reorganize two related matrices in ascending order

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Tomás Nunes
Tomás Nunes on 16 Apr 2018
Commented: dpb on 19 Apr 2018
I have two matrices, one with -1 and 1's and another with returns that are related to the number in the previous matrix. However, the -1 and 1's aren't organized, in the sense that their position varies in each row. What I pretend is a matrix with the -1's in the first couple of rows and the 1's in the last ones, given that the returns in the second matrix change positions as well. For example:
-1 1 -1 1
-1 -1 1 1
1 1 -1 -1
1 -1 1 -1
0.1 -0.3 2 0.5
-0.9 0 1 1.5
1.5 1.2 0 -0.3
0.9 0.1 1.2 0.3
and I wish to have:
-1 -1 1 1
-1 -1 1 1
-1 -1 1 1
-1 -1 1 1
0.1 2 -0.3 0.5
-0.9 0 1 1.5
0 -0.3 1.5 1.2
0.1 0.3 0.9 1.2
is this possible? I hope I was clear. Thank you

Accepted Answer

Stephen
Stephen on 19 Apr 2018
Edited: Stephen on 19 Apr 2018

This is easy with sub2ind:

>> A0 = [-1,1,-1,1;-1,-1,1,1;1,1,-1,-1;1,-1,1,-1]
A0 =
  -1   1  -1   1
  -1  -1   1   1
   1   1  -1  -1
   1  -1   1  -1
>> B0 = [0.1,-0.3,2,0.5;-0.9,0,1,1.5;1.5,1.2,0,-0.3;0.9,0.1,1.2,0.3]
B0 =
   0.10000  -0.30000   2.00000   0.50000
  -0.90000   0.00000   1.00000   1.50000
   1.50000   1.20000   0.00000  -0.30000
   0.90000   0.10000   1.20000   0.30000
>> [A1,idc] = sort(A0,2); % sort, get column indices
>> A1 =
  -1  -1   1   1
  -1  -1   1   1
  -1  -1   1   1
  -1  -1   1   1
>> Sz = size(A0);
>> idr = repmat(1:Sz(1),Sz(2),1).'; % get row indices
>> B1 = B0(sub2ind(Sz,idr,idc)) % convert column&row indices to linear.
B1 =
   0.10000   2.00000  -0.30000   0.50000
  -0.90000   0.00000   1.00000   1.50000
   0.00000  -0.30000   1.50000   1.20000
   0.10000   0.30000   0.90000   1.20000
  1 Comment
dpb
dpb on 19 Apr 2018
Good on ya' Stephen, for repmat; I kept trying to be excessively klever in applying sub2ind w/ just the single vector...

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More Answers (1)

dpb
dpb on 16 Apr 2018
[A,ix]=sort(A,2);
for i=1:size(A,1), B(i,:)=B(i,ix(i,:)); end

Bound to be a way to write the indexing but it's not coming to me at the moment...

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