Minimum value,row and column
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I want to find the minimum value of a matrix,the row and the column of it
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Jan
2015-4-16
编辑:Jan
2015-4-16
[value, index] = min(A(:));
[row, col] = ind2sub(size(A), index);
In opposite to the solution of Image Analyst, this is faster, but considers only one value even if the minimal value appears multiple times in the array.
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Mahbubur Rahman
2016-4-30
Hi, thanks again. Your answer is okay. But problem is how do I relate the lowest row for the lowest column.
A = [1 1 7 1 8; 2 4 5 9 5; 6 5 0 2 3; 3 7 5 1 9; 9 5 2 6 7]
[r c]= find(A==5);
You can see that the lowest number of c is 2. And for this c = 2, we get two r (3 and 5). Now I want to select the r_min for c_min (which is c=2). My answer should be (2,3). Just figured out that If I make a matrix
B=[c r];
Answer = B(1,:)
It gives me the desired value. Comment please.
Image Analyst
2016-4-30
It's the same solution I've been telling you. Now I've renamed minA to valueToFind since it appears that you're not always looking to find the min values. And since you want an (x,y) answer rather than "value of the row" like you asked for before I just concatenate them together:
A = [1 1 7 1 8; 2 4 5 9 5; 6 5 0 2 3; 3 7 5 1 9; 9 5 2 6 7]
% valueToFind = min(A(:)); % Find the min value of A
valueToFind = 5;
[row, column] = find(A == valueToFind)
MyAnswer = [column(1), row(1)] % In format [x,y] NOT [row, column]
You can call it MyAnswer, or B, or whatever you want, but I'd probably not use Answer.
Image Analyst
2015-4-15
Here's one way:
minValue = min(yourArray(:));
[row, column] = find(yourArray == minValue);
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Image Analyst
2015-4-15
You could run down row by row inverting the lines and then use findpeaks() in the Signal Processing Toolbox. Probably far less efficient than the first way.
You could use imregionalmin() in the Image Processing Toolbox.
There are other ways I'm sure, such as functions in the various optimization toolboxes.
Kaelan Wade
2017-6-11
Can't you just go
- A = some matrix
- [row,collum] = find(A == min(min(A)))
- min_val = S(row,collum)
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Image Analyst
2017-6-11
Yes, you can. In fact that's what my answer up above already said, though in a more efficient way. You don't need to use min twice if you use (:) and your way gives an array of all the same min value whereas most likely only a single value is needed, even if the min shows up multiple times.
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