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

Asked by MathWorks Support Team on 7 Oct 2009

Latest activity Commented on by amin nazari on 17 Jul 2019

Accepted Answer by MathWorks Support Team

7,553 views (last 30 days)

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.

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MATLAB

6 Answers

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Answer by MathWorks Support Team on 14 Mar 2019

Edited by MathWorks Support Team on 14 Mar 2019

Accepted Answer

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:

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

http://www.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)

2 Comments

Keqiao Li on 22 Dec 2016

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 on 29 Dec 2016

You can use find command with max

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

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Answer by Shakir Kapra on 20 Apr 2015

Edited by 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

1 Comments

Mohammed ABDELAZIZ on 28 Mar 2016

thnaks

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Answer by Roos on 10 May 2018

Edited by 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.

1 Comments

Erica Kiderman on 15 May 2018

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

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Answer by 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

2 Comments

Jia Li on 9 Apr 2018

Thanks! It worked very well.

amin nazari on 17 Jul 2019

Thanks. very usefull

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Answer by indrajit das on 3 Aug 2016

find(arr == max(arr));

0 Comments

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Answer by Konstantinos Fragkakis on 27 Aug 2018

Edited by 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