How do I find the indices of the maximum (or minimum) value of my matrix?
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MathWorks Support Team on 7 Oct 2009
The 'find' command only returns the indices of all the non-zero elements of a matrix. I would like to know how to find the indices of just the maximum (or minimum) value.
MathWorks Support Team on 28 Mar 2022
Edited: MathWorks Support Team on 17 Nov 2021
The "min" and "max" functions in MATLAB return the index of the minimum and maximum values, respectively, as an optional second output argument.
For example, the following code produces a row vector 'M' that contains the maximum value of each column of 'A', which is 3 for the first column and 4 for the second column. Additionally, 'I' is a row vector containing the row positions of 3 and 4, which are 2 and 2, since the maximums for both columns lie in the second row.
A = [1 2; 3 4];
[M,I] = max(A)
For more information on the 'min' and 'max' functions, see the documentation pages listed below:
To find the indices of all the locations where the maximum value (of the whole matrix) appears, you can use the "find" function.
maximum = max(max(A));
Sergio Novi Junior on 5 Jun 2020
Oscar, it will depend on your application. For example, suppose you are looking for a car with the highest final speed. You find that there are two cars that have the best results. You can choose the car with lowest price.
More Answers (12)
Shakir Kapra on 20 Apr 2015
Edited: Shakir Kapra on 20 Apr 2015
[M,I] = min(A)
where M - is the min value
and I - is index of the minimum value
Similarly it works for the max
indrajit das on 3 Aug 2016
find(arr == max(arr));
Andrew Teixeira on 1 Oct 2019
How about just:
A = magic(5);
[Amins, idx] = min(A);
[Amin, Aj] = min(Amins);
Ai = idx(Aj);
where your final matrix minima is located at [Ai, Aj]
Juanith HL on 8 Oct 2019
A = [8 2 4; 7 3 9]
[M,I] = max(A(:)) %I is the index maximun Here tu can change the function to max or min
[I_row, I_col] = ind2sub(size(A),I) %I_row is the row index and I_col is the column index
Steven Lord on 18 Aug 2022
If you can use linear indices, instead I would recommend:
A = [8 2 4; 7 3 9];
[X, ind] = max(A, , 'all', 'linear')
This will tell us the maximum value in A is 9 and it is located in element 6 of A.
ANKUR KUMAR on 19 Sep 2017
Use this as a function and type [x,y]=minmat(A) to get the location of the minimum of matrix. for example:
Save this as a function in your base folder and use it.
function [ a,b ] = minmat( c )
Steven Lord on 6 Sep 2020
Once you've stored this code in a file named minmat.m (see this documentation page for the basics of defining a function) you can call this like any other MATLAB function (as shown on this other documentation page.)
Roos on 10 May 2018
Edited: Roos on 10 May 2018
This apparently solved your question, however for future reference I would like to mention that there is an earier solution that does not involve declaring a function.
Lets continue with any matrix A. The first step is finding the minimum value of the complete matrix with:
The double min is needed to first find min of all columns, then find min of all those min values. (there might be an easier way for this as well).
Finding the indices of this value can be done like this:
2 lines will be easier than a complete function.
Stanley Tam on 16 Oct 2020
"=" should be replaced with "==", i.e. [x,y]=find(A == minimum);
Konstantinos Fragkakis on 27 Aug 2018
Edited: Konstantinos Fragkakis on 27 Aug 2018
Function to calculate the minimum value and its indices, in a multidimensional array - In order to find the max, just replace the min(array(:)) statement with max(array(:)).
function [ minimum,index ] = minmat( array )
% Function: Calculate the minimum value and its indices in a multidimensional array
% -------- Logic description --------
% First of all, identify the Matlab convention for numbering the elements of a multi-dimensional array.
% First are all the elements for the first dimension
% Then the ones for the second and so on
% In each iteration, divide the number that identifies the minimum with the dimension under investigation
% The remainder is the Index for this dimension (check for special cases below)
% The integer is the "New number" that identifies the minimum, to be used for the next loop
% Repeat the steps as many times as the number of dimensions (e.g for a 2-by-3-by-4-by-5 table, repeat 4 times)
neldim = size(array); % Length of each dimension
ndim = length(neldim); % Number of dimensions
[minimum,I] = min(array(:));
remaining = 1; % Counter to evaluate the end of dimensions
index = ; % Initialize index
while remaining~=ndim+1 % Break after the loop for the last dimension has been evaluated
% Divide the integer with the the value of each dimension --> Identify at which group the integer belongs
r = rem(I,neldim(remaining)); % The remainder identifies the index for the dimension under evaluation
int = fix(I/neldim(remaining)); % The integer is the number that has to be used for the next iteration
if r == 0 % Compensate for being the last element of a "group" --> It index is equal to the dimension under evaluation
new_index = neldim(remaining);
else % Compensate for the number of group --> Increase by 1 (e.g if remainder 8/3 = 2 and integer = 2, it means that you are at the 2+1 group in the 2nd position)
int = int+1;
new_index = r;
I = int; % Adjust the new number for the division. This is the group th
index = [index new_index]; % Append the current index at the end
remaining = remaining + 1;
SHIVAM SUNDRIYAL on 18 May 2020
[M,I] = max(___) returns the index into the operating dimension that corresponds to the maximum value of A for any of the previous syntaxes.
Here M will represent and hold the maximum value while Iwill hold the index of the maximum value
you can use the same method to find the minimum value and its index by using the min() functiobn
A = [1 9 -2; 8 4 -5]
[M,I] = max(A)
M = 1×3
8 9 -2
I = 1×3
2 1 1
Samar Shahin on 2 May 2020
[row,col]=find( matrix_name == min(matrix_name)) returns the indices of the minimum value in each coloumn in your matrix
Hazoor Ahmad on 28 Feb 2021
Edited: Hazoor Ahmad on 28 Feb 2021
A = randi(45,5)
[mr,mir] = min(A)
[mc,mic] = min(mr)
Piotr Franaszczuk on 13 Dec 2022
Edited: Piotr Franaszczuk on 13 Dec 2022
You can do it calling max only once and without find(which would need to got through whole 2D matrix):
M is maximum, and r and c is row and column in A where it is located.
Neeraja on 21 Feb 2023
Using find function together with min works very well.
[I,J] = find(A==min(A(:)))
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