how can I rearrange columns of a matrix according to another index matrix

I have a 2 by 3 matrix

A= [32, 9, 43 ; 6, 4, 7];

and a 2 by 3 index matrix:

Ind=[1,3,2 ; 2,3,1]; 

I would like to arrange each row of A according to the index of each row of Ind. That is, I want to get a matrix B, whose first row is arranged according to the first row of Ind:

[32, 43, 9]

and the second row is arranged according to the second row of Ind:

[4, 7, 6]

So

B= [32, 43, 9 ; 4, 7, 6]

What should i do ? many thanks ! (sorry, i forgot to say that i'd like to avoid using loops)

Answer 1

[Kevin Phung]

A= [32, 9, 43 ; 6, 4, 7];
Ind=[1,3,2 ; 2,3,1]; 
B = nan(size(A));
for i = 1:size(A,1)
    B(i,:) = A(i,Ind(i,:))    
end

[Yuan Liao]

Thank you Kevin ! is it possible to avoid loops ?

[Image Analyst]

I don't think so unless you have a 1-D vector. Maybe you could do it with ind2subs() and subs2ind() but it would be way more complicated than a simple for loop. For an array this microscopic you should just use for loops. Unless you're having more than 10 or 20 million iterations, the for loop will take just a fraction of a second. How many columns and rows do you expect your arrays to be? If you have hundreds of millions or billions of elements, we may need to see if some non-for loop way is faster.

[Yuan Liao]

There are about 1000 rows in A. But i got your point: looping for 1000 rows is not that bad. thanks !