I would like to evaluate a similarity between two objects X and Y by comparing a neighbourhood in which they're located.

I construct two sets of nine concentric and equidistant circles with centers in X and Y respectively (nine circles, but utterly not Dante's infernal ;-D )

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This considered objects X and Y are surrounded by many other objects of different types. I consider all objects located only in this nine rings. In the figure above, we have two types of objects: yellow balls and green stars.

I'm counting occurrences of those surrounding objects, separately of each type and in each ring, creating two ordinary vectors. (Below, I've splitted them by type only for clarity, but generally that are two vectors: one for X and one for Y.) In our example: $X_{YellowBalls} = [1, 0, 0, 3, 0, \dots, 0]$, $X_{GreenStars} = [0, \dots, 0, 0, 3]$, $Y_{YellowBalls} = [0, 0, 0, 1, 2, \dots, 0]$, $X_{GreenStars} = [0, \dots, 2, 0, 1]$.

Now, I'm trying to infere something about similarity of X and Y on the basis of these two vecors using many vector similarity measure, which are mainly used in text mining for document comparision.

The main problem lies here: This off-the-shelf methods are prone to differentiate a way too much vector entries obtained by counting in consecutive rings, e.g. for X three yellow balls are gathered in 4th ring whereas for Y only one yellow ball is located in 4th ring and two are placed in adjoining 5th ring. They are in pronounced proximity, but they fall into different rings only by hapenstance, because they ran into a border.

A very similar partitions concers green stars, 3 in 9th against 1 in 9th and 2 in 7th), but they're not in neighbouring rings, so that's completly different story. In my opinion, a proper method of comparision should be able to catch such nicity and threat these two situations in a sundry manner.

The ultimate question is: are there any methods for vectors similarity measures that will be more aware to neighbouring positions in vectors.

So far, I've tried several adjustments, but unfortunately, none of my ideas improved it to a satisfactory extent.

Does anyone know any publications that cope with similar problem?


1 Answer 1


Is there a reason you're not comparing the locations of the green stars and yellow balls directly? In essence, take the average position of green stars and yellow balls in X, and then do the same for Y. We now have information on X and Y based upon the green balls and yellow stars that we can directly compare using any of the common techniques (cosine similarity, Euclidean distance, etc.).

  • 1
    $\begingroup$ I'm not sure I understand your idea with taking an average position. We can imagine a situation in which in X all yellow balls are placed solely in 5th ring (or 3rd, 4th and 5th, but the average is in 5th ring) and in Y there're two centers of gravity, that is balls are divided in halves and placed in innermost and outermost ring. In the second case the average position will also be 5th ring. Maybe I don't understand your solution properly? $\endgroup$ Feb 27, 2016 at 21:07
  • $\begingroup$ I think I should explain the appliction of this similarity comparision. I'm trying to compare two genomic regions, rings denote the distance from two appointed positions (so from R^2 it's reduced to R^1) and objects are simply genes that fall into those ranges. Types of object are corresponding to gene types (they were divided into some groups by biological function). Maybe I should not use vector similarity, but rather try to redesign the whole problem to compare distributions? $\endgroup$ Feb 27, 2016 at 21:17
  • $\begingroup$ In the example in my first comment above should be rather: 5th ring (or 4th, 5th, 6th) instead (or 3rd, 4th and 5th), but I think you got the overall overtone of my example. $\endgroup$ Feb 27, 2016 at 21:25
  • $\begingroup$ Sorry for my late reply, but my idea is that you already directly have the positions of the genes in R^2, so you don't actually need to convert into R^1. Take the average position of each gene type, and compare the regions on the basis of that. $\endgroup$ Feb 28, 2016 at 17:07
  • $\begingroup$ Your point still stands that the average may end up being a bad descriptor of the data if the average is not representative. What you could try instead is calculating the best Gaussian that fits that data; you already have the mean from the idea I gave in my last comment, and computing the covariance can be done this way. (You want S.) $\endgroup$ Feb 28, 2016 at 17:10

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