K-means via scipy vs scikit-learn - Practice Questions

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of K-means clustering with scipy.cluster.vq

What is the output of the following code snippet that uses scipy's kmeans and vq functions?

SciPy

import numpy as np
from scipy.cluster.vq import kmeans, vq

np.random.seed(0)
data = np.vstack((np.random.normal(0, 1, (5, 2)), np.random.normal(5, 1, (5, 2))))
centroids, distortion = kmeans(data, 2)
labels, _ = vq(data, centroids)
print(labels.tolist())

A [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]

B [1, 1, 1, 1, 1, 0, 0, 0, 0, 0]

C [0, 1, 0, 1, 0, 1, 0, 1, 0, 1]

D [1, 0, 1, 0, 1, 0, 1, 0, 1, 0]

💡 Hint

Remember that kmeans clusters points based on proximity to centroids, and vq assigns labels accordingly.

The first five points are generated around (0,0) and the next five around (5,5). The kmeans function finds two centroids near these centers. The vq function assigns the first five points to cluster 0 and the last five to cluster 1, resulting in labels [0,0,0,0,0,1,1,1,1,1].

❓ Predict Output

intermediate

Output of KMeans from scikit-learn with fixed random state

What is the output of the following code that uses scikit-learn's KMeans to cluster the same data?

SciPy

import numpy as np
from sklearn.cluster import KMeans

np.random.seed(0)
data = np.vstack((np.random.normal(0, 1, (5, 2)), np.random.normal(5, 1, (5, 2))))
kmeans = KMeans(n_clusters=2, random_state=0).fit(data)
print(kmeans.labels_.tolist())

A [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]

B [1, 1, 1, 1, 1, 0, 0, 0, 0, 0]

C [0, 1, 0, 1, 0, 1, 0, 1, 0, 1]

D [1, 0, 1, 0, 1, 0, 1, 0, 1, 0]

💡 Hint

scikit-learn's KMeans also clusters points based on proximity and uses random_state for reproducibility.

The KMeans model finds two clusters similar to scipy's kmeans. The labels_ attribute assigns the first five points to cluster 0 and the last five to cluster 1, matching the expected output.

❓ data_output

advanced

Number of unique clusters from scipy kmeans labels

After running scipy's kmeans and vq on a dataset with 3 clusters, how many unique cluster labels will the labels array contain?

SciPy

import numpy as np
from scipy.cluster.vq import kmeans, vq

np.random.seed(1)
data = np.vstack((np.random.normal(0, 1, (4, 2)), np.random.normal(5, 1, (4, 2)), np.random.normal(10, 1, (4, 2))))
centroids, _ = kmeans(data, 3)
labels, _ = vq(data, centroids)
unique_labels = len(set(labels))
print(unique_labels)

A 4

B 2

C 3

D 1

💡 Hint

The number of clusters requested is 3, so expect 3 unique labels.

Since kmeans is run with 3 clusters, the labels assigned by vq will be from 0 to 2, resulting in 3 unique labels.

❓ visualization

advanced

Visual difference between scipy and scikit-learn K-means centroids

Which option correctly describes the difference in centroid positions when clustering the same dataset with scipy.cluster.vq.kmeans and sklearn.cluster.KMeans?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.cluster.vq import kmeans
from sklearn.cluster import KMeans

np.random.seed(42)
data = np.vstack((np.random.normal(0, 1, (10, 2)), np.random.normal(5, 1, (10, 2))))

centroids_scipy, _ = kmeans(data, 2)
kmeans_sklearn = KMeans(n_clusters=2, random_state=42).fit(data)
centroids_sklearn = kmeans_sklearn.cluster_centers_

plt.scatter(data[:,0], data[:,1], c='gray', label='Data points')
plt.scatter(centroids_scipy[:,0], centroids_scipy[:,1], c='red', marker='x', s=100, label='Scipy centroids')
plt.scatter(centroids_sklearn[:,0], centroids_sklearn[:,1], c='blue', marker='o', s=100, label='Sklearn centroids')
plt.legend()
plt.title('Centroids from scipy vs sklearn K-means')
plt.show()

A Sklearn centroids are always the mean of all data points, while scipy centroids are medians.

B Scipy centroids and sklearn centroids overlap exactly because both use the same algorithm and initialization.

C Scipy centroids are always closer to the origin than sklearn centroids.

D Scipy centroids and sklearn centroids differ slightly because sklearn uses k-means++ initialization and iterative refinement.

💡 Hint

Consider differences in initialization and algorithm details between scipy and sklearn implementations.

Scipy's kmeans uses random initialization and a single run, while sklearn's KMeans uses k-means++ initialization and multiple iterations for refinement, causing slight differences in centroid positions.

🧠 Conceptual

expert

Key difference in output between scipy.cluster.vq.kmeans and sklearn.cluster.KMeans

Which statement best describes a key difference in the outputs of scipy.cluster.vq.kmeans and sklearn.cluster.KMeans when applied to the same dataset?

A Scipy's kmeans always produces deterministic results, while sklearn's KMeans results vary randomly every run.

B Scipy's kmeans returns centroids and distortion, while sklearn's KMeans returns centroids, labels, and inertia with iterative convergence.

C Sklearn's KMeans uses hierarchical clustering internally, while scipy's kmeans uses flat clustering.

D Scipy's kmeans returns labels directly, while sklearn's KMeans only returns centroids without labels.

💡 Hint

Think about what each function returns and how the algorithms run.

Scipy's kmeans returns centroids and a distortion measure but does not assign labels directly; labels are assigned separately with vq. Sklearn's KMeans returns centroids, labels, and inertia after iterative convergence with multiple runs.