Subsets of uncorrelated features

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Kais 2021-7-10

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评论： Kais 2021-7-15

Given a N by N correlation matrix of N features, how to find ALL subsets of pariwise uncorrelated features if we assume two features are uncorrelated if their correlation score is less than a threshold Alpha. There is no restriction on the number of features making the subsets. All features making a subset need to be pairwise uncorrelated.

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Answer 1

Jeff Miller 2021-7-12

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https://ww2.mathworks.cn/matlabcentral/answers/875908-subsets-of-uncorrelated-features#answer_744638

编辑：Jeff Miller 2021-7-12

在 MATLAB Online 中打开

N = 5;
R = rand(N);  % We will ignore the lower triangular part of this array
rCutoff = 0.4;
% Make a cell array that holds all possible combinations of 2, 3, 4, ... features
combos = cell(0,0);
for i=2:N
    iCombos = nchoosek(1:N,i);
    for j=1:size(iCombos,1)
        combos{end+1} = iCombos(j,:);
    end
end
ncells = numel(combos);
% Check each cell to make sure that all of the pairwise correlations are
% less than the cutoff
qualifies = true(1,ncells);
for icell=1:ncells
    features = combos{icell};
    nfeatures = numel(features);
    for ifeature=1:nfeatures-1
        for jfeature=ifeature+1:nfeatures
            iifeature = features(ifeature);
            jjfeature = features(jfeature);
            if abs(R(iifeature,jjfeature)) > rCutoff
                qualifies(icell) = false;
            end
        end
    end
end

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Jeff Miller 2021-7-13

You may well be right, that but "if sum" line is cognitively impenetrable to me. :)

Thanks for accepting my answer.

Kais 2021-7-14

在 MATLAB Online 中打开

I am looking for a pairwise uncorrleation between ALL features iofn the "feature" variable". The line:

sum(nonzeros(triu(abs(R(features,features)),1)) > rCutoff)

will result into a matrix of logical values showing which feature pairs are correlated and which are not (triu is only there to reduce the symmetric correlation matrix). If any of the values of the matrix is true (equivalently, sum of values of the matrix is different from zero), The subset unqualifies as an uncorrelated subset.

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Answer 2

Ive J 2021-7-11

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https://ww2.mathworks.cn/matlabcentral/answers/875908-subsets-of-uncorrelated-features#answer_744398

编辑：Ive J 2021-7-12

在 MATLAB Online 中打开

Let R be the pairwise correlation matrix:

N = 10;
R = rand(N); 
R(logical(eye(N))) = 1; 
for i = 1:size(R, 1) - 1
    for j = i+1:size(R, 1)
        R(j, i) = R(i, j);
    end
end
disp(R)
cutoff = 0.4; % independent features
idx = R < cutoff;
idx = triu(idx); % R(i, j) == R(j, i) in pairwise correlation matrix
features = "feature" + (1:N); % feature names
% there may be a simpler way to do this
indepFeatures = [];
for i = 1:N
    indepFeatures = [indepFeatures, arrayfun(@(x)[x, features(i)], features(idx(i, :)), 'uni', false)];
end
indepFeatures = vertcat(indepFeatures{:});
% find all cliques of this set
nodes = zeros(size(indepFeatures, 1), 1);
[~, nodes(:, 1)] = ismember(indepFeatures(:, 1), features);
[~, nodes(:, 2)] = ismember(indepFeatures(:, 2), features);
G = graph(nodes(:, 1), nodes(:, 2));
M = maximalCliques(adjacency(G)); 
indepSets = cell(size(M, 2), 1);
for i = 1:numel(indepSets)
    indepSets{i} = features(M(:, i) ~= 0);
end
indepSets(cellfun(@numel, indepSets) < 2) = []; % this can be further unified with indepFeatures

You can find maximalCliques in FEX.

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Kais 2021-7-12

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@Ive J I tried your solution on a 10 by 10 validation correlation matrix. The subsets are correct but incomplete. There should be 22 subsets (17 made of pairs and 5 made of triples) of uncorrelated features in the matrix below. Your code leads to 11 subsets (6 paird and 5 triples).

cutoff = 0.4;
R=[
    1.0000    0.8735   -0.4498    0.7882    0.8260   -0.3310    0.7521    0.8991   -0.3789    0.7062
    0.8735    1.0000   -0.4168    0.9182    0.6789   -0.3210    0.7074    0.8244   -0.5138    0.7961
   -0.4498   -0.4168    1.0000   -0.3241   -0.4364    0.7018   -0.3825   -0.3257    0.4335   -0.3814
    0.7882    0.9182   -0.3241    1.0000    0.4857   -0.2513    0.6033    0.8936   -0.5378    0.5715
    0.8260    0.6789   -0.4364    0.4857    1.0000   -0.2878    0.6495    0.5794   -0.2881    0.8435
   -0.3310   -0.3210    0.7018   -0.2513   -0.2878    1.0000   -0.3040   -0.2360    0.2695   -0.2606
    0.7521    0.7074   -0.3825    0.6033    0.6495   -0.3040    1.0000    0.6469   -0.3280    0.6357
    0.8991    0.8244   -0.3257    0.8936    0.5794   -0.2360    0.6469    1.0000   -0.3919    0.5148
   -0.3789   -0.5138    0.4335   -0.5378   -0.2881    0.2695   -0.3280   -0.3919    1.0000   -0.3725
    0.7062    0.7961   -0.3814    0.5715    0.8435   -0.2606    0.6357    0.5148   -0.3725    1.0000]

Kais 2021-7-14

编辑：Kais 2021-7-15

@Ive J I tried your code with cutoff = .7. I get 32 pairs, 2 triples, 2 quadruples, and 5 quintuples. While the number of pairs and quintuples seem to be correct, the numbers of triples and quadruples are not. For example, the triple [6, 7, 9] is missing.

There should be a total of 112 uncorrelated features. You code finds 44 only. Any clue why is that?

Kais 2021-7-15

Never mind the last question. I figured the algorithm finds max cliques so it won't count subgraphs within larger subgraphs ([6, 7,9] won't show up because it's part of the larger subset [4, 6,7,9]), which significantly reduces the number of subsets.

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