Sorting a Matrix using Indices from another matrix !
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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 !
2 个评论
the cyclist
2013-9-13
This would be much easier to answer if you gave examples of the two matrices, and the expected result.
Dimitris M
2013-9-13
编辑：Image Analyst
2013-9-13
采纳的回答
Azzi Abdelmalek
2013-9-13
[m,n]=size(A);
A=A(sub2ind([m n],idx,repmat(1:n,m,1)))
5 个评论
Image Analyst
2013-9-13
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
2013-9-13
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
2013-9-13
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
2013-9-13
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
2018-7-31
Very impressive; I've wondered about this for a long time. Thank you!
更多回答（2 个）
Image Analyst
2013-9-13
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
Azzi Abdelmalek
2013-9-13
编辑：Azzi Abdelmalek
2013-9-13
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)
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