How to use k-medoids?

[juliastats]

How to use k-medoids?

[kingzbauer]

This might be related. I keep getting this error when trying to run kmedoids

julia> kmedoids(mat, 8)
ERROR: AssertionError: !(isempty(grp))
 in _find_medoid at /root/.julia/v0.4/Clustering/src/kmedoids.jl:189
 in _kmedoids! at /root/.julia/v0.4/Clustering/src/kmedoids.jl:100
 in kmedoids at /root/.julia/v0.4/Clustering/src/kmedoids.jl:39

mat is a distance matrix.

[johnmyleswhite]

I think @lendle may be the main person who knows about this code.

[lendle]

The k-medoids code was rewritten by @cyocum in #22.

[lendle]

I just looked at the implementation and docs. C is actually an n x n matrix (not k x n).

The docs say "C – The cost matrix, where C[i,j] is the cost of assigning sample j to the medoid i" is a bit unclear. The ith row does correspond to the ith medoid for i = 1, ..., k, but corresponds to the cost associated with assigning each sample to a medoid defined by sample i for i = 1, ..., n.
@waTeim, would "C – The cost matrix, where C[i,j] is the cost of assigning sample j to a cluster with medoid sample i" be more clear?

[waTeim]

I don't think so. Are you sure because that approach has a lot of problems. It's not really the algorithm as documented in Wikipedia, it should be kxn and be re-calculated every iteration because the medoids can change. If it's nxn then that becomes unusable because take for instance how I was going to use it -- 90,000 data points gives rise to a 90000x90000 matrix which uses up more memory that the machine I have to run it on (40 GB RAM), and frankly that's kinda small compared to todays dataset sizes.

[lindahua]

The current approach takes as input a pre-computed pairwise cost matrix. When n is very large, one should use a different algorithm.

[waTeim]

Yea, that's more efficient if the number of rows^2 is storable, but in this case, nope. And like I said, this isn't even the largest of the datasets. I was looking around for a simple implementation of PAM to submit as a PR so there there would be options, but am stuck on the "compare cost of each non-medioid with a mediod cost and swap if lower" seems too inefficient to implement directly like that.

[diegozea]

@waTeim If your matrix is symmetric, maybe https://github.com/diegozea/PairwiseListMatrices.jl can be useful to you.

[alyst]

This is an old question, that does not apply to the new distance-based API.