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Below is a symmetric matrix $A$ with distances between observation $i$ and $j$.

$$ \begin{matrix} 0 & 9 & 8 & 6 & 3\\ 9 & 0 & 1 & 7 & 8\\ 8 & 1 & 0 & 6 & 9\\ 6 & 7 & 6 & 0 & 7\\ 3 & 8 & 9 & 7 & 0\\ \end{matrix} $$

My goal is to assign these into separate groups/clusters such the distance between observations within the group is minimized.

For example, the distance between observation 2 and 3 is 1 ($A_{23}$)

The distance between observation 1 and 5 is 3 ($A_{15}$)

According to that, observation 2 and 3 are likely to be part of the same "cluster". 1 and 5 also have a small distance of 3 between them, which also mean they should be part of the same "cluster". As you can see, observation 4 is very far from any other observation, which means it should be assigned to another "cluster".

The types of groups I initially trying to achieve according to the above example is as follows:

Cluster 1: observations 1, 5
Cluster 2: observations 2, 3
Cluster 3: observation 4

Do you know of an algorithm that can answer this kind of a problem?

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  • $\begingroup$ cs.cmu.edu/~aarti/Class/10701/readings/Luxburg06_TR.pdf $\endgroup$
    – Sycorax ♦
    Nov 29, 2017 at 14:46
  • $\begingroup$ This is not a distance matrix! The zeros at positions (2,5) and (5,2) indicate that the corresponding objects are co-located. Therefore they must exhibit identical distances to all other objects: this would be manifested as identical columns 2 and 5 and identical rows 2 and 5, but that's far from the case. Any attempt at clustering that assumes these are distances would therefore be invalid, in the off chance it actually succeeded. $\endgroup$
    – whuber ♦
    Nov 29, 2017 at 16:00
  • $\begingroup$ @whuber indeed you're right, I did not think of that when I created this example. It is meant to be a distance matrix - I will change the distances to remove the co-located points at (2,5). $\endgroup$
    – Flika205
    Nov 29, 2017 at 16:16
  • $\begingroup$ @whuber Also as far as that's relevant distance here is more conceptual. i.e. "distance between observations" not distance in the sense of kilometers on a 2d plane. $\endgroup$
    – Flika205
    Nov 29, 2017 at 16:22
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    $\begingroup$ It's not valid to call it a distance unless it satisfies the triangle inequality. My previous comment provides a simple demonstration that your matrix violates the triangle inequality. Many clustering methods assume the triangle inequality is satisfied, so it's important to be clear concerning whether your actual problem concerns a true distance matrix or not. $\endgroup$
    – whuber ♦
    Nov 29, 2017 at 16:39

1 Answer 1

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Use hierarchical clustering.

This gives you the best control over how distances of groups are computed. Because there is more than one way of linking.

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  • $\begingroup$ Hey Anony-Mousse, thank you for your suggestion, I'm currently looking at the documentation of it. May I ask to see an example of how this can be used with symmetric distance matrix? Thanks $\endgroup$
    – Flika205
    Nov 30, 2017 at 9:26
  • $\begingroup$ You call agnes(X) $\endgroup$
    – Has QUIT--Anony-Mousse
    Dec 1, 2017 at 1:13
  • $\begingroup$ The mentioned function is in R, right? I'm working on a project in Python, do you know what is the equivalent function in Python maybe? Thanks. $\endgroup$
    – Flika205
    Dec 1, 2017 at 9:07
  • $\begingroup$ Yes, that also exist in scipy and takes about 10 second to find. $\endgroup$
    – Has QUIT--Anony-Mousse
    Dec 2, 2017 at 10:20

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