ismember for string matrix

30 Ansichten (letzte 30 Tage)
Sam
Sam am 26 Nov. 2014
Kommentiert: Sean de Wolski am 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

Antworten (2)

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

Melden Sie sich an, um zu kommentieren.


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

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Data Type Conversion finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by