Another Issue with Join

Hello everybody,
I'm trying to merge two datasets. One calculated, and one retrieved from data, given by:
A=dataset(Output(:,2),Output(:,1),'VarNames',{'x' 'v'});
B=dataset(datarest(:,x),datarest(:,v),'VarNames',{'x' 'v'});
|which results in something in the form of:
A = x v
1 5
2 7
4 8
B = x v
2 6
3 7
In this, I want to use the x as keys, and merge it to get a dataset, which states v for every x. For this i'm using
C=join(A,B,'key',{'x' ' v' },'Type','outer','MergeKeys',true);
And the result is
C = x v
1 5
2 6
2 7
3 7
4 8
Where I want it to be:
C = x v
1 5
2 6
3 7
4 8
Thus specifying that if there are values with the same key present in A and B, I want the value of B to be in the matrix of C. I hope you understand the issue, and can help me with the answer.
Thanks in advance!

 Réponse acceptée

Tom Lane
Tom Lane le 14 Juin 2012

1 vote

Your desired result has v=6 for x=2, but A and B provide different values for v in that case. One thing you could do is
join(A,B,'key','x','type','outer','mergekeys',true)
but this would leave you with one output v variable for each input, and you'd have to reconcile how you wanted to choose from one or the other.

2 commentaires

Roy Veldhuizen
Roy Veldhuizen le 14 Juin 2012
Thank you for your answer! I used this method, removing values that i did not wanted to return in the C matrix beforehand. This did the job, but from your answer I conclude it is not possible to use the join function with a ' preference' for a certain dataset?
Tom Lane
Tom Lane le 14 Juin 2012
Play with this, and see if it gets you anywhere:
[C,IA,IB] = join(A,B,'key','x','type','outer','mergekeys',true,'leftvars','v','rightvars',{})
C.v(IB>0) = B.v(IB(IB>0))
I take v from the A dataset, but then I overwrite values based on the B dataset. I'm not sure if this is exactly what you want, but maybe you can work with it.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

Produits

Community Treasure Hunt

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

Start Hunting!

Translated by