intersection of multiple arrays

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Ananya Malik 2017-9-14

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评论： Ananya Malik 2017-9-14

I have multiple 1D arrays with integers. Eg. A=[1,2,3,4] B=[1,4,5,6,7] C=[1,5,8,9,10,11]. Here 1 is repeated in all the arrays and 4 in A&B, 5 in B&C. I want these repeated values to be assigned to a single array, based on the size of the array, and removed from other arrays. How can I achieve this. Please help. TIA.

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KL 2017-9-14

So you only wanto to remove 1 from all the matrices in the above example?

Ananya Malik 2017-9-14

编辑：Ananya Malik 2017-9-14

No. The common elements are [1,4,5]. Upon removing them from all arrays, we get A=[2,3] B=[6,7] and C=[8,9,10,11]. 1 can be present in either A or B. Suppose we put 1 in A ==> A=[1,2,3] B=[6,7] C=[8,9,10,11]. 4 can be present in A or B. Size(B)<size(A), therefore A=[1 2 3] B=[4,6 7] c=[8,9,10,11]. Finally 5 can be present in B or C, but size(B)<size(C). Therefore the final output should be A=[1,2,3] B=[4,5,6,7] and C=[8,9,10,11].

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Answer 1

Guillaume 2017-9-14

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This is in no way guaranteed to give you a perfectly balanced output. After that you need to look at optimisation algorithm which is beyond my field. This is also a lot less efficient than my other solution:

C = {[1 2 3 4 5 6], [1 4 5 6 7], [1 5 8 9 12 20]}
[uvals, ~, bin] = unique(cell2mat(C));
origarray = repelem(1:numel(C), cellfun(@numel, C));
dist = table(uvals', accumarray(bin, 1), accumarray(bin, origarray, [], @(x) {x}), 'VariableNames', {'value', 'repetition', 'arrayindices'});
dist = sortrows(dist, 'repetition');

The above builds a table of all the unique values, how many times they're repeated and where they come from originally. You can then iterate through that to distribute the values:

%distribute non-repeated values first
newC = accumarray(cell2mat(dist.arrayindices(dist.repetition == 1)), ...
                  dist.value(dist.repetition == 1), [], @(x){x'});
%and remove them from table
dist(dist.repetition == 1, :) = [];
%distribute remaining values one at a time
for row = 1:height(dist)
    destarrays = dist.arrayindices{row};  %which array originally had the current value?
    [~, destidx] = min(cellfun(@numel, newC(destarrays)));  %find which is currently smallest
    newC{destarrays(destidx)} = [newC{destarrays(destidx)}, dist.value(row)];  %append value to that smallest array
end
celldisp(newC)

I'm using a table here, which is not the most efficient container in matlab, but that make it easier to understand the code.

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Ananya Malik 2017-9-14

Thank you so much. Not exactly what I wanted, but pretty close. Will tweak it further for my use. Cheers

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Answer 2

KL 2017-9-14

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A=[1,2,3,4];
B=[1,4,5,6,7];
C=[1,5,8,9,10,11];
M = {A,B,C};
N = {[B C], [A C], [A B]};
CommonElements = unique(cell2mat(cellfun(@intersect, M,N,'UniformOutput',false)));
NewM = cellfun(@(x) removeEl(x,CommonElements),M,'UniformOutput',false);
%and then
function x = removeEl(x,a)
for el=1:length(a)
   x(x==a(el))=[];
end
end

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Answer 3

Guillaume 2017-9-14

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This is how I'd do it:

C = {[1 2 3 4], [1 4 5 6 7], [1 5 8 9 10 11]};
%sort cell array by increasing vector size:
[~, order] = sort(cellfun(@numel, C));
C = C(order);
%get all unique values
uvals = unique(cell2mat(C));
%go through all arrays, keeping all values in uvals, then removing them from uvals so they can't be used by the next array
for iarr = 1:numel(C)
  C{iarr} = intersect(C{iarr}, uvals);  %only keep the values in uvals
  uvals = setdiff(uvals, C{iarr});  %and remove the one we've just used
end
celldisp(C)

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Ananya Malik 2017-9-14

编辑：Ananya Malik 2017-9-14

I am trying to balance the number of elements in the arrays. So if I change my input to A=[1 2 3 4 5 6], B=[1 4 5 6 7] and C=[1 5 8 9 10 11]. Using the code you provided gives me A=[1 4 5 6 7] B=[2 3] and C=[8 9 10 11]. Whereas ideal case would be A=[2 3 4 6] B=[1 7 5] and C=[8 9 10 11] or A=[1 2 3 4] B=[5 6 7] C=[8 9 10 11] or something along this line.

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