Create a Cartesian product with user-specified constraints

Hello,

I'm trying to figure out how to code a simple and flexible cartesian product function

To abstract how it works conceptually, let's imagine we have a 4 X 10 matrix, with each column having ordered values 1 through 4, like this

1 1 1 1 1 1 etc

2 2 2 2 2 2 etc

3 3 3 3 3 3 etc

4 4 4 4 4 4 etc

The user specifies the number of rows and columns to be used to make the cartesian product. For example, if they specify 2 rows and 3 columns, the result should be (although the order doesn't matter)

1 1 1

1 1 2

1 2 1

2 1 1

1 2 2

2 1 2

2 2 1

2 2 2

The closest function that I've found is

function C = cartesian(varargin)

% This function creates cartesian products from input arrays

args = varargin;

n = nargin;

[F{1:n}] = ndgrid(args{:});

for i=n:-1:1

G(:,i) = F{i}(:);

end

C = unique(G , 'rows');

end

However, it asks for a series of arrays as input. In my case, I need the input to be a matrix of variable dimensions. Thank you for any help!

[Walter Roberson]

The question is a bit ambiguous as to what portion of the array is to be selected. Is it leading rows and leading columns, or tailing rows and trailing columns? It matters if for example

M = magic(5)

M = 5×5

17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

[Emmanuel]

Thank you for your help. Selection always begins from the top left. Also, the columns always begin with '1' and increment down.

[Stephen23]

"I just had to transpose my 'cartesian_index' matrix before splitting it."

You don't have to transpose anything, just split on the correct dimension:

nR = 2;
nC = 3;
X = repmat(1:nR,nC,1);
C = num2cell(X,2);
D = cartesian1(C{:})

D = 8×3

1 1 1 2 1 1 1 2 1 2 2 1 1 1 2 2 1 2 1 2 2 2 2 2

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

The function is simpler and likely more efficient without UNIQUE and the FOR loop:

function M = cartesian1(varargin)
[varargin{:}] = ndgrid(varargin{:});
M = reshape(cat(nargin+1, varargin{:}), [], nargin);
end

As posed creating that full matrix is superfluous anyway, you don't even need to duplicate all of those vectors, so you could make the whole thing simpler with one single vector:

nR = 2;
nC = 3;
D = cartesian2(1:nR,nC)

D = 8×3

1 1 1 2 1 1 1 2 1 2 2 1 1 1 2 2 1 2 1 2 2 2 2 2

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

●

function M = cartesian2(V,nC)
[C{1:nC}] = ndgrid(V);
M = reshape(cat(nC+1, C{:}), [], nC);
end

Answer 1

[Walter Roberson]

M = [1    1    1    1    1    1
2    2    2    2    2    2
3    3    3    3    3    3
4    4    4    4    4    4]

M = 4×6

1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Split = num2cell(M,2)

Split = 4×1 cell array
    {[1 1 1 1 1 1]}
    {[2 2 2 2 2 2]}
    {[3 3 3 3 3 3]}
    {[4 4 4 4 4 4]}

cartesian(Split{:})

[Emmanuel]

Thank you so much for your help. I like the idea of breaking up the matrix using num2cell. I tried implementing your suggestion with

search_rank = 2;
search_depth = 3;
cartesian_index = repmat((1:search_rank)', 1, search_depth)
cart_split = num2cell(cartesian_index,2)
cartesian(cart_split{:})

but the result was

ans =

1 2

whereas, I'd like to get back (the order doesn't matter)

ans =

1 1 1

1 1 2

1 2 1

2 1 1

1 2 2

2 1 2

2 2 1

2 2 2

[Emmanuel]

I got it to work! I just had to transpose my 'cartesian_index' matrix before splitting it. Thanks again for your great suggestion - I was stumped all morning on this.