基于给定初始簇的K-Means聚类实现问询：欧氏距离与指定迭代次数

Hey there! Let's work through your K-Means clustering problem step by step. First, I spotted a couple of tiny mistakes in your code:

• The y column in your DataFrame has a typo: the point (2,2) was incorrectly entered as 23 instead of 2
• You had a partial import (sklea...) — the correct library name is sklearn

Let's start by clarifying the problem details first:

• Raw data points: (1,1), (1,2), (2,1), (2,2), (3,3), (8,8), (9,8), (8,9), (9,9)
• Initial cluster centers: (1,1) and (2,1)
• Constraints: Use Euclidean distance, run exactly 3 iterations

Full Working Code (With Iteration Tracking) #

I'll implement this both with a manual iteration approach (for full transparency) and using sklearn (for quick execution), so you can see exactly how centroids update each time:

import pandas as pd
import numpy as np
from sklearn.cluster import KMeans

# Fix the raw data typo first
data = {'x': [1,1,2,2,3,8,9,8,9], 'y': [1,2,1,2,3,8,8,9,9]}
df = pd.DataFrame(data)

# Define initial centers as a numpy array
initial_centers = np.array([[1, 1], [2, 1]])

### Option 1: Manual Iteration (Full Transparency)
print("=== Manual 3-Iteration K-Means ===")
current_centers = initial_centers.copy()

for iter_num in range(3):
    print(f"\n--- Iteration {iter_num + 1} ---")
    
    # Calculate Euclidean distance from each point to both centers
    distances = np.sqrt(((df - current_centers[:, np.newaxis])**2).sum(axis=2))
    
    # Assign each point to the closest cluster
    cluster_labels = np.argmin(distances, axis=0)
    df[f'cluster_iter_{iter_num+1}'] = cluster_labels
    
    # Show current cluster assignments
    print("Cluster Assignments:")
    print(df[['x', 'y', f'cluster_iter_{iter_num+1}']])
    
    # Update centroids (average of points in each cluster)
    new_centers = []
    for cluster in range(2):
        cluster_points = df[cluster_labels == cluster][['x', 'y']]
        new_centroid = cluster_points.mean().values if len(cluster_points) > 0 else current_centers[cluster]
        new_centers.append(new_centroid)
    
    current_centers = np.array(new_centers)
    print(f"Updated Centroids: {current_centers.round(2)}")

### Option 2: Using sklearn KMeans (Quick Implementation)
print("\n=== sklearn K-Means Result ===")
kmeans = KMeans(
    n_clusters=2,
    init=initial_centers,
    max_iter=3,
    n_init=1,  # Only use the given initial centers once
    random_state=42
)
kmeans.fit(df)

print(f"Final Centroids: {kmeans.cluster_centers_.round(2)}")
print(f"Final Cluster Labels: {kmeans.labels_}")

Step-by-Step Iteration Breakdown #

Let's walk through exactly what happens in each iteration:

Iteration 1 #

• Initial Centroids: [[1, 1], [2, 1]]
• Cluster Assignments:
  • Cluster 0: (1,1), (1,2) (closer to (1,1))
  • Cluster 1: (2,1), (2,2), (3,3), (8,8), (9,8), (8,9), (9,9) (closer to (2,1))
• Updated Centroids:
  • Cluster 0: (1.0, 1.5) (average of its two points)
  • Cluster 1: (4.71, 5.71) (average of its seven points)

Iteration 2 #

• Current Centroids: [[1.0, 1.5], [4.71, 5.71]]
• Cluster Assignments:
  • Cluster 0: (1,1), (1,2), (2,1), (2,2), (3,3) (now closer to the updated Cluster 0 centroid)
  • Cluster 1: (8,8), (9,8), (8,9), (9,9) (still closer to Cluster 1 centroid)
• Updated Centroids:
  • Cluster 0: (1.8, 1.8) (average of its five points)
  • Cluster 1: (8.5, 8.5) (average of its four points)

Iteration 3 #

• Current Centroids: [[1.8, 1.8], [8.5, 8.5]]
• Cluster Assignments: No changes from Iteration 2 — all points stay in their current clusters
• Updated Centroids: Identical to Iteration 2 (since cluster membership didn't change, the average remains the same)

Final Clustering Result #

After 3 iterations, the clusters stabilize to:

• Cluster 0: (1,1), (1,2), (2,1), (2,2), (3,3)
• Cluster 1: (8,8), (9,8), (8,9), (9,9)

内容的提问来源于stack exchange，提问作者Shivam