How to determine agreement between clustering methods?

Question

Let's say you want to compare the outcome of KMeans and KMedoids. How to determine if cluster 1 from KMeans can be compared with cluster 1 with KMedoids. Or, in other words, let's say KMeans labels the clusters as {0, 1, 2, 3} and KMedoids as {a, b, c, d}: how does one make the correct comparisons? In high dimensions, one cannot visually address which clusters overlap.

Kind regards, Sean

This is not easy, there is no guarantee that they will find similar clusters, the clusters could be completely different and not comparable. — user2974951, Jun 21 '22 at 10:30
Hmm, the same holds for addressing stability. Is cluster 1 from run 1 from KMeans comparible to cluster 1 from run 2 from KMeans? — Sean_TBI_Research, Jun 21 '22 at 11:11
Go please to my web-page and download "Compare partitions". Pay attention to !KO_CLUAGREE and !KO_GRMATCH macros - their descriptions. The topic concerned in your question is "external cluster validation". — ttnphns, Jun 21 '22 at 11:16
@ttnphns Great! Thanks! I'll have to dive into the documentation. But does this compare outcomes such that the specific label becomes irrelevant? As basically, label 1 cluster outcome 1 =/ label 1 cluster outcome 2 — Sean_TBI_Research, Jun 21 '22 at 14:14
There is no meaningful 'cluster 1-ness'. What gets called 'cluster 1' in one output could be called 'cluster 2' in another output. I illustrate comparing clusterings from different methods in my answer to How to use both binary and continuous variables together in clustering? See also: How to select a clustering method? How to validate a cluster solution (to warrant the method choice)? — gung - Reinstate Monica, Jun 21 '22 at 19:14
I don't know but I believe there sure must be packages in Python and R which compute statistics comparing cluster solutions without need to know cluster labels, - similar to !KO_cluagree. — ttnphns, Jun 22 '22 at 01:31

score 0 · Answer 1 · answered Jun 21 '22 at 19:06

0

A solution can be found in the following explanation:

Does one know whether a Python implementation of above recursion exists?

It would look something in line with:

grouped_together = pd.DataFrame()
for (i in 1:N){
  for (j in (i+1):(N-1)){
    for (k in 1:2){
      grouped_together[i,j,k]=A[i,k]==A[j,k]
    }
  }
}

answered Jun 21 '22 at 19:06

Sean_TBI_Research

21

You can find a good recipe to compute the comembership confusion matrix here https://stats.stackexchange.com/q/548778/3277 – ttnphns Jun 22 '22 at 03:33

How to determine agreement between clustering methods?

1 Answers1