Asked by Florent Hannard
on 29 May 2019

Hi everyone,

Given a matrix, I would like to find all pairs of lines which have a matching pair of values.

So for :

A = [1 2 3 4;

2 5 7 8;

4 5 7 9;

4 6 7 13;

6 8 13 15]

I would like to get a matrix with similar values :

Pairs = [5 7;

4 7;

6 13]

And for which lines these values are found :

Index = [2 3;

3 4;

4 5]

In other words : intersection of A(Index(i,1),:) and A(Index(i,2),:) is Pairs(i,:)

One simple way to do it would be to use the intersect function :

for i=1:length(A(:,1))

for j=i+1:length(A(:,1))

intersect(A(i,:),A(j,:))

end

end

But this is not efficient for a very large matrix A. Could someone suggest a more efficient way to do it ?

Thanks!

Florent

Answer by Jan
on 31 May 2019

Sort the input array to avoid some overhead in intersect:

A = [1 2 3 4;

2 5 7 8;

4 5 7 9;

4 6 7 13;

6 8 13 15]

As = sort(A, 2);

nA = size(A, 1);

m = nA * (nA - 1) / 2;

Pairs = zeros(m, 2); % Pre-allocate!

Index = zeros(m, 2);

count = 0;

for i1 = 1:nA

Ai1 = A(i1, :);

for i2 = i1 + 1:nA

match = ismembc(Ai1, A(i2, :)); % C-Mex function, undocumented

if any(match)

count = count + 1;

Pairs(count, :) = Ai1(match);

Index(count, 1) = i1;

Index(count, 2) = i2;

end

end

end

Pairs = Pairs(1:count, :);

Index = Index(1:count, :);

Answer by Kevin Phung
on 29 May 2019

Edited by Kevin Phung
on 29 May 2019

A = [1 2 3 4; 2 5 7 8; 4 5 7 9; 4 6 7 13; 6 8 13 15];

Pairs = [5 7; 4 7; 6 13];

index =zeros(size(Pairs,1),1);

for i = 1:size(Pairs,1)

loc = ismember(A,Pairs(i,:));

index(i,:)=find(sum(loc,2)==numel(Pairs(i,:)))';

end

Guillaume
on 29 May 2019

Jan
on 29 May 2019

Pairs is wanted as output of the function.

Florent Hannard
on 29 May 2019

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Answer by Luna
on 29 May 2019

Edited by Luna
on 29 May 2019

Basically you can use this:

Finds the first row of the pairs' elements seperately as logical indexes. Then adds them together.

If it matches at the same time on that row, sum of rows will be 2. So these are your new indexes that are found. Repeats them for 2nd and 3rd row also.

IndexesNew = zeros(size(Pairs,1),2);

for i = 1:size(Pairs,1)

IndexesNew(i,:) = find(sum(ismember(A,Pairs(i,1)) + ismember(A,Pairs(i,2)),2) == 2)';

end

Florent Hannard
on 29 May 2019

Thank you but this is not what I need.

Pairs and index are wanted as outputs and not just the location of already indentified paires.

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Answer by Guillaume
on 29 May 2019

I don't see how you can avoid doing size(A, 1) * (size(A, 1)-1) / 2 comparisons of the rows one way or another. The problem with using intersect in a loop is that you're repeatedly sorting the rows that you may already have sorted previously so certainly taking that out of the loops would be useful.

The following may or may not be faster than using intersect in a loop. I haven't tested.

%If size(A, 1) is fixed, this part can be precomputed before applying the rest to each A

%It simply precomputes row indices for comparison

[row2, row1] = ndgrid(1:size(A, 1));

row1 = nonzeros(tril(row1, -1));

row2 = nonzeros(tril(row2, -1));

%finding pairs and locations

[uA, ~, destcol] = unique(A); %only one call to unique on the whole matrix

hasnumber = zeros(size(A, 1), numel(uA)); %columns of hasnumber correspond to values in uA, rows to rows of A. Will contain one when the corresponding row contains uA

hasnumber(sub2ind(size(hasnumber), repmat(1:size(A, 1), 1, size(A, 2))', destcol)) = 1;

ismatch = hasnumber(row1, :) + hasnumber(row2, :) == 2; %find which numbers are common to both row1 and row2

haspair = sum(ismatch, 2) == 2; %pair when exactly two numbers are common

uA = repmat(uA, 1, sum(haspair));

RowIndices = [row1(haspair), row2(haspair)] %indices of row that have a matching pair

PairValues = reshape(uA(ismatch(haspair, :)'), 2, []).' %Actual values of pair

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## 3 Comments

## Guillaume (view profile)

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## Guillaume (view profile)

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## Florent Hannard (view profile)

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