Rapid Vector Matching/Search Problem

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Kyle Johnston on 20 Aug 2014
Commented: Oleg Komarov on 23 Aug 2014
I have a matrix (we'll call A) of size m x n, and a vector (we'll call B) that is a 1 x n, and I am interested in finding the one unique index (and I know there is just one) where A(index,:) equals B, is there a way for me to quickly make the determination in MATLAB besides using the following code:
for i = 1 : m
if ((isequal(A(i,:), B)))
indexIntrest = i;
break;
end
end
Thanks
Oleg Komarov on 23 Aug 2014
Have you tried the bsxfun() solution?

Joseph Cheng on 20 Aug 2014
Edited: Joseph Cheng on 20 Aug 2014
you'll need to use the function ismember()
such as:
found_row = ismember(A,B,'rows')
which will return a logical array where a 1 will represent a match. then by doing something like this
A= randi(2,100,4);
indexIntrest=find(ismember(A,[ 2 1 1 1],'rows'))
will return the indexes that match.
Joseph Cheng on 21 Aug 2014
Alright finally grasped how the perms() function lays out the permutations. Needed to eat a cookie and it came to me.
clc
clear all
n=10;
A= perms(1:n);
B=A(randi(length(A),1),:);
space = factorial(n-1);
tic
for i = (space*(n- B(1)) + 1) : space*(n- B(1) + 1)
if (isequal(A(i,:),B))
index3 = i
break;
end
end
looptime = toc;
index4 = 1;
tic
seq = [n:-1:1];
for i =1:n-1
space=factorial(n-i);
Subind = find(seq==B(i)); %determine which section
index4= index4+space*(Subind-1);
seq = A(index4,i+1:end); %contains order of next column
end
calctime = toc;
disp([A(index4,:);B])
disp(['loop time: ' num2str(looptime)])
disp(['calc time: ' num2str(calctime)])
disp(['loop - calc: ' num2str(looptime-calctime)])

Oleg Komarov on 21 Aug 2014
You can try to use bsxfun():
find(all(bsxfun(@eq, A,B),2))
What happens is every row of A is tested with @eq for element-wise equality against B, producing a m x n logical matrix. Then, all() tests each row if all elements are true, and find() returns the position.