Sorting a matrix according to another one

15 vues (au cours des 30 derniers jours)
Mattia Salomone
Mattia Salomone le 31 Mai 2023
Commenté : Mattia Salomone le 1 Juin 2023
Dear all,
I have the following problem to solve.
clear
clc
% A matrix defined as follows
A=[11 0.001 3
11 0.001 4
12 0.003 5
9 0.002 6
8 0.000 7
10 0.004 8
8 0.000 9
9 0.002 10];
% B matrix is the reference one
B=[8 0.000
8 0.000
9 0.002
10 0.004
9 0.002
11 0.001
11 0.001
12 0.003];
I want to sort A such that the first two columns coincide with B (and sort the other A column in agreement with this, of course). I used this code, but the problem is that since in B I have repeted conditions the result is not the one I desire.
[~,Y]=ismember(A(:,1:2),B,'rows');
[~,Z]=sort(Y);
C=A(Z,:);
% C=8 0.000 7
% 8 0.000 9
% 9 0.002 6
% 9 0.002 10
% 10 0.004 8
% 11 0.001 3
% 11 0.001 4
% 12 0.003 5
But what I want is
% C=8 0.000 7
% 8 0.000 9
% 9 0.002 6
% 10 0.004 8
% 9 0.002 10
% 11 0.001 3
% 11 0.001 4
% 12 0.003 5
Do you have any suggestions that don't involve using nested loops and if conditions?
Thank you very much for your attention!
Mattia
  2 commentaires
Torsten
Torsten le 31 Mai 2023
What if B has some equal rows (like e.g. [8 0] or [11 0.001]) ?
Mattia Salomone
Mattia Salomone le 31 Mai 2023
Thanks for your answer. Well in this case the order in which they appear in C should be the one they have in A. In this case with [8 0], for example, I want
[8 0 7
8 0 9]
Because in A they appear with this order.

Connectez-vous pour commenter.

Réponse acceptée

Torsten
Torsten le 31 Mai 2023
Modifié(e) : Torsten le 31 Mai 2023
A = [11 0.001 3
11 0.001 4
12 0.003 5
9 0.002 6
8 0.000 7
10 0.004 8
8 0.000 9
9 0.002 10];
% B matrix is the reference one
B = [8 0.000
8 0.000
9 0.002
10 0.004
9 0.002
11 0.001
11 0.001
12 0.003];
[~,idx] = sortrows(A,[1 2]);
[~,jdx] = sortrows(B,[1 2]);
C = A(idx(jdx),:)
C = 8×3
8.0000 0 7.0000 8.0000 0 9.0000 9.0000 0.0020 6.0000 10.0000 0.0040 8.0000 9.0000 0.0020 10.0000 11.0000 0.0010 3.0000 11.0000 0.0010 4.0000 12.0000 0.0030 5.0000
  3 commentaires
Torsten
Torsten le 31 Mai 2023
Modifié(e) : Torsten le 31 Mai 2023
A=[11 0.001 3
11 0.001 4
12 0.003 5
9 0.002 6
8 0.000 7
10 0.004 8
8 0.000 9
9 0.002 10];
% B matrix is the reference one
B=[8 0.000
8 0.000
9 0.002
10 0.004
9 0.002
11 0.001
11 0.001
12 0.003];
Acopy = A;
C = zeros(size(A));
for i = 1:size(A,1)
idx = find(Acopy(:,1:2)==B(i,:),1);
C(i,:) = Acopy(idx,:);
Acopy(idx,:) = [];
end
C
C = 8×3
8.0000 0 7.0000 8.0000 0 9.0000 9.0000 0.0020 6.0000 10.0000 0.0040 8.0000 9.0000 0.0020 10.0000 11.0000 0.0010 3.0000 11.0000 0.0010 4.0000 12.0000 0.0030 5.0000
Mattia Salomone
Mattia Salomone le 1 Juin 2023
Thank you very much for your help!

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Creating and Concatenating Matrices dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by