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

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.

Answer 1

[MathWorks Support Team]

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:

https://www.mathworks.com/help/matlab/ref/max.html

https://in.mathworks.com/help/matlab/ref/min.html

To find the indices of all the locations where the maximum value (of the whole matrix) appears, you can use the "find" function.

https://www.mathworks.com/help/matlab/ref/find.html

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

[Keqiao Li]

what if I have two max/min values in one matrix. What I mean is if the matrix is like A = [1,2,3;4,5,6;7,8,8]. In this case, obviously, the max value is 8, but there are two "8". What should I do? Thanks!

[Chirag Parekh]

You can use find command with max

find(A == max(A(:)))

[Oscar Espinosa]

I got two indices, both have the same value. How can I choose just one of them?

[Sergio Novi Junior]

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.

Answer 2

[Shakir Kapra]

[M,I] = min(A) 
where M - is the min value
and I - is index of the minimum value
Similarly it works for the max

[Mohammed ABDELAZIZ]

thnaks

[Shahab Forghani]

Five years later, but thank you for this Mr. Kapra

[Peipei Zhai]

in this way, how to save all the index of the minimum in a matrix?

Answer 3

[indrajit das]

find(arr == max(arr));

[Hasan Salim Kanmaz]

Simplest solution. Thanks

Answer 4

[Andrew Teixeira]

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]

Answer 5

[Juanith HL]

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

[Martin Grden]

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

[Andrew Davies]

This is the most useful for working with images. :)

[Steven Lord]

If you can use linear indices, instead I would recommend:

A = [8 2 4; 7 3 9];
[X, ind] = max(A, [], 'all', 'linear')

X = 9

ind = 6

This will tell us the maximum value in A is 9 and it is located in element 6 of A.

Answer 6

[ANKUR KUMAR]

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

[Jia Li]

Thanks! It worked very well.

[amin nazari]

Thanks. very usefull

[Anup Negi]

how to use this code after saving the file , i am bignner of code

[Steven Lord]

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.)

Answer 7

[Roos]

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.

[Erica Kiderman]

Thank you for your feedback. The answer above has been corrected to address your comments.

[Stanley Tam]

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

"=" should be replaced with "==", i.e. [x,y]=find(A == minimum);

Answer 8

[Konstantinos Fragkakis]

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

[Ammaa AlSada]

find the min(or max) of the 2nd row of an unkown matrix?

how to solve it

Answer 9

[SHIVAM SUNDRIYAL]

[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

Answer 10

[Samar Shahin]

[row,col]=find( matrix_name == min(matrix_name)) returns the indices of the minimum value in each coloumn in your matrix

Answer 11

[Hazoor Ahmad]

A = randi(45,5)
[mr,mir] = min(A)
[mc,mic] = min(mr)
[mir(mic), mic]