order the rows of the 3 matrices starting from the third coordinate of the point P

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Hi! I have to order the rows of the 3 matrices starting from the third coordinate of the point P.
A = [28.0445 -17.2717 83.972 1 2 3
27.8199 -16.986 83.3748 4 5 6
27.2756 -16.339 81.871 7 8 9
25.3805 -14.0329 67.019 10 11 12];
B = [25.3805 -14.0329 67.019 13 14 15
25.3425 -14.0698 66.0921 16 17 18
25.2731 -14.1203 65.1073 19 20 21
25.1719 -14.1617 64.1436 22 23 24];
C = [25.3805 -14.0329 67.019 25 26 27
26.2863 -14.4669 67.2675 28 29 30
27.0031 -14.7142 67.6157 31 32 33
27.7458 -14.9233 68.1745 34 35 36];
P = [25.3805 -14.0329 67.019];
So the final result would be:
A_new = [25.3805 -14.0329 67.019 10 11 12
27.2756 -16.339 81.871 7 8 9
27.8199 -16.986 83.3748 4 5 6
28.0445 -17.2717 83.972 1 2 3];
B_new = [25.3805 -14.0329 67.019 13 14 15
25.3425 -14.0698 66.0921 16 17 18
25.2731 -14.1203 65.1073 19 20 21
25.1719 -14.1617 64.1436 22 23 24]; % same as B
C_new = [25.3805 -14.0329 67.019 25 26 27
26.2863 -14.4669 67.2675 28 29 30
27.0031 -14.7142 67.6157 31 32 33
27.7458 -14.9233 68.1745 34 35 36]; % same as C
P = [25.3805 -14.0329 67.019];
  2 件のコメント
Fabio Freschi
Fabio Freschi 2023 年 10 月 5 日
Are you sure that P(3) is always the lowest value in A(:,3)?
Alberto Acri
Alberto Acri 2023 年 10 月 5 日
No, P(3) is not necessarily the lowest

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採用された回答

Dyuman Joshi
Dyuman Joshi 2023 年 10 月 5 日
編集済み: Dyuman Joshi 2023 年 10 月 5 日
Use sort
A = [28.0445 -17.2717 83.972 1 2 3
27.8199 -16.986 83.3748 4 5 6
27.2756 -16.339 81.871 7 8 9
25.3805 -14.0329 67.019 10 11 12];
B = [25.3805 -14.0329 67.019 13 14 15
25.3425 -14.0698 66.0921 16 17 18
25.2731 -14.1203 65.1073 19 20 21
25.1719 -14.1617 64.1436 22 23 24];
C = [25.3805 -14.0329 67.019 25 26 27
26.2863 -14.4669 67.2675 28 29 30
27.0031 -14.7142 67.6157 31 32 33
27.7458 -14.9233 68.1745 34 35 36];
P = [25.3805 -14.0329 67.019];
%Sort according to distance from the 3rd coordinate of P
[~,idx1]=sort(abs(A(:,3)-P(3)));
[~,idx2]=sort(abs(B(:,3)-P(3)));
[~,idx3]=sort(abs(C(:,3)-P(3)));
A_new = A(idx1,:)
A_new = 4×6
25.3805 -14.0329 67.0190 10.0000 11.0000 12.0000 27.2756 -16.3390 81.8710 7.0000 8.0000 9.0000 27.8199 -16.9860 83.3748 4.0000 5.0000 6.0000 28.0445 -17.2717 83.9720 1.0000 2.0000 3.0000
B_new = B(idx2,:)
B_new = 4×6
25.3805 -14.0329 67.0190 13.0000 14.0000 15.0000 25.3425 -14.0698 66.0921 16.0000 17.0000 18.0000 25.2731 -14.1203 65.1073 19.0000 20.0000 21.0000 25.1719 -14.1617 64.1436 22.0000 23.0000 24.0000
C_new = C(idx3,:)
C_new = 4×6
25.3805 -14.0329 67.0190 25.0000 26.0000 27.0000 26.2863 -14.4669 67.2675 28.0000 29.0000 30.0000 27.0031 -14.7142 67.6157 31.0000 32.0000 33.0000 27.7458 -14.9233 68.1745 34.0000 35.0000 36.0000

その他の回答 (1 件)

Fabio Freschi
Fabio Freschi 2023 年 10 月 5 日
If you want to sort A according to the third column you can simply do the following
% your data
A = [28.0445 -17.2717 83.972 1 2 3
27.8199 -16.986 83.3748 4 5 6
27.2756 -16.339 81.871 7 8 9
25.3805 -14.0329 67.019 10 11 12];
B = [25.3805 -14.0329 67.019 13 14 15
25.3425 -14.0698 66.0921 16 17 18
25.2731 -14.1203 65.1073 19 20 21
25.1719 -14.1617 64.1436 22 23 24];
C = [25.3805 -14.0329 67.019 25 26 27
26.2863 -14.4669 67.2675 28 29 30
27.0031 -14.7142 67.6157 31 32 33
27.7458 -14.9233 68.1745 34 35 36];
% engine
[~,idx] = sort(A(:,3))
A_new = A(idx,:);
similarly for B and C

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