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 :-)

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Andrei Bobrov
Andrei Bobrov 2014년 1월 28일
편집: Andrei Bobrov 2014년 1월 29일

3 개 추천

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)))

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roberto verrelli
roberto verrelli 2014년 1월 28일
편집: roberto verrelli 2014년 1월 28일
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 2014년 1월 28일
see ADD part in this answer
roberto verrelli
roberto verrelli 2014년 1월 28일
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 2014년 1월 29일
corrected
roberto verrelli
roberto verrelli 2014년 1월 29일
Now it's perfect!Thank you very much Andrei ;-)

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추가 답변 (1개)

Pramod Kumar Jangir
Pramod Kumar Jangir 2019년 10월 20일

0 개 추천

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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