# How do I find the indices of the maximum (or minimum) value of my matrix?

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### Accepted Answer

MathWorks Support Team
on 11 Oct 2021

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));

[x,y]=find(A==maximum)

##### 4 Comments

Sergio Novi Junior
on 5 Jun 2020

### More Answers (11)

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

##### 3 Comments

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]

##### 0 Comments

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

##### 1 Comment

Martin Grden
on 12 Oct 2020

Like this one best. It should be more efficient than those traversing the matrix two times.

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:

>> A=magic(5)

>> [a,b]=minmat(A)

a =

1

b =

3

Save this as a function in your base folder and use it.

function [ a,b ] = minmat( c )

as=size(c);

total_ele=numel(c);

[~,I]=min(c(:));

r=rem(I,as(1));

a=r;

b=((I-a)/as(1))+1;

if a==0

a=as(1);

b=b-1;

else

a=r;

b=b;

end

end

##### 4 Comments

Steven Lord
on 6 Sep 2020

Roos
on 10 May 2018

Edited: Roos
on 10 May 2018

https://www.mathworks.com/matlabcentral/answers/100813-how-do-i-find-the-indices-of-the-maximum-or-minimum-value-of-my-matrix#answer_282157

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:

minimum=min(min(A));

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:

[x,y]=find(A=minimum);

2 lines will be easier than a complete function.

##### 2 Comments

Stanley Tam
on 16 Oct 2020

Above: [x,y]=find(A=minimum);

"=" 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;

end

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;

end

end

##### 1 Comment

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

Example :

A = [1 9 -2; 8 4 -5]

[M,I] = max(A)

Output:

M = 1×3

8 9 -2

I = 1×3

2 1 1

##### 0 Comments

Samar Shahin
on 2 May 2020

##### 0 Comments

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)

[mir(mic), mic]

##### 0 Comments

Tanvirul Abedien
on 18 Jan 2022 at 17:51

A = randi(45,5)

[mr,mir] = min(A)

[mc,mic] = min(mr)

[mir(mic), mic]

##### 0 Comments

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