K-means via scipy vs scikit-learn - Hands-On Comparison

K-means Clustering with SciPy and scikit-learn

📖 Scenario: You work as a data analyst for a small retail company. You want to group customers based on their shopping habits to create better marketing strategies. You will use K-means clustering to find groups of similar customers.

🎯 Goal: Build a simple K-means clustering model using both scipy and scikit-learn libraries. Compare how to set up the data, run the clustering, and get the cluster centers.

📋 What You'll Learn

Create a dataset of customer shopping data as a list of lists.

Set the number of clusters to 2 using a variable.

Use scipy.cluster.vq.kmeans to find cluster centers.

Use sklearn.cluster.KMeans to fit the same data and get cluster centers.

Print the cluster centers from both methods.

💡 Why This Matters

🌍 Real World

K-means clustering helps businesses group customers or products based on features to target marketing or improve services.

💼 Career

Data scientists and analysts often use clustering to find patterns in data without labels, helping in customer segmentation and recommendation systems.

1

Create the customer data

Create a variable called data that holds this exact list of lists: [[1.0, 2.0], [1.5, 1.8], [5.0, 8.0], [8.0, 8.0], [1.0, 0.6], [9.0, 11.0]].

SciPy

# Create the variable data with the exact list of lists
# Your code here

Hint

Use a variable named data and assign the list exactly as shown.

2

Set the number of clusters

Create a variable called num_clusters and set it to 2.

SciPy

data = [[1.0, 2.0], [1.5, 1.8], [5.0, 8.0], [8.0, 8.0], [1.0, 0.6], [9.0, 11.0]]
# Create num_clusters and set it to 2
# Your code here

Hint

Use a variable named num_clusters and assign the value 2.

3

Run K-means with SciPy and scikit-learn

Import kmeans from scipy.cluster.vq and KMeans from sklearn.cluster. Use kmeans with data and num_clusters to get centroids_scipy. Then create a KMeans object with n_clusters=num_clusters, fit it to data, and get centroids_sklearn from its cluster_centers_ attribute.

SciPy

data = [[1.0, 2.0], [1.5, 1.8], [5.0, 8.0], [8.0, 8.0], [1.0, 0.6], [9.0, 11.0]]
num_clusters = 2
# Import kmeans and KMeans
# Run kmeans from scipy to get centroids_scipy
# Run KMeans from sklearn to get centroids_sklearn
# Your code here

Hint

Use the exact variable names and imports as shown. Remember to unpack the result of kmeans into centroids_scipy and a second value you can ignore.

4

Print the cluster centers

Print the string "SciPy centroids:" followed by centroids_scipy. Then print the string "scikit-learn centroids:" followed by centroids_sklearn.

SciPy

from scipy.cluster.vq import kmeans
from sklearn.cluster import KMeans


data = [[1.0, 2.0], [1.5, 1.8], [5.0, 8.0], [8.0, 8.0], [1.0, 0.6], [9.0, 11.0]]
num_clusters = 2

centroids_scipy, _ = kmeans(data, num_clusters)

kmeans_model = KMeans(n_clusters=num_clusters, random_state=0)
kmeans_model.fit(data)
centroids_sklearn = kmeans_model.cluster_centers_

# Print the centroids from scipy and sklearn
# Your code here

Hint

Use two print statements exactly as described to show the centroids.

Practice

1. What is the main difference between K-means clustering in scipy and scikit-learn?

easy

A. scikit-learn does not support K-means clustering.

B. scikit-learn requires manual centroid initialization, but scipy does not.

C. scipy automatically plots clusters, but scikit-learn does not.

D. scipy requires separate steps for centroid calculation and label assignment, while scikit-learn combines them.

Solution

1. Step 1: Understand K-means steps in scipy

   In scipy, you first find centroids using kmeans, then assign labels with vq.

2. Step 2: Understand K-means in scikit-learn

   scikit-learn combines these steps in one KMeans class that fits and predicts labels together.

3. Final Answer:

   scipy requires separate steps for centroid calculation and label assignment, while scikit-learn combines them. -> Option D

4. Quick Check:

   K-means steps differ: separate in scipy, combined in scikit-learn [OK]

Hint: Remember: scipy splits steps, scikit-learn combines [OK]

Common Mistakes:

• Thinking scikit-learn lacks K-means
• Assuming scipy auto-assigns labels
• Confusing plotting features with clustering steps

2. Which of the following is the correct way to import K-means functions from scipy for clustering?

easy

A. import scipy.kmeans as km

B. from scipy.kmeans import cluster

C. from scipy.cluster.vq import kmeans, vq

D. from sklearn.cluster import kmeans

Solution

1. Step 1: Recall scipy K-means import syntax

   The correct import for K-means in scipy is from scipy.cluster.vq importing kmeans and vq.

2. Step 2: Check other options

   Options A and B use incorrect module names, and D is from scikit-learn, not scipy.

3. Final Answer:

   from scipy.cluster.vq import kmeans, vq -> Option C

4. Quick Check:

   Correct scipy import = from scipy.cluster.vq import kmeans, vq [OK]

Hint: Use scipy.cluster.vq for K-means imports [OK]

Common Mistakes:

• Confusing sklearn imports with scipy
• Using wrong module names like scipy.kmeans
• Trying to import cluster from scipy directly

3. Given the code below, what will be the output of labels?

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

data = np.array([[1, 2], [1, 4], [1, 0], [10, 2], [10, 4], [10, 0]])
centroids, _ = kmeans(data, np.array([[1, 2], [10, 2]]))
labels, _ = vq(data, centroids)
print(labels.tolist())

medium

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

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

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

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

Solution

1. Step 1: Understand data and centroids

   Data has two groups: points near (1, y) and points near (10, y). Kmeans with 2 clusters finds centroids near these groups.

2. Step 2: Assign labels with vq

   Points near (1, y) get label 0, points near (10, y) get label 1. So first three points labeled 0, last three labeled 1.

3. Final Answer:

   [0, 0, 0, 1, 1, 1] -> Option A

4. Quick Check:

   Clusters split by x-coordinate: left=0, right=1 [OK]

Hint: Group points by centroid proximity for labels [OK]

Common Mistakes:

• Assuming labels are reversed
• Mixing up label order
• Expecting labels to be random

4. What is wrong with this code snippet using scipy for K-means clustering?

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

data = np.array([[1, 2], [3, 4], [5, 6]])
centroids, labels = kmeans(data, 2)
print(labels)

medium

A. kmeans returns centroids and distortion, not labels.

B. Data array shape is invalid for kmeans.

C. kmeans requires 3 clusters, not 2.

D. Missing import for vq function.

Solution

1. Step 1: Check kmeans return values

   kmeans returns centroids and distortion value, not labels.

2. Step 2: Identify correct label assignment

   Labels must be assigned using vq with data and centroids after kmeans.

3. Final Answer:

   kmeans returns centroids and distortion, not labels. -> Option A

4. Quick Check:

   kmeans output ≠ labels; use vq for labels [OK]

Hint: Remember: kmeans returns centroids, not labels [OK]

Common Mistakes:

• Expecting kmeans to return labels
• Not using vq to assign labels
• Confusing distortion with labels

5. You want to cluster a dataset using K-means and compare results between scipy and scikit-learn. Which approach correctly ensures comparable cluster labels?

hard

A. Run scipy's kmeans only, then run scikit-learn's KMeans without setting random_state, compare labels directly.

B. Run scipy's kmeans and vq, then run scikit-learn's KMeans with same n_clusters and random_state, compare labels directly.

C. Run scikit-learn's KMeans only, then assign labels manually using scipy's vq with random centroids.

D. Run scipy's kmeans and assign labels randomly, then run scikit-learn's KMeans with default settings.

Solution

1. Step 1: Understand label consistency

   To compare cluster labels, both methods must use the same number of clusters and fixed random seed for reproducibility.

2. Step 2: Apply correct procedure

   Use scipy's kmeans and vq with fixed initialization, and scikit-learn's KMeans with same n_clusters and random_state. Then compare labels.

3. Final Answer:

   Run scipy's kmeans and vq, then run scikit-learn's KMeans with same n_clusters and random_state, compare labels directly. -> Option B

4. Quick Check:

   Matching clusters need same params and fixed seed [OK]

Hint: Fix random_state and n_clusters to compare labels [OK]

Common Mistakes:

• Not fixing random_state causing label mismatch
• Assigning labels randomly in scipy
• Comparing labels without same cluster count