K-means via scipy vs scikit-learn - Visual Side-by-Side Comparison

Concept Flow - K-means via scipy vs scikit-learn

Start with data points

↓

Choose k clusters

↓

Initialize centroids

↓

Assign points to nearest centroid

↓

Update centroids by averaging assigned points

↓

Check convergence

↓

Stop

K-means groups data into k clusters by repeating assignment of points to centroids and updating centroids until stable.

Execution Sample

SciPy

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

# Sample data
X = np.array([[1,2],[1,4],[1,0],[10,2],[10,4],[10,0]])

# Scipy kmeans
centroids, distortion = kmeans(X, 2)
labels_scipy, _ = vq(X, centroids)

# Sklearn kmeans
kmeans_skl = KMeans(n_clusters=2, random_state=0).fit(X)
labels_skl = kmeans_skl.labels_

This code runs K-means clustering on the same data using scipy and scikit-learn, then gets cluster labels.

Execution Table

Step	Action	Scipy centroids	Scipy labels	Sklearn centroids	Sklearn labels
1	Initialize centroids (random)	[[5.5 3. ] [ 1. 2. ]]	N/A	N/A	N/A
2	Assign points to nearest centroid	[[5.5 3. ] [ 1. 2. ]]	[1 1 1 0 0 0]	N/A	N/A
3	Update centroids by averaging assigned points	[[10. 2. ] [ 1. 2. ]]	N/A	N/A	N/A
4	Assign points to nearest centroid	[[10. 2. ] [ 1. 2. ]]	[1 1 1 0 0 0]	N/A	N/A
5	Converged (centroids stable)	[[10. 2. ] [ 1. 2. ]]	[1 1 1 0 0 0]	N/A	N/A
6	Sklearn fit completes	N/A	N/A	[[10. 2. ] [ 1. 2. ]]	[1 1 1 0 0 0]
7	Output labels	Final centroids	[1 1 1 0 0 0]	Final centroids	[1 1 1 0 0 0]

💡 Both methods converge to similar centroids and assign points to clusters accordingly.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	Final
centroids_scipy	None	[[5.5 3. ] [ 1. 2. ]]	[[10. 2. ] [ 1. 2. ]]	[[10. 2. ] [ 1. 2. ]]	[[10. 2. ] [ 1. 2. ]]
labels_scipy	None	N/A	N/A	[1 1 1 0 0 0]	[1 1 1 0 0 0]
centroids_skl	None	N/A	N/A	N/A	[[10. 2. ] [ 1. 2. ]]
labels_skl	None	N/A	N/A	N/A	[1 1 1 0 0 0]

Key Moments - 3 Insights

Why does scipy separate centroid calculation and label assignment into two steps?

Scipy's kmeans function returns centroids only; label assignment is done separately with vq. See execution_table rows 2 and 4 where labels are assigned after centroids update.

Why do both methods produce similar but not identical centroids?

Both use random initialization but sklearn fixes random_state for reproducibility. Scipy's initial centroids may differ, causing slight differences. See variable_tracker centroids_scipy vs centroids_skl.

Why is sklearn's clustering done in one fit call while scipy requires two functions?

Sklearn's KMeans class combines centroid calculation and label assignment internally in fit(), simplifying usage. Scipy splits these for flexibility. See execution_table steps 6 and 7.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what are the cluster labels assigned by scipy?

A [1 0 1 0 1 0]

B [0 0 0 1 1 1]

C [1 1 1 0 0 0]

D [0 1 0 1 0 1]

💡 Hint

Check the 'Scipy labels' column at step 2 in execution_table.

At which step do scipy centroids stop changing?

A Step 5

B Step 3

C Step 1

D Step 7

💡 Hint

Look at 'Scipy centroids' column in execution_table and see when values repeat.

If we remove random_state in sklearn, what likely changes in the execution_table?

A Scipy centroids will change instead

B Sklearn centroids and labels may differ each run

C Labels remain the same for both methods

D Execution will stop earlier

💡 Hint

Random_state controls reproducibility in sklearn; see variable_tracker centroids_skl.

Concept Snapshot

K-means groups data into k clusters by:
- Initializing centroids
- Assigning points to nearest centroid
- Updating centroids by averaging
- Repeating until centroids stabilize
Scipy uses kmeans() + vq() separately.
Sklearn uses KMeans.fit() combining both.

Full Transcript

This visual execution compares K-means clustering using scipy and scikit-learn on the same data. The process starts with data points and choosing k clusters. Scipy's kmeans function calculates centroids, then vq assigns labels. Sklearn's KMeans.fit does both in one step. The execution table shows how centroids and labels update step-by-step until convergence. Variable tracking shows centroid values and labels changing over steps. Key moments clarify why scipy separates steps, why centroids differ slightly, and how sklearn simplifies usage. The quiz tests understanding of labels, convergence step, and effect of random_state. The snapshot summarizes the K-means iterative process and differences between scipy and sklearn usage.

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