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

Question

Keyre 2018-3-5

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/386337-hello-everyone-i-m-looking-for-your-help-for-the-following-problem

评论： Keyre 2018-3-6

采纳的回答： Akira Agata

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

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Keyre 2018-3-5

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 2018-3-5

@Keyre: I do not understand your explanations.

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Answer 1

Akira Agata 2018-3-5

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https://ww2.mathworks.cn/matlabcentral/answers/386337-hello-everyone-i-m-looking-for-your-help-for-the-following-problem#answer_308365

在 MATLAB Online 中打开

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

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Akira Agata 2018-3-6

在 MATLAB Online 中打开

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 2018-3-6

Dear Akira Agata, really helpful. Thanks!

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Answer 2

Fangjun Jiang 2018-3-5

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https://ww2.mathworks.cn/matlabcentral/answers/386337-hello-everyone-i-m-looking-for-your-help-for-the-following-problem#answer_308430

在 MATLAB Online 中打开

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 =

   1     1     2     1     1
   2     3     1     3     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.