Count all pairwise interactions in a matrix

Question

Luis Cascao 2023-4-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1951578-count-all-pairwise-interactions-in-a-matrix

评论： Luis Cascao 2023-4-23

Suppose I have the following matrix:

[23 51 42; 42 23 18]

how can I count all pairwise terms occurring per row where order is not important? In this case, for row 1 the pairwise terms are (23,51),(23,42),(42,51) and for the second row (23,42),(18,42),(18,23).

Therefore the answer is:

{(23,51),1}, {(23,42),2}, {(42,51),1}, {(18,42),1}, {(18,23),1}

Many thanks

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

John D'Errico 2023-4-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1951578-count-all-pairwise-interactions-in-a-matrix#answer_1221253

编辑：John D'Errico 2023-4-23

在 MATLAB Online 中打开

If you have only one row, then it is trivial. Just use nchoosek.

Row = [2 3 5 7];
Row(nchoosek(1:4,2))
ans = 6×2
     2     3
     2     5
     2     7
     3     5
     3     7
     5     7

But you have multiple rows. We can still use nchoosek. But now we need to be a little more creative. For example:

Rows = [1 2 4 8;2 3 5 7]
Rows = 2×4
     1     2     4     8
     2     3     5     7
[nrows,ncols] = size(Rows);
combs = nchoosek(ncols,2)
combs = 6
permute(reshape(Rows(:,nchoosek(1:ncols,2)),nrows,combs,2),[2 3 1])
ans = 
ans(:,:,1) =

     1     2
     1     4
     1     8
     2     4
     2     8
     4     8


ans(:,:,2) =

     2     3
     2     5
     2     7
     3     5
     3     7
     5     7

The result is a 3-dimensional array, as it arguably should be.

As you can see, no loops were needed.

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Luis Cascao 2023-4-23

Thank you so much, exactly what I was looking for!

请先登录，再进行评论。

Answer 2

LeoAiE 2023-4-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1951578-count-all-pairwise-interactions-in-a-matrix#answer_1221238

在 MATLAB Online 中打开

Hi,

great question! Here is my attempt to solve this problem and if you find it helpful please accept the answer.. Thanks

% Input matrix
matrix = [23 51 42; 42 23 18];
% Number of rows
nrows = size(matrix, 1);
% Initialize an empty cell array to store the result
pairwise_terms = {};
% Iterate through each row
for row = 1:nrows
    % Get all unique pairs in the current row
    pairs = nchoosek(matrix(row, :), 2);
    
    % Iterate through each pair
    for i = 1:size(pairs, 1)
        % Sort the pair to ignore order
        sorted_pair = sort(pairs(i, :));
        
        if isempty(pairwise_terms)
            % If pairwise_terms is empty, add the first pair with a count of 1
            pairwise_terms(end + 1, :) = {sorted_pair, 1};
        else
            % Create a logical index array to check if the sorted_pair exists in pairwise_terms
            logical_index = cellfun(@(x) isequal(x, sorted_pair), pairwise_terms(:, 1));
    
            % Find the index of the existing pair in pairwise_terms
            idx = find(logical_index, 1);
            
            if isempty(idx)
                % If the pair does not exist, add it to pairwise_terms with a count of 1
                pairwise_terms(end + 1, :) = {sorted_pair, 1};
            else
                % If the pair exists, increment the count
                pairwise_terms{idx, 2} = pairwise_terms{idx, 2} + 1;
            end
        end
    end
end
% Display the result
pairwise_terms