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Pseudo Element by Element Vector Subtraction

조회 수: 2(최근 30일)
I am trying to subtract vectors stored in two different arrays. I want to subtract every row vector in matrix B from every row vector in matrix A such that the results of row vector 1 of A minus each of the row vectors of B is stored in a an array, the results of row vector 2 of A minus all row vectors B is stored in a different array...and so on until the end. Matrix A is a 63x2 array while matrix B is a 61x2 array. My final goal is to comapre all the results and identify and store all vectors that have duplicates.
A = [0 0
-5.22988566179964 6.05378360734033
5.82743694158990 -5.48096512411759
-7.71867012579003 13.6568061570767
-6.41307857048505 -4.78251223195356
-12.8832160382871 3.72429135615449
13.0437173772689 -8.93426740941455
-3.95560029264112 20.7165020025679
-14.9311651399118 17.1180076350708
-12.8261571409701 -9.56502446390712
-20.5365464147745 1.39479910496865
20.2599978129478 -12.3875696947115
-1.17023252546704 28.2159504090567
-22.1436601540336 20.5792091130649
-17.3827161983918 -16.1405677537219
-28.1898767912619 -0.934693146217198
23.2616199344730 -19.8031117948934
4.31055626391175 34.0435531963286
27.4257074918059 -8.83051968124259
-29.3561551681553 24.0404105910590
-19.3637625207825 -23.8914035453345
-35.8432071677493 -3.26418539740304
23.0577732626258 -27.8005142815998
9.79134505329053 39.8711559836006
10.4567903913941 28.9226347191244
34.5914171706640 -5.27346966777369
-31.5385456076644 31.7369794350011
-21.3448088431732 -31.6422393369470
-43.4965375442368 -5.59367764858889
25.6191596676504 -35.3793862084867
16.3972779876710 44.3835444937432
-35.1703582141579 38.8650883841765
-23.3258551655640 -39.3930751285596
-51.1498679207242 -7.92316989977473
28.1805460726749 -42.9582581353737
23.0032109220515 48.8959330038858
37.7741257337733 2.06617379783899
-29.3310994911081 -31.1740922588744
17.7048992093770 -36.5474963910512
-38.8021708206513 45.9931973333520
-22.3878218988045 -47.3378906450113
35.8328768192608 -45.2910320517349
29.7346992697278 53.2187867632789
18.1810613102127 55.2792607505272
40.9568342968826 9.40581726345168
-42.4339834271448 53.1213062825274
-22.9362553876826 -55.3190698002663
43.4852075658468 -47.6238059680961
36.8803922974807 49.6216947220428
-46.6615356421781 1.75362029283338
-32.2411590073453 50.5706555784546
-30.3866778744434 -47.4731795247682
32.3995579608740 -52.5168414608354
13.3589116983739 61.6625884971687
48.2086638919891 6.02780490060624
-49.3011842938828 49.0174602503434
-17.7011930022771 -61.3683773491444
46.9026731045917 -40.3904852029502
44.0260853252335 46.0246026808068
-49.8265337401194 9.10091823425565
-38.3629147966204 -48.0893325587636
55.4604934870957 2.64979253776079
51.1717783529864 42.4275106395707];
B = [5.82743694158990 -5.48096512411759
-2.48878446399039 7.60302254973639
-6.41307857048505 -4.78251223195356
-7.65333037648742 -2.32949225118584
7.21628043567895 -3.45330228529695
3.76306983314891 7.05969584549112
-7.21249501412177 3.46120147799410
-6.41307857048505 -4.78251223195356
-7.65333037648742 -2.32949225118584
7.21628043567895 -3.45330228529695
2.78536776717408 7.49944840648882
-7.21249501412177 3.46120147799410
-4.55655905742170 -6.57554328981479
-7.65333037648742 -2.32949225118584
3.00162212152522 -7.41554210018192
5.48078878937879 5.82760278727195
7.16570967885811 3.55705001346891
-7.21249501412177 3.46120147799410
-1.98104632239072 -7.75083579161255
-7.65333037648742 -2.32949225118584
-0.203846671847199 -7.99740248670635
5.48078878937879 5.82760278727195
6.14623412748235 -5.12091847720419
7.16570967885811 3.55705001346891
-2.18239043950905 7.69656884394205
-1.98104632239072 -7.75083579161255
-7.65333037648742 -2.32949225118584
2.56138640502452 -7.57887192688698
6.60593293438051 4.51238851014260
-3.63181260649347 7.12810894917545
-1.98104632239072 -7.75083579161255
-7.65333037648742 -2.32949225118584
2.56138640502452 -7.57887192688698
6.60593293438051 4.51238851014260
3.18270856310927 7.33964346561268
-7.98629064793488 0.468147078072581
-7.91426045827336 -1.16811018256442
-3.63181260649347 7.12810894917545
0.938033266759421 -7.94481551645176
7.65233074658597 -2.33277391636118
6.73148834767623 4.32285375939311
-4.82214961183880 6.38332774664147
3.18270856310927 7.33964346561268
-3.63181260649347 7.12810894917545
-0.548433488878012 -7.98117915525501
7.65233074658597 -2.33277391636118
7.14569302775288 -3.59709204123606
-3.16499809794131 7.34729794142227
6.56101181330606 4.57745824510266
-7.99885597563890 -0.135288879756905
-3.43331885838680 -7.22580940910052
-4.82214961183880 6.38332774664147
7.25182959510656 -3.37801236284544
-6.86720086673803 -4.10384603218402
5.23506238540546 -6.04930754887804
3.41746553874489 7.23332076514592
7.14569302775288 -3.59709204123606
-3.16499809794131 7.34729794142227
-7.97623692217691 -0.616153033995415
7.25182959510656 -3.37801236284544
7.14569302775288 -3.59709204123606];
SizeA = size(A,1);
SizeB = size(B,1);
for i = 1:SizeA
for j = 1:SizeB
vectortest(i,:) = vein_node_array(i,:)-Btest(j,:);
end
end

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채택된 답변

Star Strider
Star Strider 26 Oct 2020
See if this does what you want:
C1 = A(:,1).' - B(:,1); % Col #1 In ‘A’ - Col #1 In ‘B’
C2 = A(:,2).' - B(:,2); % Col #2 In ‘A’ - Col #2 In ‘B’
[Dupr,Dupc] = find((C1 == 0) & (C2 == 0));
Dupt = array2table([Dupc,Dupr], 'VariableNames',{'RowA','RowB'})
producing:
Dupt =
3×2 table
RowA RowB
____ ____
3 1
5 3
5 8
This works by creating two matrices formed by subtracting column vectors in ‘B’ from row vectors in ‘A’ (created by the transposition) for each column of both matrices. The elements of ‘C1’ and ‘C2’ are both equal to 0 if the respective rows of the ‘A’ and ‘B’ are duplicates.
In ‘Dupt’, the respective rows of matrix ‘A’ and matrix ‘B’ are duplicates, as demonstrated by:
A_dup = A(Dupc,:)
B_dup = B(Dupr,:)
producing:
A_dup =
5.827436941589900 -5.480965124117590
-6.413078570485050 -4.782512231953560
-6.413078570485050 -4.782512231953560
B_dup =
5.827436941589900 -5.480965124117590
-6.413078570485050 -4.782512231953560
-6.413078570485050 -4.782512231953560
These are by design exact matches. It might be possible to use a tolerance to find near matches, however I will defer that for the time being.

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표시 이전 댓글 수: 6
James Peach
James Peach 27 Oct 2020
Plot them using this command. They are points whose centers align in the shape of a knock off constellation.
figure
axis square
plot(A(:, 1), A(:, 2), '*', 'color', 'k', 'markersize', 12);
Star Strider
Star Strider 27 Oct 2020
O.K.
I’m beginning to understand what you’re doing. I need to think about this a bit.
Star Strider
Star Strider 29 Oct 2020
My apologies for the delay. It took me a while to figure out an acceptable approach to this problem.
This seems to work, at least with the ‘A’ matrix to identify an approximate graph of the connections between the stars:
Mdl = ExhaustiveSearcher(A);
[Idx,D] = rangesearch(Mdl, A, 9);
figure
hold on
for k = 1:numel(Idx)
xv = [repmat(A(Idx{k}(1),1),numel(Idx{k}),1), A(Idx{k}(1,:),1)];
yv = [repmat(A(Idx{k}(1),2),numel(Idx{k}),1), A(Idx{k}(1,:),2)];
plot(xv.', yv.','-p')
end
hold off
grid
axis('equal')
producing:
It works by using a -nearest-neighbour ExhaustiveSearcher exhaustive search, then uses the rangesearch function to identify the elements within a given distance (I empirically chose 9), returning the indices (in ‘Idx’ and the distances (in ‘D’ that I returned simply for informaiton). I then plotted them in the loop. The ‘xv’ and ‘yv’ matrices use the ‘Idx’ cell array elements to index into ‘A’ to plot each segment. Those take the ‘x’ vectors of each ‘Idx’ element for the first and subsequent columns of the indexed elements of ‘A’ and do the same for the ‘y’ vectors, so that the first value is replicated in the first column and the second column are the referenced neighbours. (This is a bit difficult to describe. Looking at the ‘xv’ and ‘yv’ matrices will help in understanding them.) They are then transposed so that they plot individual lines rather than one continuous line.
I did not try it on ‘B’.
I am not certain what you are doing, however this should make it easier for you to identiffy the branch-points and the properties of the stars connected to them.

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