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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the KMeans class from scikit-learn.

SciPy

from sklearn.cluster import [1]

Drag options to blanks, or click blank then click option'

AKmean

BKMeans

Ckmeans

DCluster

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'kmeans' which is not recognized.

Misspelling the class name.

The correct class name to import for K-means clustering in scikit-learn is KMeans with capital K and M.

2fill in blank

medium

Complete the code to run K-means clustering with 3 clusters using scikit-learn.

SciPy

model = KMeans(n_clusters=[1])
model.fit(data)

Drag options to blanks, or click blank then click option'

A3

B0

C1

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Setting n_clusters to 0 or 1 which is invalid or trivial.

Using a number different from the intended cluster count.

To cluster data into 3 groups, set n_clusters=3.

3fill in blank

hard

Fix the error in the code to compute K-means using scipy's kmeans function.

SciPy

from scipy.cluster.vq import kmeans
centroids, distortion = kmeans(data, [1])

Drag options to blanks, or click blank then click option'

A[3]

B'3'

C(3,)

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the number as a string like '3'.

Passing the number inside a list or tuple.

The kmeans function expects the number of clusters as an integer, not a string or list.

4fill in blank

hard

Fill both blanks to create a dictionary of cluster labels and their counts using scikit-learn.

SciPy

labels = model.[1]
counts = {label: list(labels).[2](label) for label in set(labels)}

Drag options to blanks, or click blank then click option'

Alabels_

Bcount

Dlabel_

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'labels' instead of 'labels_' attribute.

Using 'count' as a variable instead of a method.

In scikit-learn, cluster labels are accessed with labels_. To count occurrences, use the list method count.

5fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length if length is greater than 3.

SciPy

result = {word:: len(word) for word in words if len(word) [1] 3 and word [2] 'data'}

Drag options to blanks, or click blank then click option'

A:

B>

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of ':' in dictionary comprehension.

Using '==' instead of '!=' to exclude 'data'.

The dictionary comprehension syntax uses ':' between key and value. We check if length is greater than 3 and word is not equal to 'data'.

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