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
il 16 Apr 2015
Modificato: Jan
il 16 Apr 2015
[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.
12 Commenti
Mahbubur Rahman
il 30 Apr 2016
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
il 30 Apr 2016
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
il 15 Apr 2015
Here's one way:
minValue = min(yourArray(:));
[row, column] = find(yourArray == minValue);
3 Commenti
Image Analyst
il 15 Apr 2015
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
il 11 Giu 2017
Can't you just go
- A = some matrix
- [row,collum] = find(A == min(min(A)))
- min_val = S(row,collum)
1 Commento
Image Analyst
il 11 Giu 2017
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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