Inertia in k-means clustering

[frederik-f]

Dear all,

I'm using the k-means algorithm for clustering a set of curves. Does anybody know, how the inertia is exactly calculated in this case? Suppose I have the grid_points as [0, 1, 2, 3] and the data_matrix as [[1, 2, 3, 4], [6, 5, 4, 3], [5, 3, 1, -1]]. If I'm using 2 clusters, I would e.g. retrieve the cluster_centers = [[5.5, 1],[4, 2],[2.5, 3],[1, 4]] and an inertia of 10.5, which is the sum of the squared distances. The single distances read [0., 2.29128785, 2.29128785]. I would really appreciate, if anybody could tell me how these distances are calculated exactly, ideally for this specific example.

Thanks a lot in advance!

[vnmabus]

In your case you have the discretization grid $\mathbf{t}=(0,1,2,3)$ and the functions $x_0,x_1,x_2$ with $x_0(\mathbf{t})=(1,2,3,4)$, $x_1(\mathbf{t})=(6,5,4,3)$ and $x_2(\mathbf{t})=(5,3,1,-1)$.
The cluster centers are $c_0,c_1$ with $c_0(\mathbf{t})=(5.5,4,2.5,1)$ and $c_1(\mathbf{t})=(1,2,3,4)$.
The distance between two functions $f$ and $g$ is configurable using the metric parameter. By default is the $L^2$ distance, that is $d(f,g)=\sqrt{{\int }_{\mathcal{T}}|f(t)-g(t){|}^{2}dt}$ (thus, analog to the Euclidean distance).
The distances appearing in the inertia are those between each observation and the cluster it belongs to (the closest one). Thus:
$d_0=d(x_0,c_1)=0$ (because both are equal)
$d_1=d(x_1,c_0)\sim 2.29128785$
$d_2=d(x_2,c_0)\sim 2.29128785$
In all cases the integral is approximated using a Simpson quadrature rule, as implemented in scipy.integrate.simpson.

[frederik-f]

Thanks a lot for your quick response and detailed explanation, which fully resolved my question!