实现三支k-means算法
时间: 2023-10-29 16:07:16 浏览: 54
以下是三种不同的k-means算法的实现:
1. 基本k-means算法
```
# 输入数据集X和聚类数k
# 输出聚类结果clusters和聚类中心centroids
def k_means(X, k):
# 初始化聚类中心
centroids = X[:k]
# 初始化聚类结果
clusters = [[] for i in range(k)]
# 迭代聚类过程
while True:
# 将每个数据点分配到最近的聚类中心
for x in X:
distances = [(x - c) ** 2 for c in centroids]
cluster_index = distances.index(min(distances))
clusters[cluster_index].append(x)
# 更新聚类中心
new_centroids = []
for cluster in clusters:
if cluster:
new_centroids.append(sum(cluster) / len(cluster))
else:
new_centroids.append(centroids[clusters.index(cluster)])
# 判断是否收敛
if new_centroids == centroids:
break
else:
centroids = new_centroids
clusters = [[] for i in range(k)]
return clusters, centroids
```
2. 加速k-means算法(使用k-d树)
```
import numpy as np
from sklearn.neighbors import KDTree
# 输入数据集X和聚类数k
# 输出聚类结果clusters和聚类中心centroids
def k_means_kd(X, k):
# 初始化聚类中心
centroids = X[:k]
# 初始化聚类结果
clusters = [[] for i in range(k)]
# 构建k-d树
tree = KDTree(X)
# 迭代聚类过程
while True:
# 将每个数据点分配到最近的聚类中心
for x in X:
distances, indices = tree.query([x], k=1)
cluster_index = np.argmin(distances)
clusters[cluster_index].append(x)
# 更新聚类中心
new_centroids = []
for cluster in clusters:
if cluster:
new_centroids.append(sum(cluster) / len(cluster))
else:
new_centroids.append(centroids[clusters.index(cluster)])
# 判断是否收敛
if new_centroids == centroids:
break
else:
centroids = new_centroids
clusters = [[] for i in range(k)]
return clusters, centroids
```
3. mini-batch k-means算法
```
import numpy as np
# 输入数据集X、聚类数k和批次大小batch_size
# 输出聚类结果clusters和聚类中心centroids
def mini_batch_k_means(X, k, batch_size):
# 初始化聚类中心
centroids = X[:k]
# 初始化聚类结果
clusters = [[] for i in range(k)]
# 迭代聚类过程
while True:
# 随机选择一个批次
batch = np.random.choice(X, batch_size, replace=False)
# 将每个数据点分配到最近的聚类中心
for x in batch:
distances = [(x - c) ** 2 for c in centroids]
cluster_index = distances.index(min(distances))
clusters[cluster_index].append(x)
# 更新聚类中心
new_centroids = []
for cluster in clusters:
if cluster:
new_centroids.append(sum(cluster) / len(cluster))
else:
new_centroids.append(centroids[clusters.index(cluster)])
# 判断是否收敛
if new_centroids == centroids:
break
else:
centroids = new_centroids
clusters = [[] for i in range(k)]
return clusters, centroids
```