Python实现使用遗传算法确定KMeans聚类簇的个数
时间: 2023-07-11 09:11:53 浏览: 127
kmeans聚类算法python实现
下面是一个Python实现使用遗传算法确定KMeans聚类簇的个数的示例代码:
```python
import numpy as np
from sklearn.cluster import KMeans
from sklearn.metrics import silhouette_score
import random
# 定义适应度函数
def fitness(X, k):
kmeans = KMeans(n_clusters=k).fit(X)
labels = kmeans.labels_
score = silhouette_score(X, labels)
return score
# 遗传算法参数设置
pop_size = 50 # 种群大小
chrom_length = 10 # 基因长度
pc = 0.8 # 交叉概率
pm = 0.1 # 变异概率
max_iter = 100 # 最大迭代次数
# 初始化种群
pop = np.random.randint(0, 2, size=(pop_size, chrom_length))
# 迭代过程
for i in range(max_iter):
# 计算适应度
fitness_values = [fitness(X, int(''.join(map(str, chrom)), 2)) for chrom in pop]
# 选择操作
fitness_sum = sum(fitness_values)
prob = [fitness_values[j] / fitness_sum for j in range(pop_size)]
cum_prob = np.cumsum(prob)
new_pop = []
for j in range(pop_size):
r = random.random()
for k in range(pop_size):
if r <= cum_prob[k]:
new_pop.append(pop[k])
break
# 交叉操作
for j in range(0, pop_size, 2):
if random.random() < pc:
pos = random.randint(1, chrom_length - 1)
new_pop[j][pos:], new_pop[j+1][pos:] = new_pop[j+1][pos:], new_pop[j][pos:]
# 变异操作
for j in range(pop_size):
if random.random() < pm:
pos = random.randint(0, chrom_length - 1)
new_pop[j][pos] = 1 - new_pop[j][pos]
# 更新种群
pop = new_pop
# 选择最优个体
best_chrom = max(pop, key=lambda x: fitness(X, int(''.join(map(str, x)), 2)))
k = int(''.join(map(str, best_chrom)), 2)
# KMeans聚类
kmeans = KMeans(n_clusters=k).fit(X)
labels = kmeans.labels_
```
在上述代码中,`X`是样本数据,`pop_size`、`chrom_length`、`pc`、`pm`和`max_iter`分别表示种群大小、基因长度、交叉概率、变异概率和最大迭代次数。在迭代过程中,首先计算种群中每个个体的适应度(即聚类结果的质量),然后进行选择、交叉和变异操作,最终选择适应度最高的个体作为最优解。最优解的基因表示聚类簇的个数。最后,使用最优解进行KMeans聚类,并将聚类结果存储在`labels`中。
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