Why is it wrong to compare k-means inertia between two runs built on different feature sets?

### The definition Inertia — also called the within-cluster sum of squares — is `sum over clusters, sum over points in that cluster, of squared_distance(point, centroid_of_that_cluster)`. It is exactly the quantity k-means exists to minimise: the assign-then-recentre loop lowers it at every step until assignments stop changing. Two properties follow directly from the formula and cause most of the misuse. **It is a sum, not an average or a ratio.** It has no upper bound and no normalisation. Zero is the floor, reached only when every point coincides with its centroid. There is no scale on which 340 is good and 1200 is bad; the number only means something relative to another number computed on identical geometry. **It is in squared units of the feature space.** Everything that changes those units changes the value: - *Scaling.* Multiply every feature by a constant `c` and every distance is multiplied by `c`, so every squared distance — and the total — is multiplied by `c*c`. Standardising a table therefore produces an inertia unrelated to the raw-table inertia, and usually a different partition too. - *Dimensionality.* Squared Euclidean distance is a sum over columns of squared per-column deviations. Adding columns adds non-negative terms, so a wider feature matrix generally carries a larger inertia even if it describes the same customers. - *Row count.* Each row contributes one term. Cluster the same structure on twice as many rows and the sum roughly doubles. Dividing by the number of rows to get a mean squared distance removes this one effect but leaves scale and dimensionality untouched. ### The failure in the wild The recurring version of this mistake: a first clustering run over eight behavioural columns reports an inertia of 1200; someone adds five more columns, rescales along the way, gets 340, and writes in the deck that the segmentation improved by a factor of three. Nothing of the kind was measured. The second number would be smaller or larger depending on the units of the new columns; it is not evidence about compactness relative to the first run at all. ### Where inertia is legitimately used - **Comparing restarts.** k-means converges to a local optimum that depends on initialisation. Running it many times from different seeds on identical data and keeping the lowest-inertia result is exactly the right use, because every restart shares the same geometry. - **Checking convergence.** Inertia decreases monotonically across the iterations of a single run; watching it flatten confirms the loop settled. - **As an ingredient.** It is the within-cluster dispersion term inside indices that normalise it, which is precisely what makes those indices comparable where the raw sum is not. ### What to report instead If the goal is to compare two candidate partitions built differently, use a dimensionless index rather than a raw sum: silhouette is bounded in `[-1, 1]`, and the Davies-Bouldin and Calinski-Harabasz indices are ratios, so multiplying every feature by one constant leaves them unchanged. Even then, be honest about the limit: an index computed in one feature space answers whether the partition is compact and separated *in that space*. It does not license the claim that one feature space is better than another — that judgement has to come from what the clustering is for, not from a geometry statistic.