Effacer les filtres
Effacer les filtres

How made the code as easy?

2 vues (au cours des 30 derniers jours)
VIJAY
VIJAY le 8 Nov 2018
Modifié(e) : madhan ravi le 8 Nov 2018
P=[1 32 3 78 5 6 7 8 9 10] ;P is my original array which has the size is 1*10; I have Q=[2 4 7];I want delete in Q array value in P array; My expect output is P=[ 1 3 5 6 8 9 10] by using for loop only because of the size of Q array may be change ;P(:,[Q(1) Q(2) Q(3)])=[] this command use only the size of Q array is 1*3 .So I need for loop logic which is common of any array size.

Réponse acceptée

madhan ravi
madhan ravi le 8 Nov 2018
Modifié(e) : madhan ravi le 8 Nov 2018
P(Q)=[] %edited after Jan's comment
  13 commentaires
Jan
Jan le 8 Nov 2018
You can omit the square brackets and 1:end :
P(Q)=[]
madhan ravi
madhan ravi le 8 Nov 2018
+1 @Jan cool

Connectez-vous pour commenter.

Plus de réponses (1)

Stephan
Stephan le 8 Nov 2018
Modifié(e) : Stephan le 8 Nov 2018
If you want to delete the values at the indicies in P, that are specified in Q use the answer from Madhan above. If you need to delete the values in P specified in Q use:
P(ismember(P,Q))=[]
  1 commentaire
madhan ravi
madhan ravi le 8 Nov 2018
+1 another way of approach

Connectez-vous pour commenter.

Catégories

En savoir plus sur Creating and Concatenating Matrices 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