Sort the array by template

I have two arrays A1 and B1 in which elements correspond:

A1 = [a1 a2 a3 a4 a5];
B1 = [b1 b2 b3 b4 b5];

then I change order of elements of the first array anyhow (they are all unique), for exaple like this:

A2 = [a4 a1 a3 a5 a2];

The question is how to get array B2 sorted in the same way? It should be

B2 = [b4 b1 b3 b5 b2];

I tried to write function which is:

B2 = zeros(length(B1),1);
for i = 1:length(B2)
   ind = find( A1 == A2(i) );
   B2(i) = B1(ind);   
end

but as it uses for loops speed is not fast. Maybe there is a way to do it by using MatLab builtin functions?

Answer 1

[Ameer Hamza]

For a generalized matrix a1 and a2, you can use the following code

[~,index] = ismember(a2', a1', 'row');
b2 = b1(index)

This will work on R2016b and later.

[Rostislav Teryaev]

It is absolutelly correct! Thank you! However it is not faster than code with for loops for 4e4 elements :)

b2 = zeros( length(b1), 1 );
for i = 1:length(b2)
    ind = find( all(a1 == a2(i,:), 2) );
    b2(i) = b1(ind);   
end

[Ameer Hamza]

You didn't mention the size of your matrices. The slowdown is happening because of huge logical matrices creation. For your case, try the updated answer.

[Rostislav Teryaev]

Your code is even slower. But this:

index = sparse(a1(1,:) == a2(1,:)' & a1(2,:) == a2(2,:)')*(1:length(b1))'
b2 = b1(index);

make it faster. Not really significantly but faster for sure. Thank you again

[Ameer Hamza]

Have you tried the updated answer?

[Rostislav Teryaev]

WOOOOW! Such elegant) Yes this solution is the best one. It is instant for 4e4 elements.

[Ameer Hamza]

You are welcome.

[Rostislav Teryaev]

How did you train yourself to invent such elegant solutions?

[Ameer Hamza]

You will get hold of such solution once you start to develop some intuition about these frequently used MATLAB functions. Of course, that comes with time, experience and reading a lot of documentation :)

Answer 2

[sloppydisk]

a1 = [1 4 2 5 7];
b1 = 1:5;

mySecretOrder = [1 3 2 5 4];
a2 = a1(mySecretOrder);
[~, order] = ismember(a2, a1)
b1 = b1(order)

[Rostislav Teryaev]

nice! But what if the a1 and a2 are matrices? Real example:

a1 = [5 2 3 4 2 3 6 6 6
      1 1 1 3 4 5 4 5 2]
b1 = [19 28 19 38 24 20 27 11 16]

a2 = [2 2 3 3 4 5 6 6 6
      1 4 1 5 3 1 2 4 5]
b2 = [28 24 19 20 38 19 16 27 11]

There are columns (which stand for directions from 5 to 1, from 2 to 1 and so on) then they change their order in matrix and become a2. So I need to sort b2 which is also in the order of a1 columns and turn it into b2 which is in a2 order.

Sorry for asking not very this question in the first time. I supposed I would manage to extrapolate solution of easy case for more complicated case like this.

[sloppydisk]

Please see Ameer's answer.