ismember for string matrix

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Sam
Sam el 26 de Nov. de 2014
Comentada: Sean de Wolski el 26 de Nov. de 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

Respuestas (2)

Sean de Wolski
Sean de Wolski el 26 de Nov. de 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 comentario
Sam
Sam el 26 de Nov. de 2014
that hit the nail on the head, thanks!

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dpb
dpb el 26 de Nov. de 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 comentarios
dpb
dpb el 26 de Nov. de 2014
Did you try it and see???
Sean de Wolski
Sean de Wolski el 26 de Nov. de 2014
Probably, but converting to cells is heavier weight if you don't need them for other things.

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