Hello Everyone, I'm looking for your help for the following problem

Use the matrix given on the first link.I need a matlab code that can generate on the second link. Thanks!

[Image Analyst]

Sounds like homework. See http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers

[Keyre]

Dear Image Analyst,this is totally my explanation to more illustrate, not a homework, anyways you can give me direction or best reference for this issue too.

[Akira Agata]

What is the rule to chose one value for each column?

[Keyre]

The maximum value for each column but it become the second or third maximum if the row side already two maximum values are selected

[Jan]

@Keyre: I do not understand your explanations.

Answer 1

[Akira Agata]

The second image you uploaded is not the correct maximum value array for each column... Anyway, let me try to do this without using for-loop:

A = [17 14 18 15 14 19;...
  1 5 3 30 8 14;...
  20 2 16 7 13 11];

idx = bsxfun(@eq,A,max(A));
A(~idx) = NaN;

The result is:

>> A
A =
 NaN    14    18   NaN    14    19
 NaN   NaN   NaN    30   NaN   NaN
  20   NaN   NaN   NaN   NaN   NaN

[Keyre]

Hi Sir, this is selection of maximum value per column. In addition i need only two maximum values per row.As that time the program should find the next maximum values instead. Hence 14 and 19 should not be selected as maximum value for that columns. So the second requirement is in each row two maximum column values should be appeared.

[Akira Agata]

Hi Keyre-san,

Thank you for the clarification. Here is my 2nd try. I hope this will be helpful somehow.

A = [17 14 18 15 14 19;...
1 5 3 30 8 14;...
20 2 16 7 13 11];

idx = false(size(A));
for kk = 1:size(A,2)
  [~,pt] = sort(A(:,kk),'descend');
  for ll = 1:3
      if nnz(idx(pt(ll),:)) < 2
          idx(pt(ll),kk) = true;
          break;
      end
  end
end
A(~idx) = NaN;

And here is the result:

>> A
A =
 NaN    14    18   NaN   NaN   NaN
 NaN   NaN   NaN    30   NaN    14
  20   NaN   NaN   NaN    13   NaN

[Keyre]

Dear Akira Agata, really helpful. Thanks!

Answer 2

[Fangjun Jiang]

I'll provide a lead. I think the rest should be relatively easy.

A = [17 14 18 15 14 19;...
     1  5  3  30 8  14;...
    20  2 16  7  13 11];
[~,IndexMatrix]=sort(A,'descend');

IndexMatrix =

     3     1     1     2     1     1
     1     2     3     1     3     2
     2     3     2     3     2     3

The first row of IndexMatrix is what you need to work on. Starting from the third column, do a loop, compare it with all the previous number to see if it occurs twice already. If it does, replace it with the number in the same column but next row in IndexMatrix. You need to do this in a while-loop because the next number could also occur twice already, although in this one case, it didn't happen.

[Keyre]

 Dear Fangjun Jiang, Thanks for your helpful direction