Recap: \(k\)-Means Clustering

DS701 Session 4 — Wed Sep 16, 2026

Today’s plan

  • 5 min — knowledge-check review
  • 20 min — highlights and Q&A (answer or pass — answering always earns credit)
  • 60 min — in-class activity (small groups)
  • wrap-up and cold-call check-ins

Knowledge-Check Review

KC 1: What does \(k\)-means minimize?

Name the quantity and write it as a formula.

Show answer

Answer: the Within-Cluster Sum of Squares (WCSS), a.k.a. the inertia:

\[ \underset{C_1, \dots, C_K}{\arg\min} \hspace{1em} \sum_{k=1}^K \hspace{1em} \sum_{x\in C_k} \Vert x-c_k\Vert^2. \]

For every point, the squared Euclidean distance to its own cluster center, summed over all points. In scikit-learn this is kmeans.inertia_.

KC 2: One iteration by hand

Points \(1.2,\ 1.8,\ 2.1,\ 7.3,\ 7.9,\ 8.2,\ 9.1,\ 10.5\), initial centers \(A = 4.0\), \(B = 11.0\).

Show answer

Assign — closest center wins (watch \(7.9\): \(|7.9-4.0| = 3.9\) vs. \(|7.9-11.0| = 3.1 \Rightarrow B\)):

Cluster Points
A 1.2, 1.8, 2.1, 7.3
B 7.9, 8.2, 9.1, 10.5

Update — center of mass of each cluster:

\[ A' = \frac{1.2+1.8+2.1+7.3}{4} = 3.1, \qquad B' = \frac{7.9+8.2+9.1+10.5}{4} = 8.9 \]

Next round, \(7.3\) defects to \(B\) (\(4.2\) vs. \(1.6\)) and we settle at \(A^\ast = 1.7\), \(B^\ast = 8.6\).

KC 3: Why scale the features?

Cluster customers on age (18–80) and income ($20k–$200k), unscaled. What determines the clusters?

Show answer

Answer: income, and only income.

  • Assignment uses Euclidean distance, and squared income differences are ~\(10^6\) times larger than squared age differences.
  • A one-dollar difference in income counts about as much as a whole category of a binary feature.
  • Fix: center each feature at zero and rescale to unit standard deviation, so every feature gets a comparable say in the distance.

Highlights

What are we actually asking for?

A clustering is a grouping of data objects such that objects within a group are similar and objects in different groups are dissimilar.

So we want to:

  • minimize intra-cluster distances
  • maximize inter-cluster distances

And it is unsupervised — no labels go in, and any labels we put on the clusters we invent afterwards.

Partitional setup for \(k\)-means: data are points in \(\mathbb{R}^d\), every object belongs to one and only one cluster (mutually exclusive), the clusters cover everything (exhaustive), and \(K\) is given in advance.

Hard problem, easy algorithm

  • Minimizing WCSS exactly is NP-hard for \(d \ge 2\) — no efficient exact algorithm is expected.
  • \(k=1\) and \(k=n\) are the easy corner cases (global mean; every point its own center, WCSS \(=0\)).
  • Nonetheless: Lloyd’s algorithm (1957) is a simple heuristic that works very well in practice.
  • Voted a top-10 data mining algorithm, and independently rediscovered several times.
  • Careful with the wording: the \(k\)-means problem \(\neq\) the \(k\)-means algorithm.

Lloyd’s algorithm

  1. Pick \(K\) cluster centers \(\{c_1, \dots, c_K\}\)
  2. Assign: define \(C_j\) as the points closest to center \(c_j\)
  3. Update: set \(c_j\) to the center of mass of \(C_j\)
  4. Repeat step 2 until convergence
    • centers move below a threshold, or inertia changes below a threshold, or max iterations

Neither step 2 nor step 3 can increase WCSS \(\Rightarrow\) it always converges — but only to a local minimum.

Initialization matters

  • Same data, same \(K\), different random start \(\Rightarrow\) different answer.
  • A bad start gets stuck: the algorithm can never climb back out, because it never increases WCSS.
  • Strategy: pick initial centers that are far apart — this is \(k\)-means++.
  • It is scikit-learn’s default, and n_init restarts the whole thing several times and keeps the best WCSS.

Choosing \(K\)

\(K\) is our first hyperparameter: a parameter that must be set before the model parameters (the centers) can be learned.

WCSS alone can’t pick it — WCSS decreases monotonically in \(K\), all the way to \(0\) at \(K = n\).

So: sweep \(K = 1, 2, 3, \dots\), fit each, and score it.

  • Elbow method — plot WCSS vs. \(K\) and look for the bend where the marginal gain flattens.

We’ll also talk about the silhouette coefficient in an upcoming lecture.

Where \(k\)-means breaks

Assignment is “nearest center”, so cluster regions are straight-sided Voronoi cells and clusters are implicitly assumed to be spheres around their center.

  • Non-spherical — two half-moons get sliced by a straight boundary.
  • Anisotropic — stretched/rotated blobs, same failure after a linear transformation.
  • Unequal variance — squared distances make the wide cluster expensive, so it gets split.
  • Unequal sizes — the objective prefers roughly balanced, equal-radius cells (500/100/10 will not survive).
  • Bad init — a local minimum of WCSS.
  • Wrong \(K\) — ask for 2 and you get 2, whatever the data says.

Feature scaling

Because \(k\)-means looks for spherical clusters in Euclidean distance, the units of each feature are part of the model.

\[ \begin{bmatrix}\text{age } 27\\\text{income } 75000\\\text{gender } 0\end{bmatrix}, \qquad \begin{bmatrix}\text{age } 45\\\text{income } 42000\\\text{gender } 1\end{bmatrix} \]

Standard remedy, per feature and independently:

  1. subtract the mean (centers the feature at zero), then
  2. divide by the standard deviation (puts every feature on a comparable range).

Whenever you build or pick a distance metric, think about feature scale first.

Where it works well

\(k\)-means is a workhorse when the assumptions roughly hold:

  • Customer segmentation — spending / order value / order count \(\rightarrow\) personas (scale first: dollars vs. counts).
  • Iris — recovers the three species almost exactly; setosa is cleanly separated, versicolor/virginica overlap a little.
  • Image compression — cluster pixels in \((R,G,B)\) space, replace each pixel with its centroid. \(k=16\) means a 4-bit palette index instead of 24 bits: \(24/4 = 6\times\).

Original, 24-bit color

\(k=16\) centroids, \(6\times\) smaller

In-Class Activity

Activity: \(k\)-Means from scratch and in scikit-learn

Goal: implement one Lloyd’s iteration yourself, check it against sklearn.cluster.KMeans, then the stretch — sweep \(k\), and settle a case where the elbow is not clear — and finally break \(k\)-means on purpose and diagnose why.

  • Work in groups of 2–3.
  • Parts marked (autograded) are submitted to Gradescope; the rest is participation.
  • Staff will circulate — be ready to explain any part of your work.
  • Dependencies: numpy, matplotlib, scikit-learn only.

  

The activity notebook goes live on the day of the lecture. Colab is optional — you can also open it on GitHub and run it locally.

Going deeper

Full lecture notes: \(k\)-Means Clustering

Also see:

Other clustering evaluation metrics:

Also see k-means++ initialization

Next up: hierarchical clustering (04) — clusters without choosing \(k\) up front.

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