Rapid Vector Matching/Search Problem

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]

Have you tried the bsxfun() solution?

Answer 1

[Joseph Cheng]

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.

[Kyle Johnston]

I am familiar with "ismember" and I was hoping for something faster then that. I have also tried implementing fitctree, assigning each row a "class" and then using the resulting tree to quickly find the location of the matching row. But that is also too slow.

if it helps, A is resulting from the function perms(1:n) in MATLAB, and B is a member of the set that A represents.

I've settled on the following for right now:

space = factorial(n)/n;
for i = (space*(n- B(1)) + 1) : space*(n- B(1) + 1)
   if (isequal(A(i,:),B))
    indexIntrest = i
    break;
   end
end

And by the way, the example with ismember will yield:

[found_row, indexInterest] = ismember(A,B,'rows')

making the find given in the answer above not necessary

[Joseph Cheng]

I did it that way just in case you had more than 1. When i do your ismember(A,B,'rows') it does not come back with the index of which row.. that is why i used find to come up with the index.

[Joseph Cheng]

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)])

Answer 2

[Oleg Komarov]

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.