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

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Sameer
Sameer 2014-5-23
编辑： Cedric 2014-5-23
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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Cedric
Cedric 2014-5-23
编辑：Cedric 2014-5-23
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
Cedric 2014-5-23
编辑：Cedric 2014-5-23
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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George Papazafeiropoulos
编辑：George Papazafeiropoulos 2014-5-23
A=[1 1 0 -1 0 0 1 0 1 1];
[~,c]=histc(A,unique(A));
out=A(c==1);
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Sameer
Sameer 2014-5-23
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
Cedric 2014-5-23
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 2014-5-23
If you're looking specifically for the value of -1, you can use the following:
index1=find(A==-1)
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Mahdi
Mahdi 2014-5-23
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 2014-5-23
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
编辑：George Papazafeiropoulos 2014-5-23
A=[1 1 -1 0 0 0 1 0 1 1];
[~,c]=histc(A,unique(A));
out=find(c==1);
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George Papazafeiropoulos
George Papazafeiropoulos 2014-5-23
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 2014-5-23
编辑：Sameer 2014-5-23
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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