How to tell which centroid is most common/ about skmeans HOT 1 CLOSED

Hi @brandonros ,

I don't know if I really understand the question, but I'll try to answer: This algorythm implementation doesn't explicitly give you the most common cluster, but it's easily obtainable, simply counting the elements of each cluster and then getting the one with more items. Suppose, given this result;

clusters = {
	it: 2,
	k: 3,
	idxs: [ 2, 2, 0, 2, 1, 1, 1, 2, 0, 2, 0, 2, 1, 1, 0 ],
	centroids: [ 13, 23, 3 ]
}

You can do a reduce to count the elements of each cluster:

count = clusters.idxs.reduce((map,idx)=>{
    map[a.centroids[idx]] = map[a.centroids[idx]] || 0;
    map[a.centroids[idx]]++;
    return map;
},{});

So, count will be;

{3: 6, 13: 4, 23: 5}

Finally, you can sort the keys (cluster centroids) by its length:

res = Object.keys(count).sort((a,b)=>count[b] - count[a])

Will give yoy the array of centroids sorted by its size:

["3", "23", "13"]

from skmeans.