I am looking for a symbol to represent the operation of taking unique values from a vector. So, say the symbol was $\theta$:

$v = [0, 0, 0, 1, 1, 1, 3, 1, 2, 0]$

$\theta(v) = [0, 1, 3, 2]$

Or is this something that just isn't defined?



Thanks all for your help here. Your answers and comments have led me to realise that doing this operation on a vector is not really what I want to do; using a set makes much more sense logically.

(Background: I have many coordinates $\boldsymbol{p}_i = [x_i, y_i, z_i]$ in 3D space, which I then perform a rounding function on so they are equally sampled. However, there will be many duplicates points at each new coordinate, so I remove the duplicated $\boldsymbol{p}$ points to get the final set of points. So it makes sense to define a set $\boldsymbol{P} = \{\boldsymbol{p}_1, ... \boldsymbol{p}_i, \cdot \boldsymbol{p}_n\}$).

  • $\begingroup$ I don't know of any universal notation for this kind of operation, so you'll probably have to define $\theta$ yourself (like you just did). At least $\theta$ is linear, so it'll have nice properties. $\endgroup$ Dec 20, 2010 at 15:01
  • 4
    $\begingroup$ @Gunnar Magnusson: Linearity does not make sense here, because we are dealing with integer vectors. Also, $\theta$ is not additive: while $[0,1]$ is mapped to $[0,1]$ and $[1,0]$ is mapped to $[1,0]$, $[0,1]+[1,0]=[1,1]$ is mapped to $[1]$. $\endgroup$
    – Rasmus
    Dec 20, 2010 at 15:12
  • 1
    $\begingroup$ @Rasmus How do you know that the OP is only interested in integer vectors? This restriction doesn't appear anywhere in the post. $\endgroup$
    – Alex B.
    Dec 20, 2010 at 15:45
  • $\begingroup$ @Rasmus: So is $\theta$ supposed to return the set of coordinates of a vector with some ordering? I thought the OP just wanted to pick out certain coordinates, like he does in $\theta$ where he picks (for example) the 1st, 4th, 7th and 9th ones. Then the function is linear from $\mathbb R^{10}$ to $\mathbb R^4$. $\endgroup$ Dec 20, 2010 at 16:48
  • $\begingroup$ @Alex Bartel: I deduced this from the fact that the OP uses only integers as coordinates. That's of course not a well-founded logical argument, I admit. ;) $\endgroup$
    – Rasmus
    Dec 20, 2010 at 17:05

2 Answers 2


A common way to express that you are throwing away duplicates is to translate a sequence into a set. A set is commonly understood as an unordered collection of distinct objects. It is usually denoted using curly braces. So you could write $\theta(v) = \{v_i\}_{i\in\{1,\ldots,n\}}$, where $v=[v_1,\ldots,v_n]$.

  • $\begingroup$ I think the value of $\theta$ is supposed to be a vector again, not just a set. That is, I think the OP wants to endow the set of entries of the input vector with the total order given by "appearing earlier". $\endgroup$
    – Rasmus
    Dec 20, 2010 at 17:10
  • $\begingroup$ Aha. So am I right in thinking that taking the unique elements of a set S would always return S? $\endgroup$ Dec 20, 2010 at 17:27
  • $\begingroup$ @Bill Yes, that's right. $\endgroup$
    – Alex B.
    Dec 20, 2010 at 17:46

There is no standard notation for the set of coefficients of a vector. Indeed, there's not even any standard notation for the set of coefficients of a polynomial (which would be useful when taking the content, i.e. the gcd / ideal generated by the coefficients).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.