Clustering, PCA, Text and Production Practice

Practise k-means, PCA, anomaly detection, TF-IDF text classification, cosine similarity, saving models and drift checks.

Intermediate7 problemsUpdated October 24, 2026

Review first: k-Means Clustering, Principal Component Analysis (PCA), Anomaly Detection, Text Classification, Embeddings and Semantic Search, Saving and Loading Models, MLOps: Models in Production

1.Cluster centres
Intermediate
Cluster the points [[1, 2], [1.5, 1.8], [1, 1], [8, 8], [9, 9.5], [8.5, 9]] into 2 groups with KMeans(n_clusters=2, n_init=10, random_state=0). Print the labels and the cluster centres rounded to 2 decimals, sorted by their first coordinate.
Show hint
kmeans.labels_ and kmeans.cluster_centers_; sort the centres with sorted(..., key=lambda c: c[0]).
Show solution
Python
import numpy as np from sklearn.cluster import KMeans points = np.array([[1, 2], [1.5, 1.8], [1, 1], [8, 8], [9, 9.5], [8.5, 9]]) kmeans = KMeans(n_clusters=2, n_init=10, random_state=0).fit(points) print(kmeans.labels_) print(sorted(kmeans.cluster_centers_.round(2).tolist(), key=lambda c: c[0]))
Output
[1 1 1 0 0 0] [[1.17, 1.6], [8.5, 8.83]]
2.Best k by silhouette
Intermediate
Generate data with make_blobs(n_samples=500, centers=4, random_state=2). For k from 2 to 6, fit k-means (n_init=10, random_state=0) and print k with its silhouette score (3 decimals). Which k scores highest?
Show hint
silhouette_score(X, km.labels_) from sklearn.metrics.
Show solution
Python
from sklearn.cluster import KMeans from sklearn.datasets import make_blobs from sklearn.metrics import silhouette_score X, _ = make_blobs(n_samples=500, centers=4, random_state=2) for k in range(2, 7): km = KMeans(n_clusters=k, n_init=10, random_state=0).fit(X) print(k, round(silhouette_score(X, km.labels_), 3))
Output
2 0.556 3 0.66 4 0.625 5 0.567 6 0.506
The data was generated with 4 blobs, yet k = 3 scores highest: two of the blobs lie close together, so treating them as one cluster gives better separation. The silhouette score measures separation, not the true number of groups.
3.Components for 90% variance
Intermediate
How many principal components keep 90% of the variance of the scaled digits data (load_digits)? Print the number and the exact share of variance they keep (3 decimals).
Show hint
PCA(n_components=0.9); then n_components_ and explained_variance_ratio_.sum().
Show solution
Python
from sklearn.datasets import load_digits from sklearn.decomposition import PCA from sklearn.preprocessing import StandardScaler X = StandardScaler().fit_transform(load_digits().data) pca = PCA(n_components=0.9).fit(X) print(pca.n_components_, round(pca.explained_variance_ratio_.sum(), 3))
Output
31 0.9
4.Unusual transactions
Intermediate
Transaction amounts are normally around ₹1,500 ± 300. Simulate 400 normal amounts (rng = np.random.default_rng(5)), append [9000, 50, 7500], fit IsolationForest(contamination=0.01, random_state=0) on the amounts as one column, and print the flagged amounts rounded to whole rupees, sorted.
Show hint
Reshape with .reshape(-1, 1); predict returns -1 for flagged rows.
Show solution
Python
import numpy as np from sklearn.ensemble import IsolationForest rng = np.random.default_rng(5) amounts = np.r_[rng.normal(1500, 300, 400), [9000, 50, 7500]].reshape(-1, 1) flags = IsolationForest(contamination=0.01, random_state=0).fit_predict(amounts) print(np.sort(amounts[flags == -1].ravel()).round(0))
Output
[ 50. 2308. 2332. 7500. 9000.]
With 403 amounts, a contamination of 1% flags 5 of them: the three planted amounts plus the two largest normal ones. contamination sets how many points are flagged, so ranking by score is usually more useful than a fixed rate.

5.Review sentiment

Intermediate

Train a TF-IDF + logistic regression pipeline on these reviews (1 = positive) and predict the labels of "great food" and "slow and cold".

Python

reviews = ["great food and friendly staff", "food was cold and bland", "loved the great service",
           "very slow service", "friendly and fast", "cold, slow and rude"]
labels = [1, 0, 1, 0, 1, 0]

Show hint

make_pipeline(TfidfVectorizer(), LogisticRegression()).fit(reviews, labels).

Show solution

Python

from sklearn.feature_extraction.text import TfidfVectorizer
from sklearn.linear_model import LogisticRegression
from sklearn.pipeline import make_pipeline

reviews = ["great food and friendly staff", "food was cold and bland", "loved the great service",
           "very slow service", "friendly and fast", "cold, slow and rude"]
labels = [1, 0, 1, 0, 1, 0]
model = make_pipeline(TfidfVectorizer(), LogisticRegression()).fit(reviews, labels)
print(model.predict(["great food", "slow and cold"]))

Output

[1 0]

6.Most similar document

Intermediate

Represent these documents with TF-IDF and find which one is most similar (by cosine similarity) to the query "refund for my broken phone". Print the document and the similarity (3 decimals).

Python

docs = ["phone screen broken, need repair", "how to get a refund", "refund for a broken phone charger", "delivery is late"]

Show hint

Fit the vectoriser on docs, transform the query, and use cosine_similarity from sklearn.metrics.pairwise.

Show solution

Python

from sklearn.feature_extraction.text import TfidfVectorizer
from sklearn.metrics.pairwise import cosine_similarity

docs = ["phone screen broken, need repair", "how to get a refund", "refund for a broken phone charger", "delivery is late"]
tfidf = TfidfVectorizer().fit(docs)
scores = cosine_similarity(tfidf.transform(["refund for my broken phone"]), tfidf.transform(docs))[0]
best = scores.argmax()
print(docs[best], round(scores[best], 3))

Output

refund for a broken phone charger 0.861

7.Check features for drift
Intermediate
With rng = np.random.default_rng(3), simulate training data for two features (clicks: normal mean 20, sd 5, 2000 rows; basket: normal mean 800, sd 150, 2000 rows) and recent data (clicks unchanged with 500 rows; basket mean 900, sd 150, 500 rows). For each feature print the KS statistic (3 decimals) and whether the p-value is below 0.01.
Show hint
ks_2samp(train, recent) from scipy.stats returns statistic and pvalue.
Show solution
Python
import numpy as np from scipy.stats import ks_2samp rng = np.random.default_rng(3) train = {"clicks": rng.normal(20, 5, 2000), "basket": rng.normal(800, 150, 2000)} recent = {"clicks": rng.normal(20, 5, 500), "basket": rng.normal(900, 150, 500)} for name in train: result = ks_2samp(train[name], recent[name]) print(name, round(result.statistic, 3), bool(result.pvalue < 0.01))
Output
clicks 0.056 False basket 0.229 True
Only basket has shifted (mean 800 to 900). The clicks feature has a small KS statistic and its p-value stays above 0.01, as expected for data from the same distribution.

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