Issue with finding the indices of the minimum element in a 3-Dimensional Array

For a 2D array A, the command

[row,column]=find(A==min(min(A)))

outputs the row and column indices of the minimum respectively. But using a similar technique isn't working for a 3D array. For a 10x10x11 array variance_matrix (file attached), the command

find(variance_matrix==min(min(min(variance_matrix))))

outputs the location "151" which I presume represents the element in 2nd page/sheet, 5th column and 1st row.

But, the command

[row,col,depth] = find(variance_matrix==min(min(min(variance_matrix))));

outputs an absurd answer. "depth" is shown to be a logical operator while "col" (16) exceeds the actual column size (10).

I'm hoping someone would explain what's going wrong and if there's a right way to do it for multidimensional (>=3) arrays.

Answer 1

[Torsten]

A = rand(10,10,11);
B = A(:);
index = find(A==min(B))

index = 990

B(index)

ans = 2.4262e-04

[i j k] = ind2sub([10 10 11],index)

i = 10

j = 9

k = 10

A(i,j,k)

ans = 2.4262e-04

[PASUNURU SAI VINEETH]

Thank you, this works! But I'm curious what went wrong with the approach I've tried.

[Bruno Luong]

@PASUNURU SAI VINEETH The find command is not designed for more than 2D array.

So when you try on A_3D of size (m x n x p)

[i,j,page] = find(A_3d);

Matlab reshape A_3D to 2D array:

A_2D = A_3D(:,:)

or more precisely

A_2D = reshape(A_3D, [m, n*p])

The user command

[i,j,page] = find(A_3d);

is acutally performed like following

[i,j,value] = find( reshape(A_3D, [m, n*p]) );
page = value;

Therefore

• j can take values from 1 to n*p (and not n)
• and what you think page returned is actually the value of A_3D that are not equal to 0, in your case class logical (all TRUE) since your array A_3D is of class logical.

To check it you could fix your command (beside Torsen's solution) by

[i,j] = find(A_3d);
s = size(A_3d);
s = s(2:end); % n x p, p must be > 1
[j,page] = ind2sub(s, j);

[PASUNURU SAI VINEETH]

@Bruno Luong Thanks a lot for the detailed explanation.

[Torsten]

@Bruno Luong Thank you.

Answer 2

[Steven Lord]

Use the 'all' dimension argument and the 'linear' index argument to obtain the linear index of the maximum of the array considering the data in all dimensions.

A = reshape(randperm(24), [3 2 4])

A = 
A(:,:,1) =

     5    13
     9     3
    23     6


A(:,:,2) =

    22    15
     8    24
    19    16


A(:,:,3) =

    11    21
     1    14
     2    10


A(:,:,4) =

    18    12
    20     7
     4    17

[value, ind] = min(A, [], 'all', 'linear')

value = 1

ind = 14

A(ind)

ans = 1

If you need the subscripts instead of the linear index, use ind2sub.

[row, column, page] = ind2sub(size(A), ind)

row = 2

column = 1

page = 3