Sorting a Matrix using Indices from another matrix !

Hello

I have a pretty simple question !

I have a matrix A (lets assume has a size 5 x 5) and I have another matrix of same size (5 x 5) called IDX that contains the values for sorting A in a colum-wise manner.

So how can I easily sort each column of A using matrix IDX without using a for loop or extracting each column of IDX as an individual vector ?

Thank you in advance !

[the cyclist]

This would be much easier to answer if you gave examples of the two matrices, and the expected result.

[Dimitris M]

Ok sure

A = [1 1 1 0 0 0; 
     0 0 0 1 1 1; 
     0 1 0 1 0 1]'
IDX = [4 5 6 3 2 1; 
       1 2 3 6 5 4; 
       6 5 4 3 2 1]'

And I want to sort A using the indices of IDX in a columnwise manner!

Is it clear enough now?

Answer 1

[Azzi Abdelmalek]

[m,n]=size(A);
A=A(sub2ind([m n],idx,repmat(1:n,m,1)))

[Image Analyst]

Perhaps I misunderstood what you want, because Azzi's and my code give completely different results. I thought you wanted to take each column of IDX to be the sort order for the corresponding column of A. So that if IDX was

IDX =
     4     1     6
     5     2     5
     6     3     4
     3     6     3
     2     5     2
     1     4     1

as you gave it. So then the result should be

new A=
     A(4,1)     A(1, 2)     A(6, 3)
     A(5,1)     A(2, 2)     A(5, 3)
     A(6,1)     A(3, 2)     A(4, 3)
     A(3,1)     A(6, 2)     A(3, 3)
     A(2,1)     A(5, 2)     A(2, 3)
     A(1,1)     A(4, 2)     A(1, 3)

where you take the A from the row where IDX told you to, in a column-by-column basis. You rxample wasn't good because with so many 0's and 1's, it's hard to tell what went where. However if

A = [1:6;7:12;13:18]'
A =
     1     7    13
     2     8    14
     3     9    15
     4    10    16
     5    11    17
     6    12    18

Then my code gives

A =
     4     7    18
     5     8    17
     6     9    16
     3    12    15
     2    11    14
     1    10    13

while Azzi's gives:

A =
     3     7    13
     2     8    14
     1     9    15
     6    10    16
     5    11    17
     4    12    18

As you can see they're totally different. My columns of A are re-ordered according to the numbers in the columns of IDX. I can't really figure out what algorithm Azzi's follows. True, mine uses a for loop while his doesn't but you shouldn't be afraid of for loops for arrays this tiny - it's only a factor when you get up in to the tens of millions of iterations, not just 3 iterations. But whatever - if his gives you the order that you want, then fine, I must have misunderstood.

[Azzi Abdelmalek]

Image Analyst, The results are the same

IDX =[4 1 6;5 2 5;6 3 4;3 6 3;2 5 2;1 4 1]
A = [1:6;7:12;13:18]'
%----------Azzi's code------------------
[m,n]=size(A);
A1=A(sub2ind([m n],idx,repmat(1:n,m,1)))
%---------Analyst's code----------------
[rows, columns] = size(A)
% Sort columns of A according to the same column of IDX
for col = 1 : columns
  A(:,col) = A(IDX(:,col), col);
end
%-------------------------------------------------------
isequal(A1,A)

[Image Analyst]

Sorry - you're right. I ran your code after mine and forgot to set A back to the original A. I still don't follow what yours does, but it works though.

[Azzi Abdelmalek]

idx =[4 1 6;5 2 5;6 3 4;3 6 3;2 5 2;1 4 1];
A = [1:6;7:12;13:18]';
idx1=idx  % indicate line index
   4     1     6
   5     2     5
   6     3     4
   3     6     3
   2     5     2
   1     4     1
[m,n]=size(A);
idx2=repmat(1:n,m,1) %indicate column index
   1     2     3
   1     2     3
   1     2     3
   1     2     3
   1     2     3
   1     2     3
idx=sub2ind([m n],idx1,idx2) % to get linear index
   4     7    18
   5     8    17
   6     9    16
   3    12    15
   2    11    14
   1    10    13
A=A(idx)

[Aaron Thode]

Very impressive; I've wondered about this for a long time. Thank you!

Answer 2

[Image Analyst]

Try this:

% Create random sample data so it will be easy
% for us to see if the sorting worked.
A = randi(9, [6,3]) 
IDX = [4 5 6 3 2 1; 
       1 2 3 6 5 4; 
       6 5 4 3 2 1]'
[rows, columns] = size(A)
% Sort columns of A according to the same column of IDX
for col = 1 : columns
  A(:,col) = A(IDX(:,col), col);
end
% Print out to command window.
A

In the command window:

A =
     7     7     7
     3     8     7
     9     2     3
     1     5     7
     4     5     6
     4     6     2
IDX =
     4     1     6
     5     2     5
     6     3     4
     3     6     3
     2     5     2
     1     4     1
rows =
     6
columns =
     3
A =
     1     7     2
     4     8     6
     4     2     7
     9     6     3
     3     5     7
     7     5     7

Answer 3

[Azzi Abdelmalek]

Edit2

A = [1 1 1 0 0 0; 0 0 0 1 1 1; 0 1 0 1 0 1]'
idx= [4 5 6 3 2 1; 1 2 3 6 5 4; 6 5 4 3 2 1]'
[m,n]=size(A);
idx=bsxfun(@plus,idx,(0:m:(n-1)*m))
A=A(idx)

[Azzi Abdelmalek]

Look at Edit2