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How to find the indices of element occuring once in a vector?

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Sameer
Sameer on 23 May 2014
Edited: Cedric Wannaz on 23 May 2014
Hello all
I want to know...How can I get the indices of a value that is occuring only once in a vector...please guide.
Example: A=[1 1 0 -1 0 0 1 0 1 1]
Task: To identify the indices of -1 (as it is occuring only once) in A.
Please Help!!!
Regards

Accepted Answer

Cedric Wannaz
Cedric Wannaz on 23 May 2014
Edited: Cedric Wannaz on 23 May 2014
There are ways to solve your problem based on HISTC or ACCUMARRAY. However, the simplest approach if you really have only two situations (unique 1 or unique -1) is probably the following:
if sum( A == 1 ) == 1
pos = find( A == 1 ) ;
else
pos = find( A == -1 ) ;
end
value = A(pos) ;
  4 Comments
Cedric Wannaz
Cedric Wannaz on 23 May 2014
Well, I would personally go for clarity .. otherwise there is even a one liner actually:
[~,pos,value] = find( A .* (A == -1 + 2*(sum(A==1)==1)) )

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More Answers (3)

George Papazafeiropoulos
George Papazafeiropoulos on 23 May 2014
Edited: George Papazafeiropoulos on 23 May 2014
A=[1 1 0 -1 0 0 1 0 1 1];
[~,c]=histc(A,unique(A));
out=A(c==1);
  2 Comments
Cedric Wannaz
Cedric Wannaz on 23 May 2014
Sagar, you should take the time to understand his example. In particular, see what c is, what c==1 is, etc. Maybe read about logical indexing, and if you cannot use the latter and really need the position of unique element(s), read about FIND.

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Mahdi
Mahdi on 23 May 2014
If you're looking specifically for the value of -1, you can use the following:
index1=find(A==-1)
  3 Comments
Sameer
Sameer on 23 May 2014
Thank you....but unique command returns the values that are present in the vector so here it is -1 0 1. But I am looking for the single value that is either 1 or -1 and then the indices of that particular value.
Regards

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George Papazafeiropoulos
George Papazafeiropoulos on 23 May 2014
Edited: George Papazafeiropoulos on 23 May 2014
A=[1 1 -1 0 0 0 1 0 1 1];
[~,c]=histc(A,unique(A));
out=find(c==1);
  3 Comments
Sameer
Sameer on 23 May 2014
Unfortunately its not working as in the attached image you can see that 1 is the unique and its index should be 7 but the code is showing for -1 instead that is 1 2 3. where A=[-1 -1 -1 0 0 0 1 0].
Awaiting your response.
Regards

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