Hello Everyone, I'm looking for your help for the following problem
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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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Akira Agata
2018-3-5
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
3 个评论
Akira Agata
2018-3-6
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
更多回答(1 个)
Fangjun Jiang
2018-3-5
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.
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