## Find all lines of a matrix which have a matching pair of values

### Florent Hannard (view profile)

on 29 May 2019
Latest activity Commented on by Luna

### Luna (view profile)

on 10 Jun 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

Guillaume

### Guillaume (view profile)

on 29 May 2019
Note that:
length(A(:, 1))
is more simply written as
size(A, 1)
which will probably execute faster as well (since it doesn't need to allocate memory for the temporary A(:, 1)).
Guillaume

### Guillaume (view profile)

on 29 May 2019
What if a pair is common to more than two rows? How should Index be constructed in that case?
"But this is not efficient for a very large matrix A"
What is considered large in this context?
Florent Hannard

### Florent Hannard (view profile)

on 29 May 2019
That's right thanks, but what I hope is to get rid of (at least) one of the loop (with something similar to ismember).
Large could be something like [500x10] or more, and I will need to perform this operation many times (which is why I am trying to optimize it).
EDIT : I know that a pair will never be common to more than two rows.

### Jan (view profile)

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, :);

Luna

on 10 Jun 2019
+1

### Kevin Phung (view profile)

on 29 May 2019
Edited by Kevin Phung

### Kevin Phung (view profile)

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

### Guillaume (view profile)

on 29 May 2019
Is the question about identifying/ the pairs (the hard part) or just the location of already identified pairs?
Jan

### Jan (view profile)

on 29 May 2019
Pairs is wanted as output of the function.
Florent Hannard

### Florent Hannard (view profile)

on 29 May 2019
Indeed, Pairs and index are wanted as outputs and not just the location of already indentified paires.

on 29 May 2019
Edited by Luna

### Luna (view profile)

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

### Florent Hannard (view profile)

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.

### Guillaume (view profile)

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