How do I iterate through each element in an n-dimensional matrix in MATLAB?

You can use linear indexing to access each element.

for idx = 1:numel(array)
    element = array(idx)
    ....
end

This is useful if you don't need to know what i,j,k, you are at. However, if you don't need to know what index you are at, you are probably better off using arrayfun()

The idea of a linear index for arrays in matlab is an important one. An array in MATLAB is really just a vector of elements, strung out in memory. MATLAB allows you to use either a row and column index, or a single linear index. For example,

A = magic(3)
A =
     8     1     6
     3     5     7
     4     9     2

A(2,3)
ans =
     7

A(8)
ans =
     7

We can see the order the elements are stored in memory by unrolling the array into a vector.

A(:)
ans =
     8
     3
     4
     1
     5
     9
     6
     7
     2

As you can see, the 8th element is the number 7. In fact, the function find returns its results as a linear index.

find(A>6)
ans =
     1
     6
     8

The result is, we can access each element in turn of a general n-d array using a single loop. For example, if we wanted to square the elements of A (yes, I know there are better ways to do this), one might do this:

B = zeros(size(A));
for i = 1:numel(A)
  B(i) = A(i).^2;
end

B
B =
    64     1    36
     9    25    49
    16    81     4

There are many circumstances where the linear index is more useful. Conversion between the linear index and two (or higher) dimensional subscripts is accomplished with the sub2ind and ind2sub functions.

The linear index applies in general to any array in matlab. So you can use it on structures, cell arrays, etc. The only problem with the linear index is when they get too large. MATLAB uses a 32 bit integer to store these indexes. So if your array has more then a total of 2^32 elements in it, the linear index will fail. It is really only an issue if you use sparse matrices often, when occasionally this will cause a problem. (Though I don't use a 64 bit MATLAB release, I believe that problem has been resolved for those lucky individuals who do.)

As pointed out in a few other answers, you can iterate over all elements in a matrix A (of any dimension) using a linear index from 1 to numel(A) in a single for loop. There are also a couple of functions you can use: arrayfun and cellfun.

Let's first assume you have a function that you want to apply to each element of A (called my_func). You first create a function handle to this function:

fcn = @my_func;

If A is a matrix (of type double, single, etc.) of arbitrary dimension, you can use arrayfun to apply my_func to each element:

outArgs = arrayfun(fcn, A);

If A is a cell array of arbitrary dimension, you can use cellfun to apply my_func to each cell:

outArgs = cellfun(fcn, A);

The function my_func has to accept A as an input. If there are any outputs from my_func, these are placed in outArgs, which will be the same size/dimension as A.

One caveat on outputs... if my_func returns outputs of different sizes and types when it operates on different elements of A, then outArgs will have to be made into a cell array. This is done by calling either arrayfun or cellfun with an additional parameter/value pair:

outArgs = arrayfun(fcn, A, 'UniformOutput', false);
outArgs = cellfun(fcn, A, 'UniformOutput', false);

One other trick is to use ind2sub and sub2ind. In conjunction with numel and size, this can let you do stuff like the following, which creates an N-dimensional array, and then sets all the elements on the "diagonal" to be 1.

d = zeros( 3, 4, 5, 6 ); % Let's pretend this is a user input
nel = numel( d );
sz = size( d );
szargs = cell( 1, ndims( d ) ); % We'll use this with ind2sub in the loop
for ii=1:nel
    [ szargs{:} ] = ind2sub( sz, ii ); % Convert linear index back to subscripts
    if all( [szargs{2:end}] == szargs{1} ) % On the diagonal?
        d( ii ) = 1;
    end
end

You could make a recursive function do the work

• Let L = size(M)
• Let idx = zeros(L,1)
• Take length(L) as the maximum depth
• Loop for idx(depth) = 1:L(depth)
• If your depth is length(L), do the element operation, else call the function again with depth+1

Not as fast as vectorized methods if you want to check all the points, but if you don't need to evaluate most of them it can be quite a time saver.