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Question:

I have a problem. I need to iterate through every element in an n-dimensional matrix in MATLAB. The problem is, I don't know how to do this for an arbitrary number of dimensions. I know I can say

`for i = 1:size(m,1) for j = 1:size(m,2) for k = 1:size(m,3) `

and so on, but is there a way to do it for an arbitrary number of dimensions?

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Solution:1

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()

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Solution:2

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.)

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Solution:3

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 a couple of other tricks 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); `

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Solution:4

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 `

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Solution:5

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.

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Solution:6

these solutions are more faster (about 11%) than using `numel`

;)

`for idx = reshape(array,1,[]), element = element + idx; end `

or

`for idx = array(:)', element = element + idx; end `

UPD. tnx @rayryeng for detected error in last answer

# Disclaimer

The timing information that this post has referenced is incorrect and inaccurate due to a fundamental typo that was made (see comments stream below as well as the edit history - specifically look at the first version of this answer). Caveat Emptor.

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Solution:7

If you look deeper into the other uses of `size`

you can see that you can actually get a vector of the size of each dimension. This link shows you the documentation:

www.mathworks.com/access/helpdesk/help/techdoc/ref/size.html

After getting the size vector, iterate over that vector. Something like this (pardon my syntax since I have not used Matlab since college):

`d = size(m); dims = ndims(m); for dimNumber = 1:dims for i = 1:d[dimNumber] ... `

Make this into actual Matlab-legal syntax, and I think it would do what you want.

Also, you should be able to do Linear Indexing as described here.

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Solution:8

You want to simulate n-nested for loops.

Iterating through n-dimmensional array can be seen as increasing the n-digit number.

At each dimmension we have as many digits as the lenght of the dimmension.

Example:

Suppose we had array(matrix)

`int[][][] T=new int[3][4][5]; `

in "for notation" we have:

`for(int x=0;x<3;x++) for(int y=0;y<4;y++) for(int z=0;z<5;z++) T[x][y][z]=... `

to simulate this you would have to use the "n-digit number notation"

We have 3 digit number, with 3 digits for first, 4 for second and five for third digit

We have to increase the number, so we would get the sequence

`0 0 0 0 0 1 0 0 2 0 0 3 0 0 4 0 1 0 0 1 1 0 1 2 0 1 3 0 1 4 0 2 0 0 2 1 0 2 2 0 2 3 0 2 4 0 3 0 0 3 1 0 3 2 0 3 3 0 3 4 and so on `

So you can write the code for increasing such n-digit number. You can do it in such way that you can start with any value of the number and increase/decrease the digits by any numbers. That way you can simulate nested for loops that begin somewhere in the table and finish not at the end.

This is not an easy task though. I can't help with the matlab notation unfortunaly.

**Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com**

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