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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

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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) ;

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Cedric Wannaz
Cedric Wannaz on 23 May 2014
Note that you can contract this into a 2 liners
pos = find( A == (-1 + 2 * (sum(A==1) == 1)) ) ;
value = A(pos) ;
but you lose in clarity.
Mahdi
Mahdi on 23 May 2014
Very nice Cedric! I didn't even think of doing that way!
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

Sameer
Sameer on 23 May 2014
Hi...Thanks for replying.
I want to know the index of an entry which is unique to a particulat set of values. lets say
if A=[1 1 1 -1 0 0] then I want the indices of -1. if A=[-1 -1 -1 0 1 0] then I want the indices of 1.
I hope I made it clear what I am looking for.
Awaiting your Reply.
Regards
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
Hi...Thanks for replying.
I want to know the index of an entry which is unique to a particulat set of values. lets say
if A=[1 1 1 -1 0 0] then I want the indices of -1. if A=[-1 -1 -1 0 1 0] then I want the indices of 1.
I hope I made it clear what I am looking for.
Awaiting your Reply.
Regards
Mahdi
Mahdi on 23 May 2014
You can use the unique function.
[C, ia, ic]=unique(A)
Where the matrix C gives the unique value (1 or -1 in this case), and ia gives the indices where these are found.
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
Thank you...I appreciate your help. But the above solution is confined to a particular A(i guess) and its returing the indicies of only -1 if the elements of A changes.
Actually I have large number of different vectors and in each either 1 or -1 is unique (either only 1 or -1 in a vector like A=[1,1,1,-1,0,0] (-1 is occuring once),B=[-1,-1,-1,1,0,0](1 is occuring once)) so for A i should have the indices of -1 for B indices of 1. so I am looking for a code which will determine the unique number in a vector and return its corresponding indices and only of that particular number...not like unique function.
Awaiting your reply.
Regards
George Papazafeiropoulos
George Papazafeiropoulos on 23 May 2014
Try the above code for different A. Define A as you want and then execute the two last lines of the code. I think it works...
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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