How to construct a binary matrix reporting 1 in case of equal rows of two arrays of different dimensions?

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MRC
MRC on 6 Aug 2014
Edited: Ashish Gudla on 6 Aug 2014
Hi, I have a matrix nxk
A=[ 1 2; 3 4; 5 6; 7 8]
and a matrix B mxk (m can be > = or < n)
B=[ 2 3; 4 5; 1 2; 5 6; 10 23; 7 8]
Each row of B is different.
I want to construct a matrix C nxm in which the hi-th element is 1 if A(h,:)=B(i,:), i.e.
C=[0 0 1 0 0 0; 0 0 0 0 0 0; 0 0 0 1 0 0; 0 0 0 0 0 1];
without looping.

Accepted Answer

Ashish Gudla
Ashish Gudla on 6 Aug 2014
Edited: Ashish Gudla on 6 Aug 2014
You could create the 'C' matrix with all zeros and then, find the positions where its supposed to be '1' and replace it.
To find the positions where there would be '1', you can use "ismember" function with rows( see doc ) to get the lowest index of each row in A that appears in B.
[a1,~] = size(A);
[b1,~] = size(B);
[~,t2]=ismember(A,B,'rows');
t1 = 1:a1;
t1= t1(t2~=0); %ignore zero indices
t2 = t2(t2~=0);
C = zeros(a1,b1);
i = sub2ind(size(C),t1,t2');
C(i) =1;

More Answers (1)

Chris Turnes
Chris Turnes on 6 Aug 2014
Edited: Chris Turnes on 6 Aug 2014
One way to do this without looping would be to exploit the property that two vectors w and v are equal if and only if
Using this, you could construct each term and compare:
>> IP = A*B.';
>> An = sum(abs(A).^2, 2); % abs only necessary if data is complex
>> Bn = sum(abs(B).^2, 2);
>> C = (IP == An*ones(1, length(Bn))) & (IP == ones(length(An), 1)*Bn.');
This is a way of vectorizing the operation, but it will use much more memory than looping would.

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