remove repeated rows to produce two new matrices

1 vue (au cours des 30 derniers jours)
JL
JL le 12 Sep 2019
Modifié(e) : Stephen23 le 12 Sep 2019
Hi everyone, I have 2 matrices, A and B. As you can see, everyrow in A is for a particular row in B. What I want to to do is for a repetitive, I want to remove them and take the highest value in A. For example, 3 to 10 appears twice (3 4 5 7 8 10 & 3 4 5 6 9 8 10 with corresponding A values of 0.213157025 and 0.207067988 respectively). From this two, I would take 0.213157025 because its higher. I then produce C and D accordingly by producing C in terms of putting the first and last number in each row with their corresponding A values in D.
a = [0.359883241
0.277815278
0.247562838
0.26249033
0.226989582
0.213157025
0.207067988
0.204762201
];
b =[3 4 0 0 0 0 0
3 4 5 0 0 0 0
3 4 5 7 6 0 0
3 4 5 7 0 0 0
3 4 5 7 8 0 0
3 4 5 7 8 10 0
3 4 5 6 9 8 10
3 4 5 7 8 10 11
];
I would like to produce C and D
C =[3 4
3 5
3 6
3 7
3 8
3 10
3 11]
D = [0.359883241
0.277815278
0.247562838
0.26249033
0.226989582
0.213157025
0.204762201
];

Réponse acceptée

Adam Danz
Adam Danz le 12 Sep 2019
% Convert b to [first,last] non-zero per row
b2 = splitapply(@(x)x([find(x~=0,1,'first'),find(x~=0,1,'last')]),b,(1:size(b,1)).');
% Get unique rows of b2
[C, Cidx] = unique(b2,'rows');
% Get the corresponding 'a' values
D = a(Cidx);

Plus de réponses (1)

Stephen23
Stephen23 le 12 Sep 2019
Modifié(e) : Stephen23 le 12 Sep 2019
An old-fashioned way:
a = [0.359883241;0.277815278;0.247562838;0.26249033;0.226989582;0.213157025;0.207067988;0.204762201]
b = [3,4,0,0,0,0,0;3,4,5,0,0,0,0;3,4,5,7,6,0,0;3,4,5,7,0,0,0;3,4,5,7,8,0,0;3,4,5,7,8,10,0;3,4,5,6,9,8,10;3,4,5,7,8,10,11]
% Indices of last non-zero:
tmp = b.';
tmp(end+1,:) = 0;
idx = 1==diff(cumprod(tmp,1)==0);
idx(end+1,:) = false;
%
[T,X] = sortrows([b(:,1),tmp(idx),a]);
[C,Y] = unique(T(:,1:2),'rows','last');
D = a(X(Y));
Giving:
>> C
C =
3 4
3 5
3 6
3 7
3 8
3 10
3 11
>> D
D =
0.35988
0.27782
0.24756
0.26249
0.22699
0.21316
0.20476

Catégories

En savoir plus sur MATLAB 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