Building a connectivity matrix from one-to-many columns

Hi All.

For a simpler example of what I am doing - I have two columns of data. They are both numbers, both 'n' in size. Say column A and column B.

Each element in column A, maps to the corresponding location in column B. The elements are not unique - one number will occur many times in column A, for instance, each time mapping to a unique value in column B (there are no repetitions).

For a given element in A, I want to find all of the corresponding mapping points in B.

I then want to build a matrix, where rows and columns correspond to IDs from A. The (i,j) position will then be equal to 1, if the i,j'th elements in A have a common mapping point in B.

For a small example,

A = [1 1 2 2 2 3 3] B = [a b a c d b d]

Then, '1' has a map in common (1 goes to a and b, and so on) with '2' - 'a' and '3' - 'b'. and '2' has a common point with '3' - 'd'. Then, the matrix would be:

(1,2) = 1, (2,3)=1, (1,3)=1. (it will be symmetric, so the order (1,2) or (2,1) doesn't matter).

I can do this with lots of loops, but if someone has a better way I would be thankful.

Neil.

[Sean de Wolski]

I don't understand why 1,3 = 1. Can you provide the _actual_ resulting matrix and use numbers instead of letters in B (or provide scalars for those letters).

[Walter Roberson]

1,3 = 1 because 1 maps to b (position 2) and 3 maps to b (position 6)

[Neil]

As Walter says.

I thought I had an answer with a plethora of loops, but it seems I keep making mistakes. Hopefully I can still find an answer!

I think whether or not B contains letters or number isn't important - it is a one to many relationship from A to B that I want to express as a matrix. I could have numbers instead of letters in B, but all it all those values are simply labels for a set.

The numbers in A are the important ones - these are the values I will use as indices for my matrix. Therefore, the matrix is square with dimensions equal to the number of unique entries in A.

Answer 1

[Sean de Wolski]

So something like:

a = pi; %numbers;
b = 2.1;
c = 1.2;
d = 42;

A = [1 1 2 2 2 3 3]'; %column vectors
B = [a b a c d b d]';

Aacc = accumarray(A,B,[],@(x){x}); %build cell array of parts
[r c] = find(triu(true(numel(Aacc)),1)); %non diagonal coords
connmat = diag(cellfun('prodofsize',Aacc)); %make diagonal
for ii = 1:numel(r);
    n = sum(ismember(Aacc{r(ii)},Aacc{c(ii)})); %number intersecting
    connmat(r(ii),c(ii)) = n; %save
    connmat(c(ii),r(ii)) = n;
end

?