how can obtain min of matrix

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sara
sara el 25 de En. de 2015
Editada: David Young el 26 de En. de 2015
  1 comentario
Matz Johansson Bergström
Matz Johansson Bergström el 25 de En. de 2015
Editada: Matz Johansson Bergström el 26 de En. de 2015
Please write the question in text and not as a attached image. The description of your question will not be indexed by the Mathworks search engine if you provide an image.

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Respuestas (5)

Matt J
Matt J el 26 de En. de 2015
[row,col]=find(matrix==min(nonzeros(matrix)));
  2 comentarios
sara
sara el 26 de En. de 2015
thanks Matt
Matz Johansson Bergström
Matz Johansson Bergström el 26 de En. de 2015
Ah nonzeros , didn't even know it existed. This is the shortest and best solution and should be accepted as the answer. Good job Matt.

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Matz Johansson Bergström
Matz Johansson Bergström el 25 de En. de 2015
Editada: Matz Johansson Bergström el 25 de En. de 2015
This solution is maybe a little ugly, but it works Say that A is the matrix.
tmp = A; %we will destroy elements, so we store A in tmp
tmp(tmp==0) = []; %get rid of 0-elements
val = min(min(tmp)); %find value of the smallest element
[u,v] = find(A==val, 1); %find the position
u and v is the (first) row and column of the index of the smallest element in A. The smallest element could occur several times in the matrix.

David Young
David Young el 25 de En. de 2015
Editada: David Young el 26 de En. de 2015
Another approach:
tmp = A; % avoid destroying A
tmp(tmp == 0) = Inf; % make zero elements bigger than non-zeros
[minVal, minIndex] = min(tmp(:)); % find min value, linear index
[minRow, minCol] = ind2sub(size(A), minIndex); % convert to subscripts
or if you prefer
tmp = A;
tmp(tmp == 0) = Inf; % as above
[colminvals, colminrows] = min(tmp); % find min in each column
[minVal, minCol] = min(colminvals); % find overall min and its column
minRow = colminrows(minCol); % select row of overall min
or my personal preferred method, avoiding copying the matrix and also avoiding a repeat scan with the find operation:
nzpos = A ~= 0;
indexes = 1:numel(A);
indnz = indexes(nzpos);
[minVal, minIndnz] = min(A(nzpos));
[minRow, minCol] = ind2sub(size(A), indnz(minIndnz));

sara
sara el 26 de En. de 2015
thanks
  6 comentarios
Matz Johansson Bergström
Matz Johansson Bergström el 26 de En. de 2015
Very good.
sara
sara el 26 de En. de 2015
Editada: sara el 26 de En. de 2015
ohhh yes I just tried it for my example. and I was in a hurry and ...thanks dear David and Matz...

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sara
sara el 26 de En. de 2015
Editada: sara el 26 de En. de 2015
Dear ImageAnalyst
I accepted your answer becuase I was in hurry and I check it just for one example...I can not do any changes if you can please edit this...
thanks
  2 comentarios
Image Analyst
Image Analyst el 26 de En. de 2015
I deleted my answer. Is there not any "Accept this answer" link for any of the others?
sara
sara el 26 de En. de 2015
thanks

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