Effacer les filtres
Effacer les filtres

ismember for string matrix

63 vues (au cours des 30 derniers jours)
Sam
Sam le 26 Nov 2014
Commenté : Sean de Wolski le 26 Nov 2014
I have two string matrices;
A=['c1' 'c ' 'b ' 'd9']'; %UNIQUELIST
B=['d9' 'c1']'; %ORIGINALLIST
I would like do find member of B in A, using:
[LIA,LOCB]=ismember(A,B);
and it returns
LOCB =
3
4
3
0
0
0
1
2
But I actually would like it to return matching row index like this:
LOCB =
2
0
0
1
Thanks for your help
Sam

Réponses (2)

Sean de Wolski
Sean de Wolski le 26 Nov 2014
You need to make A and B cell arrays so that each string piece is a separate element (rather than a 1xn string. The fix is simple use {} instead of []
A={'c1' 'c ' 'b ' 'd9'}';
B={'d9' 'c1'}'
[LIA,LOCB]=ismember(A,B)
  1 commentaire
Sam
Sam le 26 Nov 2014
that hit the nail on the head, thanks!

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dpb
dpb le 26 Nov 2014
You've got string arrays here; use the 'rows' optional argument to treat them as such instead of as individual characters...
>> [~,loc]=ismember(A,B,'rows')
loc =
2
0
0
1
>>
  3 commentaires
dpb
dpb le 26 Nov 2014
Did you try it and see???
Sean de Wolski
Sean de Wolski le 26 Nov 2014
Probably, but converting to cells is heavier weight if you don't need them for other things.

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