Function matchpairs does not work in MATLAB R2023a when I use a sparse cost matrix. Please help me out ASAP!
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Ulyana Grechaniuk
le 9 Avr 2024 à 3:28
Modifié(e) : Bruno Luong
le 11 Avr 2024 à 15:56
You can run this simple code and see my problem. Am I doing something wrong?
% testing matchpairs ----------------------------------------------------------------
clear;clc
% --- using matrix in usual form (not sparse) --- WORKS
M = 99*ones(16,16);
M(1,1) = 2;
M(2,1) = 1;
M(1,2) = 1;
M(3,2) = 2;
M(2,3) = 2;
M(4,3) = 1;
M(6,3) = 4;
M(3,4) = 1;
M(4,4) = 2;
M(7,4) = 4;
M(6,5) = 2;
M(8,5) = 1;
M(3,6) = 4;
M(5,6) = 2;
M(7,6) = 1;
M(4,7) = 4;
M(6,7) = 1;
M(10,7) = 2;
M(5,8) = 1;
M(9,8) = 2;
M(8,9) = 2;
M(12,9) = 1;
M(7,10) = 2;
M(11,10) = 1;
M(13,10) = 4;
M(10,11) = 1;
M(12,11) = 2;
M(14,11) = 4;
M(9,12) = 1;
M(11,12) = 2;
M(10,13) = 4;
M(13,13) = 2;
M(14,13) = 1;
M(11,14) = 4;
M(13,14) = 1;
M(15,14) = 2;
M(14,15) = 2;
M(16,15) = 1;
M(15,16) = 1;
M(16,16) = 2;
[matchings, unassignedRows, unassignedCols] = matchpairs(M,12345)
OUTPUT IS CORRECT:
% --- using sparse matrix --- DOES NOT WORK
Ms = sparse(16,16);
Ms(1,1) = 2;
Ms(2,1) = 1;
Ms(1,2) = 1;
Ms(3,2) = 2;
Ms(2,3) = 2;
Ms(4,3) = 1;
Ms(6,3) = 4;
Ms(3,4) = 1;
Ms(4,4) = 2;
Ms(7,4) = 4;
Ms(6,5) = 2;
Ms(8,5) = 1;
Ms(3,6) = 4;
Ms(5,6) = 2;
Ms(7,6) = 1;
Ms(4,7) = 4;
Ms(6,7) = 1;
Ms(10,7) = 2;
Ms(5,8) = 1;
Ms(9,8) = 2;
Ms(8,9) = 2;
Ms(12,9) = 1;
Ms(7,10) = 2;
Ms(11,10) = 1;
Ms(13,10) = 4;
Ms(10,11) = 1;
Ms(12,11) = 2;
Ms(14,11) = 4;
Ms(9,12) = 1;
Ms(11,12) = 2;
Ms(10,13) = 4;
Ms(13,13) = 2;
Ms(14,13) = 1;
Ms(11,14) = 4;
Ms(13,14) = 1;
Ms(15,14) = 2;
Ms(14,15) = 2;
Ms(16,15) = 1;
Ms(15,16) = 1;
Ms(16,16) = 2;
[matchings, unassignedRows, unassignedCols] = matchpairs(Ms,-1)
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Ulyana Grechaniuk
le 10 Avr 2024 à 20:57
1 commentaire
Bruno Luong
le 11 Avr 2024 à 15:54
Modifié(e) : Bruno Luong
le 11 Avr 2024 à 15:56
@Ulyana Grechaniuk you are welcome.
You should move your reply to comment using the cross arrows "Move" on the top right. Thanks.
Plus de réponses (1)
Bruno Luong
le 9 Avr 2024 à 7:04
Modifié(e) : Bruno Luong
le 9 Avr 2024 à 8:46
Your sparse matrix filled default cost wit 0 not 99. Then your expectation on that the result is identocal for different cost matrices (regardless data structure using) is simply plain wrong.
If you fill sparse or dense matrix with the same values it give identical outputs:
% testing matchpairs ----------------------------------------------------------------
clear;clc
% --- using matrix in usual form (not sparse) --- WORKS
M = 99*ones(16,16);
M(1,1) = 2;
M(2,1) = 1;
M(1,2) = 1;
M(3,2) = 2;
M(2,3) = 2;
M(4,3) = 1;
M(6,3) = 4;
M(3,4) = 1;
M(4,4) = 2;
M(7,4) = 4;
M(6,5) = 2;
M(8,5) = 1;
M(3,6) = 4;
M(5,6) = 2;
M(7,6) = 1;
M(4,7) = 4;
M(6,7) = 1;
M(10,7) = 2;
M(5,8) = 1;
M(9,8) = 2;
M(8,9) = 2;
M(12,9) = 1;
M(7,10) = 2;
M(11,10) = 1;
M(13,10) = 4;
M(10,11) = 1;
M(12,11) = 2;
M(14,11) = 4;
M(9,12) = 1;
M(11,12) = 2;
M(10,13) = 4;
M(13,13) = 2;
M(14,13) = 1;
M(11,14) = 4;
M(13,14) = 1;
M(15,14) = 2;
M(14,15) = 2;
M(16,15) = 1;
M(15,16) = 1;
M(16,16) = 2;
[matchings, unassignedRows, unassignedCols] = matchpairs(M,12345)
matchings
Ms = sparse(M);
[matchings, unassignedRows, unassignedCols] = matchpairs(M,12345);
matchings
Cross check, now both filled with 0s
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% --- using sparse matrix --- DOES NOT WORK
Ms = sparse(16,16);
Ms(1,1) = 2;
Ms(2,1) = 1;
Ms(1,2) = 1;
Ms(3,2) = 2;
Ms(2,3) = 2;
Ms(4,3) = 1;
Ms(6,3) = 4;
Ms(3,4) = 1;
Ms(4,4) = 2;
Ms(7,4) = 4;
Ms(6,5) = 2;
Ms(8,5) = 1;
Ms(3,6) = 4;
Ms(5,6) = 2;
Ms(7,6) = 1;
Ms(4,7) = 4;
Ms(6,7) = 1;
Ms(10,7) = 2;
Ms(5,8) = 1;
Ms(9,8) = 2;
Ms(8,9) = 2;
Ms(12,9) = 1;
Ms(7,10) = 2;
Ms(11,10) = 1;
Ms(13,10) = 4;
Ms(10,11) = 1;
Ms(12,11) = 2;
Ms(14,11) = 4;
Ms(9,12) = 1;
Ms(11,12) = 2;
Ms(10,13) = 4;
Ms(13,13) = 2;
Ms(14,13) = 1;
Ms(11,14) = 4;
Ms(13,14) = 1;
Ms(15,14) = 2;
Ms(14,15) = 2;
Ms(16,15) = 1;
Ms(15,16) = 1;
Ms(16,16) = 2;
[matchings, unassignedRows, unassignedCols] = matchpairs(Ms,-1);
matchings
Mf = full(Ms);
[matchings, unassignedRows, unassignedCols] = matchpairs(Mf,-1);
matchings
1 commentaire
Bruno Luong
le 9 Avr 2024 à 7:17
One way to use sparse and keep filling 0 is to shift the value of the cost matrix
% testing matchpairs ----------------------------------------------------------------
Ms = sparse(16,16);
Ms(1,1) = 2;
Ms(2,1) = 1;
Ms(1,2) = 1;
Ms(3,2) = 2;
Ms(2,3) = 2;
Ms(4,3) = 1;
Ms(6,3) = 4;
Ms(3,4) = 1;
Ms(4,4) = 2;
Ms(7,4) = 4;
Ms(6,5) = 2;
Ms(8,5) = 1;
Ms(3,6) = 4;
Ms(5,6) = 2;
Ms(7,6) = 1;
Ms(4,7) = 4;
Ms(6,7) = 1;
Ms(10,7) = 2;
Ms(5,8) = 1;
Ms(9,8) = 2;
Ms(8,9) = 2;
Ms(12,9) = 1;
Ms(7,10) = 2;
Ms(11,10) = 1;
Ms(13,10) = 4;
Ms(10,11) = 1;
Ms(12,11) = 2;
Ms(14,11) = 4;
Ms(9,12) = 1;
Ms(11,12) = 2;
Ms(10,13) = 4;
Ms(13,13) = 2;
Ms(14,13) = 1;
Ms(11,14) = 4;
Ms(13,14) = 1;
Ms(15,14) = 2;
Ms(14,15) = 2;
Ms(16,15) = 1;
Ms(15,16) = 1;
Ms(16,16) = 2;
% Shift matrix
[I,J,K] = find(Ms);
fillvalue = 99;
Msshift = sparse(I,J,K-fillvalue,16,16);
[matchings, unassignedRows, unassignedCols] = matchpairs(Msshift,12345);
matchings
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