Find subscripts in a 3D matrix

I have a 3D matrix, called A, and I want to find the subscripts for all places where elements of vector B appear. The elements of B apper more than one time in the matrix A.

I tried it using function subs to obtain directly the susbsripts, but I obtained a result where many page numbers are wrong (it shows a higher number (5) than the existing pages (3)). Bellow the code:

A(:,:,1)=[1 2 ; 3 3 ; 5 7 ; 4 2];

A(:,:,2)=[9 1 ; 4 4 ; 9 2 ; 4 6];

A(:,:,3)=[2 4 ; 7 4 ; 3 1 ; 6 4];

B = [9 4 1];

[row,col,page] = ind2sub(size(A),find(A(:)==B))

Also I tried using function find to obtain the linear indexes to later convert them to subscripts; however, I only obtain the first place were the element is placed on the matrix:

n = arrayfun(@(x) find(A==x,1),B);

[row, col, page] = ind2sub(size(A), n);

Answer 1

[Walter Roberson]

A(:,:,1)=[1 2 ; 3 3 ; 5 7 ; 4 2];
A(:,:,2)=[9 1 ; 4 4 ; 9 2 ; 4 6];
A(:,:,3)=[2 4 ; 7 4 ; 3 1 ; 6 4];
B = [9 4 1];
[row,col,page] = ind2sub(size(A),find(ismember(A,B)))

row = 12×1

1 4 1 2 3 4 1 2 1 2

col = 12×1

1 1 1 1 1 1 2 2 2 2

page = 12×1

1 1 2 2 2 2 2 2 3 3

[Walter Roberson]

Your error is that A(:) == B is numel(A) by length(B). The first column each element of A to the first element of B; the second column compares each element of A to the second element of B, and so on.

Another way you could do it is

[row,col,page] = ind2sub(size(A),find(any(A(:)==B,2)))

[Fernando Montero]

That was a really fast answer. Thanks for your help Walter!

Can you help me with the correct way to convert this subscripts to linear indices?

[Walter Roberson]

find(ismember(A,B)) would be linear

sub2ind(size(A), row,col,page)

if you need to reconstruct linear

[Fernando Montero]

Thanks again!