The code

Straight from labs/course-6-r/python/13_kmeans.py, unchanged.

"""Experiment 13 (Python equivalent) -- K-Means clustering.

R version: ../13_kmeans.R  (kmeans(scale(df), centers = 3, nstart = 25))

Includes the scaling demonstration from Unit 4: the same data clustered with
and without scale(), to show that forgetting it changes the answer.
"""
import numpy as np
from sklearn.cluster import KMeans
from sklearn.preprocessing import StandardScaler
from sklearn.metrics import silhouette_score

# Step 1: Make the customer data, from a seeded generator
RNG = np.random.default_rng(42)

# Customers: annual income (rupees) and age. Deliberately different scales.
income = np.concatenate([RNG.normal(3_00_000, 40_000, 20),
                         RNG.normal(9_00_000, 60_000, 20),
                         RNG.normal(15_00_000, 80_000, 20)])
age = np.concatenate([RNG.normal(28, 4, 20), RNG.normal(45, 5, 20),
                      RNG.normal(38, 6, 20)])
X = np.column_stack([income, age])


def fit(data, k=3):
    km = KMeans(n_clusters=k, n_init=25, random_state=42)
    labels = km.fit_predict(data)
    return km, labels


if __name__ == "__main__":
    print("K-MEANS                        R: kmeans(scale(df), 3, nstart = 25)")
    print(f"  {X.shape[0]} customers, 2 features")
    print(f"  income range {income.min():,.0f} to {income.max():,.0f}")
    print(f"  age    range {age.min():.0f} to {age.max():.0f}")

    # Step 2: Cluster without scaling
    print("\nWITHOUT scaling")
    km_raw, lab_raw = fit(X)
    print(f"  silhouette = {silhouette_score(X, lab_raw):.4f}")
    for c in range(3):
        m = X[lab_raw == c]
        print(f"    cluster {c}: n={len(m):2d}  mean income={m[:,0].mean():>10,.0f}"
              f"  mean age={m[:,1].mean():5.1f}")

    # Step 3: Cluster with scaling
    print("\nWITH scaling                   R: scale() before kmeans()")
    Xs = StandardScaler().fit_transform(X)
    km_s, lab_s = fit(Xs)
    print(f"  silhouette = {silhouette_score(Xs, lab_s):.4f}")
    for c in range(3):
        m = X[lab_s == c]
        print(f"    cluster {c}: n={len(m):2d}  mean income={m[:,0].mean():>10,.0f}"
              f"  mean age={m[:,1].mean():5.1f}")

    # Step 4: Choose k by the elbow method
    print("\nELBOW METHOD                   R: sapply(1:10, ...$tot.withinss)")
    for k in range(1, 8):
        km = KMeans(n_clusters=k, n_init=10, random_state=42).fit(Xs)
        bar = "#" * int(km.inertia_ / 6)
        print(f"    k={k}  WSS={km.inertia_:7.2f}  {bar}")

    # Step 5: Compare the two clusterings, and the variances
    agree = (lab_raw == lab_s).mean()
    best_agree = max(agree, 1 - agree)
    print(f"\n  Unscaled and scaled clusterings agree on {best_agree:.0%} of points"
          if best_agree < 1 else "\n  Both clusterings agree here")
    print("  With this data the income separation is so wide that both find it,")
    print("  but the unscaled version is driven by income ALONE -- age contributes")
    print("  essentially nothing, because a 20-year age gap is 20 units against an")
    print("  income gap of 600,000. Scaling is what lets age matter at all.")

    var_ratio = income.var() / age.var()
    print(f"\n  variance ratio income:age = {var_ratio:,.0f} : 1")
    assert var_ratio > 1000, "the scale problem must be real for the point to stand"
    print("  that ratio is why scale() is mandatory, not optional ✓")

Where this sits

The same thing in R: K-Means Clustering in R.

The theory behind it is in this course’s units; the whole lab, with all 18 experiments, is on the lab page.