K-means via scipy vs scikit-learn - Performance Comparison

Time Complexity: K-means via scipy vs scikit-learn

O(n x k x i)

Understanding Time Complexity

We want to understand how the time it takes to run K-means clustering grows as we use more data points or clusters.

This helps us know how fast or slow the algorithm will be when using scipy compared to scikit-learn.

Scenario Under Consideration

Analyze the time complexity of this K-means clustering code using scipy.

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

# Generate sample data
data = np.random.rand(1000, 2)

# Run K-means clustering
centroids, distortion = kmeans(data, 5, iter=20)
    

This code runs K-means on 1000 points with 5 clusters and up to 20 iterations.

Identify Repeating Operations

Look at what repeats in the algorithm:

• Primary operation: Assigning each data point to the nearest cluster center.
• How many times: This happens every iteration, up to 20 times here.
• Also, updating cluster centers after assignments repeats each iteration.

How Execution Grows With Input

As we increase data points or clusters, the work grows like this:

Input Size (n points)	Approx. Operations
10	10 points x 5 clusters x 20 iterations = 1,000
100	100 x 5 x 20 = 10,000
1000	1000 x 5 x 20 = 100,000

Pattern observation: The operations grow roughly in a straight line with the number of points and clusters multiplied by iterations.

Final Time Complexity

Time Complexity: O(n x k x i)

This means the time grows proportionally with the number of points (n), clusters (k), and iterations (i).

Common Mistake

[X] Wrong: "The time depends only on the number of data points."

[OK] Correct: The number of clusters and iterations also multiply the work, so ignoring them misses important parts of the cost.

Interview Connect

Understanding how K-means scales helps you explain algorithm choices clearly and shows you can think about performance in real projects.

Self-Check

"What if we reduce the number of iterations by half? How would the time complexity change?"

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