How to find the 5 minimum values in a multidimensional matrix and the indices to which these entries correspond.

I have a 597 x 194 matrix to which I need to find the 5 smallest values within the entire matrix. As well as this I need the corresponding indices for those minimum values.

i.e. the smallest value (of the 5) is value 4.1146, and corresponds to indices of 336, 170. How could I find this value along with the next 4 smallest, and the indices for the entry?

[Robin]

To find the 5 smallest values in a matrix and their corresponding indices, you can use the min function along with the sort and ind2sub functions in MATLAB. Here is an example of how to do this:

%

% Define the matrix 
% (this is terrible, since you are likely to get multiple
% zeros as random values over the 115818 element matrix
A = rand(597, 194);

% Find the 5 smallest values and their indices 
[sortedValues, sortedIndices] = sort(A(:));
smallestValues = sortedValues(1:5);
smallestIndices = sortedIndices(1:5);

% Convert the linear indices to row-column indices 
[rowIndices, colIndices] = ind2sub(size(A), smallestIndices); 

% Print the results 
fprintf('Smallest value: %.4f, indices: (%d, %d)\n', ...
    smallestValues(1), rowIndices(1), colIndices(1)); 

Smallest value: 0.0000, indices: (436, 97)

fprintf('Second smallest value: %.4f, indices: (%d, %d)\n', ...
    smallestValues(2), rowIndices(2), colIndices(2)); 

Second smallest value: 0.0000, indices: (582, 173)

fprintf('Third smallest value: %.4f, indices: (%d, %d)\n', ...
    smallestValues(3), rowIndices(3), colIndices(3));

Third smallest value: 0.0000, indices: (579, 73)

fprintf('Fourth smallest value: %.4f, indices: (%d, %d)\n', ...
    smallestValues(4), rowIndices(4), colIndices(4));

Fourth smallest value: 0.0000, indices: (578, 119)

fprintf('Fifth smallest value: %.4f, indices: (%d, %d)\n', ...
    smallestValues(5), rowIndices(5), colIndices(5));

Fifth smallest value: 0.0001, indices: (175, 30)

This code first flattens the matrix into a vector using the : operator and then sorts it using the sort function. It then selects the first 5 values and their indices and converts the linear indices to row-column indices using the ind2sub function. Finally, it prints the results to the command window.

I hope this helps! Let me know if you have any questions.

[Charles Howman]

Yes this is perfect thank you! It is a much more elegant solution to one one I eventually went with which involved using min to find the smallest value, then re-assigning it to a larger value.

%initialise matrix to store values
minimums = zeros(5, 3);
%start by finding smallest value in whole matrix
[pairDmin1] = min(pairD, [],'all');
minimums(1, 1) = pairDmin1;
%then find the indices for this matrix
[X1, Y1] = find(pairD==pairDmin1);
minimums(1, 2) = X1;
minimums(1, 3) = Y1;
%now going to replace this value by a large value
pairD(X1, Y1) = 999;
%now find new smallest value which will be 2nd overall smallest
[pairDmin2] = min(pairD, [], 'all');
minimums(2, 1) = pairDmin2;
[X2, Y2] = find(pairD==pairDmin2);
minimums(2, 2) = X2;
minimums(2, 3) = Y2;
pairD(X2, Y2) = 999;
%now 3rd smallest
[pairDmin3] = min(pairD, [], 'all');
minimums(3, 1) = pairDmin3;
[X3, Y3] = find(pairD==pairDmin3);
minimums(3, 2) = X3;
minimums(3, 3) = Y3;
pairD(X3, Y3) = 999;
%now 4th smallest
[pairDmin4] = min(pairD, [], 'all');
minimums(4, 1) = pairDmin4;
[X4, Y4] = find(pairD==pairDmin4);
minimums(4, 2) = X4;
minimums(4, 3) = Y4;
pairD(X4, Y4) = 999;
%now 5th smallest
[pairDmin5] = min(pairD, [], 'all');
minimums(5, 1) = pairDmin5;
[X5, Y5] = find(pairD==pairDmin5);
minimums(5, 2) = X5;
minimums(5, 3) = Y5;
pairD(X5, Y5) = 999;
%now can view the 5 minimums and their corresponding indices
display(minimums)

Answer 1

[the cyclist]

% Some input data
rng default
M = rand(3);

% Find the sorted values of M, and the linear matrix indexing to them
[sortedM,linearIndex] = sort(M(:));

% Find the row and column indices
[rowIndex,columnIndex] = ind2sub(size(M),linearIndex);

% The first 5 of them
n = 5;

sortedM(1:n)

ans = 5×1

0.0975 0.1270 0.2785 0.5469 0.6324

rowIndex(1:n)

ans = 5×1

3 3 1 2 2

columnIndex(1:n)

ans = 5×1

2 1 3 3 2

[Charles Howman]

This is also a much more elegant solution than what I eventually came up with, thank you for answering too!

Answer 2

[William Rose]

@Charles Howman,

A=rand(3,4)

A = 3×4

0.4651 0.6844 0.3092 0.4551 0.9791 0.6301 0.3823 0.2112 0.9126 0.7742 0.4814 0.3253

[r,c]=size(A);
k=5;  %number of values to find
[minval,idx]=mink(reshape(A,[],1),k);
minrow=mod(idx,r);
minrow(minrow==0)=3;
mincol=ceil(idx/r);
disp([minval,minrow,mincol])

    0.2112    2.0000    4.0000
    0.3092    1.0000    3.0000
    0.3253    3.0000    4.0000
    0.3823    2.0000    3.0000
    0.4551    1.0000    4.0000

Try it. Good luck.