Exercises — Week 10 — Clustering: Sorting Without Labels¶
What you are building¶
K-Means on raw and scaled MRR + usage, named personas, and a note on what a cluster id is (and is not) for.
Predict before you run¶
mrr is tens-to-hundreds of dollars; n_support is 0, 1, 2. If you cluster unscaled, which column dominates distance, and what are the “personas” actually sorted by?
Before you start¶
- K=4 is a teaching default, not a discovered truth.
- Churn rate by cluster is observational: it describes a pile, it does not explain it.
Each task has three hints, closed by default. Open only as far as you need.
Task¶
Work in starter.py. Run from the repo root:
0. Predict first. mrr is in dollars (tens to low hundreds); n_support is a small integer (0, 1, 2…) — roughly a 100–1000× difference in raw scale. Before running exercise 1: if you cluster on both columns unscaled, which one do you expect to dominate the distance calculation, and what will the resulting "personas" actually be sorted by? Write your guess, then run exercise 1 to check.
1. Raw, then scaled. Cluster on raw mrr + total_usage (no scaler) and profile the clusters. Then repeat with StandardScaler and profile again. Which column sorted each version?
Hint 1 — a nudge
K-Means measures straight-line distance. A $100 gap in MRR and a 100-event gap in usage are the same distance to it. Look at each column's spread before you look at any cluster.
Hint 2 — the approach
KMeans(n_clusters=4, n_init=10, random_state=42) on df[["mrr", "total_usage"]].to_numpy(), then again on StandardScaler().fit_transform(...) of the same columns. Profile each with groupby("cluster") medians. The column whose median climbs steadily across clusters is the one doing the sorting.
Hint 3 — most of the code
from sklearn.cluster import KMeans
from sklearn.preprocessing import StandardScaler
from lib.course_data import load_customer_360
df = load_customer_360()
cols = ["mrr", "total_usage"]
print(df[cols].std().round(1).to_string(), "\n")
def profile(X, label: str) -> None:
clusters = KMeans(n_clusters=4, n_init=10, random_state=42).fit_predict(X)
table = df.assign(cluster=clusters).groupby("cluster").agg(
n=("user_id", "count"), mrr=("mrr", "median"), usage=("total_usage", "median"),
churn=("is_churned", "mean"),
)
print(label, "\n", table.sort_values("mrr").round(2).to_string(), "\n")
profile(df[cols].to_numpy(), "raw")
profile(StandardScaler().fit_transform(df[cols]), "scaled")
2. Name the personas. From the K=4 table above, write a one-line name and one marketing action per cluster. If two rows get the same name, merge them.
Hint 1 — a nudge
A persona name is a PR title for a row of medians. If you can't write one without saying "cluster 2", the row isn't a persona yet.
Hint 2 — the approach
Read each row's MRR, usage, and adoption against the overall medians — high / low / typical. Two rows that read the same are one persona, and K was too big.
Hint 3 — a skeleton
3. Peek, don't train. Churn rate by cluster is a story. Training a classifier on cluster id is usually weaker than training on the original features — the id is a lossy compression.
Hint 1 — a nudge
A cluster id takes several columns and squeezes them into one of four labels. What did the squeeze throw away?
Hint 2 — the approach
Print churn rate per scaled cluster — that's the story you're allowed to tell. Then write down why cluster would be a poor column in the /predict contract (think: what happens when you refit K-Means next month?).
Hint 3 — most of the code
The note on why the id is not an API feature is yours.Success criteria¶
- Raw and scaled cluster profiles side by side, with the column that sorted each one named.
- Four (or fewer) named personas with one action each.
- Written note: cluster id is not an API feature.
After you run¶
Personas are unsorted piles with names. They are not a /predict contract.