find combination of labels

Hello! I describe the view of my problem: I have an user that visit same locations. This location are indicated like

A=[618 574 608];

every value in A indicate a location. To every location correspond different labels that indicate the places that a user can visit in a location:

B=[574 2;574 3;608 1;618 1;618 2;618 3;618 4];

The first column of B indicates the locations and the second column indicates the label associated to every location. E.g. Location 574 corresponds labels 2,3 Location 608 corresponds label 1 Location 618 corresponds label 1,2,3,4

What I want: starting from A and considering B, in which are indicated location and associated labels, I want to find all the possible combination of label of the sequence of location in A. The aspected output is

C=[ 121; 221; 321; 421; 131; 231; 331; 431 ]

Now I explain the output: A=[618 574 608], label associated to 618 are 1,2,3,4: the first digits of values in C; label associated to 574 are 2,3: second digits of values in C; label associated to 608 is 1: third digits of values in C.

Can you help me? I hope the question is clear: if not, I can try to explain better.

[Azzi Abdelmalek]

This is not clear

[elisa ewin]

what is not clear? the output? I want to find the combination of (1,2,3,4) (2,3) 1

Answer 1

[Ced]

Hi

Your procedure has two steps:

1. find all labels for a current position

2. find all possible combinations

This is one possible way of doing it

A=[618 574 608]; 
B=[574 2;574 3;608 1;618 1;618 2;618 3;618 4];

% 1. find label for each position
N = length(A);
A_mat = cell(N,1);
for i = 1:N
    A_mat{i} = B(B(:,1) == A(i),2);
end

% 2. combine
[ X, Y, Z ] = meshgrid(A_mat{:});
C = str2num([ num2str(X(:)) num2str(Y(:)) num2str(Z(:)) ]);

NOTE: I am assuming that the order of the combinations at the end does not matter? Otherwise, you will need to reorder them

EDIT: If you have more than 3 nodes in A, you can use ndgrid instead of meshgrid. Have a look at Guillaume's answer.

[elisa ewin]

If the dimension of A are different of 3 (for example 5), meshgrid don't work: how can I replace it?

[Guillaume]

See my answer that does not depend on the size of A.

It also uses simple arithmetic to construct your final numbers instead of number to string and string to number conversions so should be much faster.

Answer 2

[Guillaume]

Here is one way to do it. Note that I'm using ndgrid to generate all the combinations. If you have too many locations and labels, ndgrid will run out of memory.

A=[618 574 608];
B=[574 2;574 3;608 1;618 1;618 2;618 3;618 4];
C = arrayfun(@(loc) B(ismember(B(:, 1), A), 2), A, 'UniformOutput', false); %find all labels for each location
[C{:}] = ndgrid(C{:}); %generate all possible combinations
C = cellfun(@(c) c(:), C, 'UniformOutput', false); %reshape into column vectors
C = [C{:}] %and concatenate all into one vector
%I would leave the output C as is, if you really want it as one number:
D = sum(bsxfun(@times, C, 10.^(size(C, 2)-1:-1:0)), 2)

edit: typo in the code

[Ced]

I believe the C = arrayfun... line should be

C = arrayfun(@(loc) B(ismember(B(:, 1), loc), 2), A, 'UniformOutput', false); %find all labels for each location

i.e. the "A" should be "loc"

Answer 3

[Jos (10584)]

A simple one-liner will do

A = [618 574 608]
B = [574 2;574 3;608 1;618 1;618 2;618 3;618 4]
C1 = arrayfun(@(k) B(B(:,1)==A(k),2), 1:numel(A), 'un', 0)

The cell C1{i} now holds the values of B(:,2) for which B(:,1) equals A(i) To get all the combinations, my function ALLCOMB (a powerful wrapper around ndgrid) makes that quite easy:

C2 = allcomb(C1{:})

ALL COMB can be downloaded from the Matlab File Exchange: http://www.mathworks.com/matlabcentral/fileexchange/10064

[Guillaume]

I'm unclear how that is different from my:

 C = arrayfun(@(loc) B(ismember(B(:, 1), A), 2), A, 'UniformOutput', false)

and how that answer the question of generating all the combinations

[Jos (10584)]

No need of ismember here. I forgot about the combinations (now added).