problem with the setdiff function

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alex
alex le 17 Mar 2014
Commenté : alex le 17 Mar 2014
Hello!
i have a problem here
A = 2:.1:15
B=[9.4000 10.4000]
C=setdiff(A,B)
the elements 9.4000 and 10.4000 are still in the C
however,if B=[9.3000 10.3000] then the setdiff works fine,and the elements are not in the C.
any idea why this happens?
  2 commentaires
per isakson
per isakson le 17 Mar 2014
Try
format hex
alex
alex le 17 Mar 2014
unfortunatelly the format hex is not working

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Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 17 Mar 2014
Modifié(e) : Azzi Abdelmalek le 17 Mar 2014
  3 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 17 Mar 2014
Modifié(e) : Azzi Abdelmalek le 17 Mar 2014
Use some tolerance
A = 2:.1:15
B=[9.4000 10.4000 1000]
tol=0.0001
s=abs(bsxfun(@minus,A',B))<tol
[ii,jj]=find(s==1);
A(ii)=[]
alex
alex le 17 Mar 2014
thank you very much!
it works excellent!

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Plus de réponses (1)

Jos (10584)
Jos (10584) le 17 Mar 2014
A = 2:.1:15
B = [9.4000 10.4000]
D = 2 ; % number of digits that should match (e.g. 1.231 = 1.229)
C = setdiff(round(A*10^D),round(B*10^D)) ./ (10^D)
  1 commentaire
alex
alex le 17 Mar 2014
thank you for your answer! very useful!

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