row ranking among multiple matrices

2 Ansichten (letzte 30 Tage)
mingcheng nie am 17 Aug. 2023
Kommentiert: mingcheng nie am 18 Aug. 2023
Hi there,
I have three matrix A, B, and C. Here A is a complex matrix of size N*M, where each row of A is generated from the same distribution. Each row of A are also related to corresponding rows of B and C, i.e., the first row will related to the first rows of B and C, so if the row position of A changes, then I want B and C's corresponding rows will be changed as well . Now, calculate the power of each row in A, i.e., the first row power will be: abs(A(1,1))^2+abs(A(1,2))^2+...+abs(A(1,M))^2, then ranking the rows of A based on row's power in descending order, i.e., the highest power row to be the first row, and the lowest poewr row. Meanwhile, I want the rows in B and C can align the ranking changes in A. Is there any efficient way to do this?
0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Bruno Luong am 17 Aug. 2023
Bearbeitet: Bruno Luong am 17 Aug. 2023
Just very standard matlab programming
% Generate test data
N=10; M=3;
A=rand(N,M)+1i*rand(N,M);
MB = 1;
MC = 2;
B=randi(10,N,MB)
B = 10×1
7 4 3 5 3 4 6 3 6 6
C=randi(10,N,MC)
C = 10×2
6 3 3 2 8 8 3 6 1 4 1 9 7 3 3 1 10 10 5 2
p = 2; % power
[Anorm,is] = sort(vecnorm(A,p,2),'descend')
Anorm = 10×1
1.8755 1.7151 1.6110 1.5849 1.4104 1.3776 1.2816 1.2354 1.1222 1.0291
is = 10×1
5 1 8 4 7 6 10 9 3 2
A = A(is,:)
A =
0.7121 + 0.3695i 0.4868 + 0.9979i 0.9602 + 0.8479i 0.9269 + 0.4427i 0.5795 + 0.9249i 0.2050 + 0.8081i 0.4052 + 0.9197i 0.9609 + 0.1776i 0.3922 + 0.6903i 0.3646 + 0.2818i 0.1953 + 0.7376i 0.9552 + 0.8972i 0.0714 + 0.7200i 0.3226 + 0.4563i 0.8747 + 0.6232i 0.1924 + 0.3216i 0.5120 + 0.6637i 0.8073 + 0.6346i 0.6943 + 0.0221i 0.5488 + 0.4397i 0.8141 + 0.0525i 0.3099 + 0.1480i 0.4871 + 0.3865i 0.8718 + 0.5114i 0.7629 + 0.4168i 0.2592 + 0.0964i 0.4537 + 0.4703i 0.3264 + 0.1703i 0.5729 + 0.3254i 0.6828 + 0.1528i
B = B(is,:)
B = 10×1
3 7 3 5 6 4 6 6 3 4
C = C(is,:)
C = 10×2
1 4 6 3 3 1 3 6 7 3 1 9 5 2 10 10 8 8 3 2
1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
mingcheng nie am 18 Aug. 2023
Thank you so much for your explaination!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Steven Lord am 17 Aug. 2023
Call sort with two outputs. Use the second output to reorder the other arrays. See the "Sort Vectors in Same Order" example on the sort documentation page; you would need to generalize it to index into matrices rather than vectors, but that's not difficult.
0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden

Melden Sie sich an, um zu kommentieren.

Jon am 17 Aug. 2023
Bearbeitet: Jon am 17 Aug. 2023
Similar to @Bruno Luong, but since I already coded up example before I saw @Bruno Luong's I will provide it as alternative here
% Make some example data
m = 5;
n = 3;
A = randn(m,n,"like",1+1i)
A =
0.0358 - 1.7015i 0.4671 - 0.2650i 0.6336 - 0.5262i 0.8548 + 0.0416i 0.2163 - 0.9266i 0.6586 - 1.1739i -1.6230 + 0.9340i 0.6844 - 1.0415i -1.2676 - 0.0650i 0.7128 - 0.1322i 1.0431 + 0.6256i -0.1131 + 1.5801i 0.7582 + 0.1192i -0.6561 + 0.3697i -0.5786 + 0.2997i
B = repmat((1:m)',1,m)
B = 5×5
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5
C = repmat(10*(1:m)',1,m)
C = 5×5
10 10 10 10 10 20 20 20 20 20 30 30 30 30 30 40 40 40 40 40 50 50 50 50 50
% Calculate power in each row of A
p = sum(abs(A).^2,2)
p = 5×1
3.8630 3.4495 6.6708 4.5145 1.5809
% Find sort index in descending order
[~,idx] = sort(p,1,'descend');
% Sort the arrays
Asort = A(idx,:)
Asort =
-1.6230 + 0.9340i 0.6844 - 1.0415i -1.2676 - 0.0650i 0.7128 - 0.1322i 1.0431 + 0.6256i -0.1131 + 1.5801i 0.0358 - 1.7015i 0.4671 - 0.2650i 0.6336 - 0.5262i 0.8548 + 0.0416i 0.2163 - 0.9266i 0.6586 - 1.1739i 0.7582 + 0.1192i -0.6561 + 0.3697i -0.5786 + 0.2997i
Bsort = B(idx,:)
Bsort = 5×5
3 3 3 3 3 4 4 4 4 4 1 1 1 1 1 2 2 2 2 2 5 5 5 5 5
Csort = C(idx,:)
Csort = 5×5
30 30 30 30 30 40 40 40 40 40 10 10 10 10 10 20 20 20 20 20 50 50 50 50 50
2 KommentareKeine anzeigenKeine ausblenden
Jon am 17 Aug. 2023
Just made a quick edit on the repmat dimensions so that the example matrices would be a little more obvious
mingcheng nie am 18 Aug. 2023

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Logical finden Sie in Help Center und File Exchange

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by