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

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MathWorks Support Team 2009 年 10 月 7 日

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.

### 採用された回答

MathWorks Support Team 2019 年 3 月 14 日

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)

#### 2 件のコメント

Keqiao Li 2016 年 12 月 22 日
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 2016 年 12 月 29 日
You can use find command with max
find(A == max(A(:)))

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### その他の回答 (7 件)

Shakir Kapra 2015 年 4 月 20 日

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

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Mohammed ABDELAZIZ 2016 年 3 月 28 日
thnaks

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indrajit das 2016 年 8 月 3 日
find(arr == max(arr));

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ANKUR KUMAR 2017 年 9 月 19 日
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 件のコメント

Jia Li 2018 年 4 月 9 日
Thanks! It worked very well.
amin nazari 2019 年 7 月 17 日
Thanks. very usefull

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Roos 2018 年 5 月 10 日

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.

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Erica Kiderman 2018 年 5 月 15 日

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Andrew Teixeira 2019 年 10 月 1 日
A = magic(5);
[Amins, idx] = min(A);
[Amin, Aj] = min(Amins);
Ai = idx(Aj);
where your final matrix minima is located at [Ai, Aj]

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Konstantinos Fragkakis 2018 年 8 月 27 日

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

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Ammaa AlSada 2018 年 12 月 15 日
find the min(or max) of the 2nd row of an unkown matrix?
how to solve it

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Juanith HL 2019 年 10 月 8 日
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

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