To find the maximum value in a matrix?

Let me have a 3X3 matrix

6   8   9
7   10  11
21  22  8

How to find the maximum value from this matrix?

[Jan]

Sorry that I mention the barely obvious, but the answer is 22.

[KHOIROM Motilal]

• clc
• close all
• clear all
• X=[99 67 65;
• 63 62 61;
• 41 40 9];
• MAX=X(1,1);
• for i=1:3
• for j=1:3
• if MAX<= X(i,j);
• MAX=X(i,j);
• end
• end
• end
• disp(MAX)

Answer 1

[Michael Völker]

Starting in R2018b, you can use the following command to find the maximum over all elements in an array A:

M = max(A, [], 'all');

For previous releases, use:

M = max(A(:));

[Image Analyst]

To get it's location as well, accept both outputs of max:

[maxValue, linearIndexesOfMaxes] = max(A(:));

Note that there can be the max value at more than one location. To get the rows and columns (instead of linear indexes), you can use ind2subs() or find():

[rowsOfMaxes colsOfMaxes] = find(A == maxValue);

[tilia margaret pelagia dhont]

what do the empty square brackets represent?

[Stephen23]

"what do the empty square brackets represent?"

See: https://www.mathworks.com/matlabcentral/answers/395391-how-could-this-be-achieved-in-one-line-of-code#comment_557719

[Steven Lord]

The max function's first two inputs are the two matrices whose values you want to compare. If you only want to compute the maximum of one matrix, you need something to use as a placeholder for that second matrix. Otherwise if you wrote something like this, are you asking for the maximum of the elements of a matrix and the value 1 or are you asking for the maximum along the 1st dimension?

max(A, 1)

To break that ambiguity, that syntax is interpreted as the former (the maximum of the elements of A and the value 1) while the following is the latter (the maximum along the 1st dimension.)

max(A, [], 1)

Answer 2

[Azzi Abdelmalek]

 max(max(A))

[Andrei Bobrov]

M = [6   8   9
7   10  11
21  22  8];

index = 1;
max1 = M(index);
for jj = 2:numel(M)
    if max1 < M(jj)
        max1 = M(jj);
        index = jj;
    end
end

:)

[José-Luis]

Or if you really want to throw efficiency out the window:

M = [6   8   9
7   10  11
21  22  8];
unikM = unique(M);
val = unikM(1);
idx = unikM > val;
while (sum(idx) > 1)
    idx = find(idx);
    val = unikM(idx(1));
    idx = unikM > val;
    maxVal = unikM(idx)
end

[Jonathan Posada]

This works for the 2D case but if ndims(A)>2, then max(max(A)) will return a matrix. I believe OP wants the maximum element along all dimensions

[DGM]

Not that this is a good idea, but for an arbitrary number of dimensions:

A = rand(100,100,100,10); % a fairly large ND array

% find global maximum of A
maxval = max(A);
for n = 2:ndims(A)
    maxval = max(maxval);
end
maxval

maxval = 1.0000

It hasn't been so for quite some time, but in my experience, this iterative approach had a significant speed advantage with larger N-D arrays in older versions (2x-3x as fast as max(A(:)) for the arrays I was using). I don't remember if that advantage still existed in R2012x, but it did in R2009b. In current versions, using vectorization or 'all' are faster for small arrays and roughly equivalent for large arrays. That's on my hardware, so I make no guarantees that it's exactly universal.

Performance aside, it's hard to justify this verbose method over the canonical techniques, if only for the sake of readability.

It's not something I'd recommend, and I doubt that the legacy performance is the typical reason that people gravitate to the approach, but I thought it was interesting to note for old time's sake.

Answer 3

[Tom]

M = max(A,[],'all') finds the maximum over all elements of A. This syntax is valid for MATLAB® versions R2018b and later.

[John Doe]

This should be upvoted and/or somehow appear closer to the chosen answer, as M = max(A,'all') seems not to work at all in R2018b+ (returning the entire matrix).

M = max(A(:)) seems to work in R2018b+ and presumably universally.

[Steven Lord]

The [] as the second input is required when you want to specify a dimension, including 'all'. The function call max(A, 'all') only works if A and 'all' are compatibly sized.

>> max(1:3, 'all')
ans =
    97   108   108
    
>> max(1:3, [], 'all')
ans =
     3    

Answer 4

[Dmaldo01]

This will work for all dimensions. If more efficient than ind2sub for less than 16000 elements.

 [M,Index] = maxEl(MatVar)

index = size(MatVar);
Index = index*0;
M = max(MatVar(:));

A = find(MatVar==max(MatVar(:)),1);
for i = 1:length(index)
   Index(i) = mod(ceil(A),index(i));
   A = A/index(i);
end
 Index(Index==0)=index(Index==0);

Answer 5

[Yokesh]

If matrix dimension is 'n', then max element can be found by:

max(max(.....maxn^2((A))...)

We have to include n^2 times max

[Steven Lord]

No, you don't need to include multiple calls to max. See the accepted Answer for approaches that call max only once regardless of how many dimensions the input argument has.

[Walter Roberson]

Also it would only be n max calls.

Answer 6

[JPS]

or you can use,

M = max(max(A));

[Adrian Brown]

hello,

There is any way for a matrix size NxM to get the k maximum element in the whole matrix not in rows or colomns but in only elements. for example matrix A = [1 3 2 5, 7 9 12 8, 12 8 9 0] for K= 3 the 3 maximum elements are 12 9 and 8 and I want to return there location in the matrix.

I really appreciate any help

[Walter Roberson]

A = [1 3 2 5; 7 9 12 8; 12 8 9 0]

A = 3×4

1 3 2 5 7 9 12 8 12 8 9 0

[best3, best3idx] = maxk(A(:),3)

best3 = 3×1

12 12 9

best3idx = 3×1

3 8 5

The three maximum values are 12, 12, and 9, not 12, 9, and 8. If you are interested in the three maximum unique values, then you need to explicitly take into account that some values occur more than once.

k = 3;
uA = unique(A, 'sorted');
nresults = min(length(uA), k);
results = cell(nresults, 1);
for K = 1 : k
   this_max = uA(end-K+1);
   results{K,1} = this_max;
   results{K,2} = find(A==this_max).';
end

disp(results)

    {[12]}    {[ 3 8]}
    {[ 9]}    {[ 5 9]}
    {[ 8]}    {[6 11]}

The output is a cell array, in which the first column gives you the value that is the maximum, and the second column gives you all the linear indices into the array. The code could be modified to give row and column outputs instead without much change.

The code does not assume that the number of occurrences is the same for each of the values (if that were known to be true then a numeric array could be created instead of a cell array.)