so find max id per row
offsets = [0;offsets(1:end-1)];
column = (id - offsets(workingA(:, 1)))*2;                                  %and subtract max of previous row from each row
result = accumarray([urows(id, 1), column], workingA(:, 3), [], @max);      %distribute max for each unique pair and row
result = result + [accumarray([urows(id, 1), column-1], workingA(:, 2), [], @(v) v(1)), zeros(size(result, 1), 1)]  %add 1st element of each pair

With a loop, a cleaner (in my opinion) version of Madhan's answer:

result = zeros(size(A));
for row = 1:size(A, 1)
    [uvals, ~, id] = unique(A(row, 1:2:end));
    result(row, 2:2:end) = accumarray(id, A(row, 2:2:end), [size(A, 2)/2, 1], @max);
    result(row, 1:2:numel(uvals)*2) = uvals;
end
result(:, all(result == 0, 1)) = []

[Ghost]

Thank you indeed.