using find(ismember) instead of a loop?

I have matrix B which is has around 1 million rows and Matrix A which has 800,000 rows. I want to match the observations of A and B and then assign to A the observations of B that matches. I tried a loop which taking forever in terms of time. I am wondering if I can do it in one shot using find(ismember). below are my codes:

for i=1:size(B,1)
x0=find(A(:,2)==B(i,2) & A(:,1)==B(i,1));
B(i,8)=A(x0,9);
end

[Matt J]

I suspect you haven't shown your actual code, making it hard to interpret what you are trying to do. In the posted code, you are comparing columns of A with rows of B which should be producing errors like the following, unless you are using R2016b

      >> [1 2 3] == [1;2;3]
    Error using  == 
    Matrix dimensions must agree.

Also, the code you've shown uses only the 1st, 2nd, and 9th column of A. So, why is it significant that "A has 800000 rows"?

[wesso Dadoyan]

I am not receiving any error. the size is significant to show that the loop takes a lot of time . i need a one step approach rather than a loop.

Answer 1

[Guillaume]

Note that the find was completely unnecessary in your original example, you could just have used the logical array returned by the comparison:

x0=A(:,2)==B(i,2) & A(:,1)==B(i,1);
B(i,8)=A(x0,9);  %would have worked exactly the same with marginal speed gain.

To answer your question:

[found, Arow] = ismember(B(:, 1:2), A(:, 1:2), 'rows');
B(found, 8) = A(Arow(found), 9);