Clustering

Cluster Inertia

Compute k-means inertia: sum of squared distances from each point to its assigned centroid. Loop over points, look up the centroid by assignment index, compute squared Euclidean distance, accumulate. Library: NumPy fancy indexing centroids[assignments] broadcasts centroids per point for a vectorized diff. RESULT: inertia (rounded).

By hand

With NumPy

centroids[assignments] selects the centroid row for each point; subtracting gives per-point delta vectors; squaring and summing over axis=1 yields squared distances; a final np.sum gives inertia.

naive.py 
points = [[1,2],[2,1],[1,3],[7,6],[8,7],[6,8]]
assignments = [0, 0, 0, 1, 1, 1]
centroids = [[1, 1], [7, 7]]
inertia = 0.0
for i in range(len(points)):
    cx = centroids[assignments[i]][0]
    cy = centroids[assignments[i]][1]
    dx = points[i][0] - cx
    dy = points[i][1] - cy
    inertia = inertia + dx*dx + dy*dy
print('RESULT:', round(inertia, 4))
library.py 
import numpy as np
from dalib.display import set_display
set_display()

points = np.array([[1,2],[2,1],[1,3],[7,6],[8,7],[6,8]], dtype=float)
assignments = np.array([0, 0, 0, 1, 1, 1])
centroids = np.array([[1.0, 1.0], [7.0, 7.0]])
deltas = points - centroids[assignments]
sq_dists = np.sum(deltas**2, axis=1)
print('sq_dists:', [round(float(v), 4) for v in sq_dists])
inertia = float(np.sum(sq_dists))
print('RESULT:', round(inertia, 4))

sq_dists: [1.0, 1.0, 4.0, 1.0, 1.0, 2.0]
RESULT: 10.0

Implementation notes

  • centroids[assignments[i]] uses the assignment as an integer index into the centroid list — no if/else branching needed per cluster.
  • Squared Euclidean distance (no sqrt) matches the metric k-means minimizes; inertia and the assignment loop are consistent.
  • Cross-reference: kmeans-one-iteration (this chapter) for centroids that minimize inertia after one full assign+update cycle.