How to find the index of the first absolute minimum entry of a matrix

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Tarek Hajj Shehadi
Tarek Hajj Shehadi 2022-5-19
评论： Stephen23 2022-5-19
I am working on a model structure optimization problem and to summarize consider this matrix:
A=[5,7,0.5,5 ; 2,0.5,4,1 ; 0.5,6,7,9]
In this matrix, 0.5 is clearly the smallest element of this matrix.
It is repeated 3 times.
Question: Can someone provide me with an algorithm that returns the index and value of this absolute minimum "only the first time it appears"
For instance:
A=[5,7,0.5,5 ; 2,0.5,4,1 ; 0.5,6,7,9];
[value,index]=somecode(A)
%returns the value of the absolute minimum
>> value = 0.5
%returns the index of the first time this absolute minimum appears
>> (3,1) % the first time it appears is in column 3 row 1 and NOT column 2 row 2
  1 个评论
Stephen23
Stephen23 2022-5-19
Your example matrix is not very good because it contains the value 0.5 in both (1,3) and (3,1) locations.
Catalytic's answer for example gives the wrong answer because of this confusion.

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采纳的回答

Catalytic
Catalytic 2022-5-19
编辑：Catalytic 2022-5-19
Ran in:
A=[5,7,0.5,5 ; 2,1,4,1 ; 1,6,7,0.5],
A = 3×4
5.0000 7.0000 0.5000 5.0000 2.0000 1.0000 4.0000 1.0000 1.0000 6.0000 7.0000 0.5000
[value,index]=min(A(:))
value = 0.5000
index = 7
[Iindex,Jindex]=ind2sub(size(A),index)
Iindex = 1
Jindex = 3

更多回答（2 个）

Stephen23
Stephen23 2022-5-19
Ran in:
A = [5,7,0.5,5;2,0.5,4,1;0.5,6,7,9];
[V,X] = min(A.',[],'all');
[C,R] = ind2sub(size(A),X)
C = 3
R = 1
Mathieu NOE
Mathieu NOE 2022-5-19
hello
here you are my friend
A=[5,7,0.5,5 ; 2,0.5,4,1 ; 0.5,6,7,9]
[val,ind_linear] = min(A,[],'all','linear'); % M = MIN(X,[],'all') returns the smallest element of X.
% [M,I] = MIN(X,[],...,'linear') returns the linear indices of the
% minimum values in vector I.
ind_linear = min(ind_linear); % take only the first one in case there are multiple answers
[row,col] = ind2sub(size(A),ind_linear) % Convert from the linear index back to its row and column form.

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