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

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]]. Sq-dists: (0+1)+(1+0)+(0+4)+(0+1)+(1+0)+(1+1) = 1+1+4+1+1+2 = 10.

naive.py
Replay: real traced execution (multi-file project)
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))

  1. points ← [[1, 2], [2, 1], [1, 3], [7, 6], [8, 7], [6, 8]]

    1points = [[1,2],[2,1],[1,3],[7,6],[8,7],[6,8]]2assignments = [0, 0, 0, 1, 1, 1]
    values this step[[1, 2], [2, 1], [1, 3], [7, 6], [8, 7], [6, 8]]points

  2. assignments ← [0, 0, 0, 1, 1, 1]

    1points = [[1,2],[2,1],[1,3],[7,6],[8,7],[6,8]]2assignments = [0, 0, 0, 1, 1, 1]3centroids = [[1, 1], [7, 7]]
    values this step[0, 0, 0, 1, 1, 1]assignments

  3. centroids ← [[1, 1], [7, 7]]

    2assignments = [0, 0, 0, 1, 1, 1]3centroids = [[1, 1], [7, 7]]4inertia = 0.0
    values this step[[1, 1], [7, 7]]centroids

  4. inertia ← 0.0

    3centroids = [[1, 1], [7, 7]]4inertia = 0.05for i in range(len(points)):
    values this step0.0inertia

  5. i ← 0, cx ← 1, cy ← 1, dx ← 0, dy ← 1, inertia ← 1.0

    pass 1 of 6
    4inertia = 0.05for i in range(len(points)):6    cx = centroids[assignments[i]][0]7    cy = centroids[assignments[i]][1]8    dx = points[i][0] - cx9    dy = points[i][1] - cy10    inertia = inertia + dx*dx + dy*dy11print('RESULT:', round(inertia, 4))
    values this step0i1cx1cy0dx1dy0.0 → 1.0inertia
    All 6 passes — pass 1 is the card above
    passicxcydxdyinertia
    1011010.0 → 1.0
    20 → 1——0 → 11 → 01.0 → 2.0
    31 → 2——1 → 00 → 22.0 → 6.0
    42 → 31 → 71 → 7—2 → -16.0 → 7.0
    53 → 4——0 → 1-1 → 07.0 → 8.0
    64 → 5——1 → -10 → 18.0 → 10.0

  6. for i in range(len(points)):

    4inertia = 0.05for i in range(len(points)):6    cx = centroids[assignments[i]][0]

  7. stdout ← RESULT: 10.0

    10    inertia = inertia + dx*dx + dy*dy11print('RESULT:', round(inertia, 4))
    values this stepRESULT: 10.0stdout

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.

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.