Find index of same values in array

Hi all,
i have to find the index of the same value in an array,see the following example
a=[1 2 3 1]
i want b=[1 4] as output..how can i do this? A solution using find is this
  • u=unique(a)
  • n=histc(a,u)
  • find(a==u(n>1))
but if in the a array there isn't 2 or more same value i get this error
??? Error using ==> eq Matrix dimensions must agree.
Thank you for your support :-)

 Respuesta aceptada

Andrei Bobrov
Andrei Bobrov el 28 de En. de 2014
Editada: Andrei Bobrov el 29 de En. de 2014
find(ismember(a,u(n>1)))
ADD
a=[nan nan 12 3 5 9 8 7 12 NaN NaN 0 0 2 1 4 4 4]
u=unique(a(~isnan(a)));
n=histc(a,u);
d = u(n > 1);
out =find(ismember(a,d(end)))

5 comentarios

roberto verrelli
roberto verrelli el 28 de En. de 2014
Editada: roberto verrelli el 28 de En. de 2014
Thank you very much Andrei,it works!! But now i have the problem issued by Walter.. if i have
  • a=[10 9 9 10]
i have to take only the indexes of the higher value in a,in this example 10 is the max value so the output that i would have is b=[1 4] How take care of this?
Andrei Bobrov
Andrei Bobrov el 28 de En. de 2014
see ADD part in this answer
roberto verrelli
roberto verrelli el 28 de En. de 2014
Thank you Andrei! I have still a problem,sometimes the a matrix is like this
  • a=[12 12 NaN NaN]
in this case n=histc(a,u) returns n=[0 0 0] so it's not working..i can replace NaN with zeros but if then i have the a matrix like this
  • a=[12 0 0 0] it's not working beacuse out =find(ismember(a,d(end))) will be
  • out=[2 3 4]
and not
  • out=[1]
Andrei Bobrov
Andrei Bobrov el 29 de En. de 2014
corrected
roberto verrelli
roberto verrelli el 29 de En. de 2014
Now it's perfect!Thank you very much Andrei ;-)

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Más respuestas (1)

Pramod Kumar Jangir
Pramod Kumar Jangir el 20 de Oct. de 2019

0 votos

Hi Andrei,
your code only wokrs for the consecutive equal values. Could you please modify it for random same values in an array?
Pramod

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