0
$\begingroup$

I'm facing problems with determining the right number of clusters for my 2D time-series data.

I have a numerical simulation that outputs a time-series of 2D grids that represent a mass density evolving in time (imagine a drop of liquid spreading from the center of the box) for a certain set of parameters (that characterize the liquid). I redo this simulation for many different parameters (sampled from a grid on the space of parameters). Just by looking at the time-series, I can make up different regimes of dynamics in this parameter space (e.g. the liquid drop spreading fast or slow, uniformly in one direction vs. isotropically, etc.).

I'm running Kernel KMeans with the GAK for a fixed number of clusters (k) on the time-series, hiding the parameters. When I reintroduce the parameters and plot the assignments in this parameter space, the clusters are well separated, indicating that the model is learning something "meaningful" about the dynamics. When I run the algorithm for different k and plot the residual loss, I don't really see an elbow, though.

Elbow Method Attempt

Has anyone faced a similar challenge or can suggest alternative methods or metrics for choosing k in this context? Any insights or suggestions would be greatly appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

You can use results from K-means to explore cluster validity using other methods, based on different distance methods. Firstly, I would recommend looking at the effect of various feature transformations (normalization, mean-zero standardization, and percentiles) on cluster validity. That is, transform feature values first, then run K-means, then generate your plot. There are also a number of different estimators one can use for cluster validity, such as Davies-Bouldin index, Dunn's index, Silhouette distance, Hubert's $\Gamma$ statistic, etc.

In the plots below, I show the average object to cluster distance as a function of $k$ using 4 distance metrics (Euclidean, Manhattan, Chebychev, Canberra) for the raw, normalized, and mean-zero standardized feature values. You can see that when feature values are standardized, the Canberra distance shows are very distinct knee at $k=3$, and so does Chebychev.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.