I have a 2d 'grid' of xy points (2000 of them). I want to find the average vector between points I have drawn the vectors on the graph, how do I find these vectors mathematically? I want to find the average vector between points for the whole set of points.

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  • $\begingroup$ hey! um, so, is there a rule to generate all these 2000 points (e.g. "$L=\{m \vec a + n\vec b\,|\, m,n \in \mathbb Z\}$ is the set of all points"), or do you have a dataset? Also, really, the average between all these points? That'd be an average about $2000!$ vectors, and I'm not sure how helpful the result would be, or do you just mean "neighbors" in the dataset? $\endgroup$ Commented Nov 7, 2018 at 19:38
  • $\begingroup$ Dataset, I'm looking for the average vector. Neighbors, I've already found the nearest set of neighbors $\endgroup$ Commented Nov 7, 2018 at 19:42
  • $\begingroup$ ok, so how do you know which "direction" to count? because if I look at just the set of the three points you've connected, then I can draw the arrows like you did, or I can draw them the opposite direction. If I average the two vectors you chose, I'd get a vector in "top-rightish direction"; if I draw them the other way around the opposite; if I consider both, then the average vector is always a zero vector, because two opposite vectors would cancel themselves out. $\endgroup$ Commented Nov 7, 2018 at 19:47
  • $\begingroup$ I'm really stumbling a bit in the dark here and feel a little stupid for not understanding your question properly :( May I ask what the higher-level goal is? My current guess is that you want to actually find a best-matching $L$ as I described it in my first comment, but I actually don't know and it's a shot in the dark. $\endgroup$ Commented Nov 7, 2018 at 19:51
  • $\begingroup$ @MarcusMüller I though there was a algorithm that I could just throw in the data and it would spit out the vectors, it's been a while for linear algebra. I now know it's an over determined set of linear equations, I suppose I could find all of the vectors between each point and average them together, I just thought there was a slicker way to compute basis vectors for the points since the are on their own R^2 space (not for the regular grid) $\endgroup$ Commented Nov 7, 2018 at 20:38


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