One possible solution:
function [Index] = GetIntersectionIndicies(m1, m2)
[~, I1] = intersect(m1, m2, 'rows');
Index = zeros(size(m1, 1), 1);
Index(I1) = 1;
By the way, I love the inventive solution of @Shai, and it is much faster than my solution if your matrices are small. But if your matrices are large, then my solution will dominate. This is because if we set T = size(m1, 1)
, then the tmp
variable in the answer of @Shai will be T*T, ie a very large matrix if T
is large. Here's some code for a quick speed test:
%# Set parameters
T = 1000;
M = 10;
%# Build test matrices
m1 = randi(5, T, 2);
m2 = randi(5, T, 2);
%# My solution
tic
for m = 1:M
[~, I1] = intersect(m1, m2, 'rows');
Index = zeros(size(m1, 1), 1);
Index(I1) = 1;
end
toc
%# @Shai solution
tic
for m = 1:M
tmp = bsxfun( @eq, permute( m1, [ 1 3 2 ] ), permute( m2, [ 3 1 2 ] ) );
tmp = all( tmp, 3 ); % tmp(i,j) is true iff m1(i,:) == m2(j,:)
imdices = any( tmp, 2 );
end
toc
Set T = 10
and M = 1000
, and we get:
Elapsed time is 0.404726 seconds. %# My solution
Elapsed time is 0.017669 seconds. %# @Shai solution
But set T = 1000
and M = 100
and we get:
Elapsed time is 0.068831 seconds. %# My solution
Elapsed time is 0.508370 seconds. %# @Shai solution