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Combining data with matching elements in the first two columns

1 vue (au cours des 30 derniers jours)
Muneer
Muneer le 18 Fév 2014
Commenté : Azzi Abdelmalek le 19 Fév 2014
Hello, I have a data table that looks like this
920 381 784 0
920 381 0 21.4375
23 388 1703 0
23 388 0 4.109375
445 487 304 0
445 487 0 15.09375
1100 506 1480 0
1100 506 0 28.234375
245 520 454 0
245 520 0 40.21875
For all the entries where the first two columns match (ie, the first two rows), I would like to combine those two rows into one. So these two rows:
920 381 784 0
920 381 0 21.4375
Will become
920 381 784 21.4375
And so on for the rest of the data set. I would appreciate any help.
Thanks
  2 commentaires
the cyclist
the cyclist le 18 Fév 2014
Is that exact pattern guaranteed? Namely, will the first zero always be in column 4, and the second zero always be in column 3? And will there always be a pair of rows like that?
Muneer
Muneer le 18 Fév 2014
No it isn't guaranteed that there will always be a pair of rows. If there is one row with no matching row, however, it will have a value in column three, not column four. Thanks

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Réponse acceptée

the cyclist
the cyclist le 18 Fév 2014
Modifié(e) : the cyclist le 19 Fév 2014
Here's one way:
[~,i,j]=unique(M(:,1:2),'rows');
[M(i,[1 2]),accumarray(j,M(:,3)),accumarray(j,M(:,4))]
where M is your original array.
EDIT: Original solution only gave the last two columns, so I fixed it to give all the columns you need.
  3 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 18 Fév 2014
This does not give the result
Azzi Abdelmalek
Azzi Abdelmalek le 19 Fév 2014
The cyclist, your code doesn't work if there are more then two duplicate rows (I mean duplicate for the two first column).

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Plus de réponses (1)

Azzi Abdelmalek
Azzi Abdelmalek le 18 Fév 2014
Modifié(e) : Azzi Abdelmalek le 18 Fév 2014
M=[920 381 784 0
920 381 0 21.4375
23 388 1703 0
23 388 0 4.109375
445 487 304 0
445 487 0 15.09375
1100 506 1480 0
1100 506 0 28.234375
245 520 454 0
245 520 0 40.21875]
[~,idx,jj]=unique(M(:,1:2),'rows','stable'); % M is your matrix
n=numel(idx)
out=zeros(n,size(M,2))
for k=1:n
out(k,:)=max(M(find(jj==k),:))
end
  4 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 18 Fév 2014
Look at edited answer
Muneer
Muneer le 18 Fév 2014
Thanks for the help!

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