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【Python机器学习】实验08 K-means无监督聚类 2

2023-10-12 94
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简介： 【Python机器学习】实验08 K-means无监督聚类

8 使用“肘部法则”选取k值

def selecte_K(X,iter_num):
    dist_arry=[]
    for k in range(1,10):
        centroids,idx=k_means(data.values,k,iter_num)
        dist_arry.append((k,metric_square(X,idx,centroids,k)))
    best_k=sorted(dist_arry,key=lambda item:item[1])[0][0]
    return dist_arry,best_k

dist_arry,best_k=selecte_K(data.values,20)

[294]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
……
[8. 2. 0. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8.
 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8.
 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8.
 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 2. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8. 8.
 8. 8. 8. 8. 3. 7. 1. 6. 1. 3. 3. 1. 6. 1. 3. 6. 6. 1. 7. 3. 7. 4. 7. 4.
 7. 1. 7. 1. 1. 3. 3. 7. 4. 3. 7. 7. 4. 3. 3. 4. 4. 4. 1. 7. 1. 7. 7. 7.
 7. 1. 7. 1. 3. 6. 3. 3. 4. 3. 6. 4. 4. 3. 7. 3. 7. 1. 4. 1. 7. 4. 4. 4.
 6. 1. 7. 1. 4. 1. 7. 4. 1. 3. 6. 6. 3. 4. 4. 7. 3. 1. 7. 1. 3. 7. 3. 7.
 4. 7. 4. 4. 1. 4. 6. 4. 0. 0. 5. 0. 0. 0. 5. 2. 2. 2. 5. 5. 0. 5. 0. 2.
 0. 5. 5. 5. 0. 0. 0. 0. 2. 2. 2. 2. 2. 0. 2. 0. 5. 0. 0. 0. 0. 2. 0. 2.
 0. 2. 2. 5. 5. 5. 2. 2. 0. 0. 0. 5. 0. 5. 0. 0. 2. 0. 5. 5. 0. 2. 5. 0.
 5. 7. 5. 2. 2. 0. 0. 2. 5. 2. 0. 0. 0. 2. 0. 2. 5. 0. 5. 5. 0. 0. 0. 5.
 2. 0. 0. 0. 0. 2. 0. 5. 5. 2. 0. 8.]
[[6.03540601 2.90876959]
 [4.17384399 0.81938216]
 [5.05532292 3.11209335]
 [3.06175515 0.75169828]
 [2.30069511 0.99667369]
 [7.04212766 3.02764051]
 [1.44137276 0.97206476]
 [3.35444384 1.35441459]
 [1.95399466 5.02557006]]

9 画张图来可视化选择K

dist_arry
[(1, 1957.654720625167),
 (2, 913.319271474709),
 (3, 266.65851965491936),
 (4, 224.19062208578683),
 (5, 174.7886069689041),
 (6, 157.8269861031051),
 (7, 108.35021878232398),
 (8, 95.72271918958536),
 (9, 139.32120386214484)]

y=[item[1] for item in dist_arry]
plt.plot(range(1,len(dist_arry)+1),y)
[<matplotlib.lines.Line2D at 0x1d875d6e1c0>]

10 对任意样本来预测其所属的聚类

### 10 对任意样本来预测其所属的聚类
def predict_cluster(x,centroids):
    lst_dist=[]
    for i in range(centroids.shape[0]):
        dist=np.sum(np.power(x-centroids[i],2))
        lst_dist.append((i,dist))
#     print(lst_dist)
    return sorted(lst_dist,key=lambda item:item[1])[0][0],lst_dist

centroids.shape[0]

3
x=np.array([[1.0,2.0],[1.5,2.5]])
for item in x:
    print(predict_cluster(item,centroids))

(2, [(0, 26.33885755045021), (1, 10.064180005080523), (2, 5.146008608810406)])
(2, [(0, 20.80466508259735), (1, 6.584615280794261), (2, 4.586927008121937)])

试试Sklearn

#导入包
from sklearn.cluster import KMeans
#实例化一个聚类对象
estimator = KMeans(n_clusters=3)
#用训练数据来拟合该模型
estimator.fit(data.values)
#对任意一个点进行预测
print(estimator.labels_)
estimator.predict(np.array(x))

[0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
 2 2 2 0]
array([1, 1])

#查看每个聚类元素到中心点的距离和
estimator.inertia_

266.6585196549193
estimator.cluster_centers_

array([[1.95399466, 5.02557006],
       [3.04367119, 1.01541041],
       [6.03366736, 3.00052511]])

from sklearn.cluster import KMeans
# '利用SSE选择k'
SSE = []  # 存放每次结果的误差平方和
for k in range(1, 11):
    estimator = KMeans(n_clusters=k)  # 构造聚类器
    estimator.fit(data.values)
    SSE.append(estimator.inertia_)
X = range(1, 11)
plt.figure(figsize=(8, 6))
plt.xlabel('k')
plt.ylabel('SSE')
plt.plot(X, SSE, 'o-')
plt.show()

d:\Users\DELL\anaconda3\lib\site-packages\sklearn\cluster\_kmeans.py:1036: UserWarning: KMeans is known to have a memory leak on Windows with MKL, when there are less chunks than available threads. You can avoid it by setting the environment variable OMP_NUM_THREADS=2.
  warnings.warn(

实验：K-means实现无监督聚类

重新利用鸢尾花数据集来实现无监督的聚类，只读取鸢尾花数据集的前两列数据

from sklearn.datasets import load_iris
iris=load_iris()
data=iris.data[:,:2]
data

array([[5.1, 3.5],
       [4.9, 3. ],
       [4.7, 3.2],
       [4.6, 3.1],
       [5. , 3.6],
       [5.4, 3.9],
       [4.6, 3.4],
       [5. , 3.4],
       [4.4, 2.9],
       [4.9, 3.1],
       [5.4, 3.7],
       [4.8, 3.4],
       [4.8, 3. ],
       [4.3, 3. ],
       [5.8, 4. ],
       [5.7, 4.4],
       [5.4, 3.9],
       [5.1, 3.5],
       [5.7, 3.8],
       [5.1, 3.8],
       [5.4, 3.4],
       [5.1, 3.7],
       [4.6, 3.6],
       [5.1, 3.3],
       [4.8, 3.4],
       [5. , 3. ],
       [5. , 3.4],
       [5.2, 3.5],
       [5.2, 3.4],
       [4.7, 3.2],
       [4.8, 3.1],
       [5.4, 3.4],
       [5.2, 4.1],
       [5.5, 4.2],
       [4.9, 3.1],
       [5. , 3.2],
       [5.5, 3.5],
       [4.9, 3.6],
       [4.4, 3. ],
       [5.1, 3.4],
       [5. , 3.5],
       [4.5, 2.3],
       [4.4, 3.2],
       [5. , 3.5],
       [5.1, 3.8],
       [4.8, 3. ],
       [5.1, 3.8],
       [4.6, 3.2],
       [5.3, 3.7],
       [5. , 3.3],
       [7. , 3.2],
       [6.4, 3.2],
       [6.9, 3.1],
       [5.5, 2.3],
       [6.5, 2.8],
       [5.7, 2.8],
       [6.3, 3.3],
       [4.9, 2.4],
       [6.6, 2.9],
       [5.2, 2.7],
       [5. , 2. ],
       [5.9, 3. ],
       [6. , 2.2],
       [6.1, 2.9],
       [5.6, 2.9],
       [6.7, 3.1],
       [5.6, 3. ],
       [5.8, 2.7],
       [6.2, 2.2],
       [5.6, 2.5],
       [5.9, 3.2],
       [6.1, 2.8],
       [6.3, 2.5],
       [6.1, 2.8],
       [6.4, 2.9],
       [6.6, 3. ],
       [6.8, 2.8],
       [6.7, 3. ],
       [6. , 2.9],
       [5.7, 2.6],
       [5.5, 2.4],
       [5.5, 2.4],
       [5.8, 2.7],
       [6. , 2.7],
       [5.4, 3. ],
       [6. , 3.4],
       [6.7, 3.1],
       [6.3, 2.3],
       [5.6, 3. ],
       [5.5, 2.5],
       [5.5, 2.6],
       [6.1, 3. ],
       [5.8, 2.6],
       [5. , 2.3],
       [5.6, 2.7],
       [5.7, 3. ],
       [5.7, 2.9],
       [6.2, 2.9],
       [5.1, 2.5],
       [5.7, 2.8],
       [6.3, 3.3],
       [5.8, 2.7],
       [7.1, 3. ],
       [6.3, 2.9],
       [6.5, 3. ],
       [7.6, 3. ],
       [4.9, 2.5],
       [7.3, 2.9],
       [6.7, 2.5],
       [7.2, 3.6],
       [6.5, 3.2],
       [6.4, 2.7],
       [6.8, 3. ],
       [5.7, 2.5],
       [5.8, 2.8],
       [6.4, 3.2],
       [6.5, 3. ],
       [7.7, 3.8],
       [7.7, 2.6],
       [6. , 2.2],
       [6.9, 3.2],
       [5.6, 2.8],
       [7.7, 2.8],
       [6.3, 2.7],
       [6.7, 3.3],
       [7.2, 3.2],
       [6.2, 2.8],
       [6.1, 3. ],
       [6.4, 2.8],
       [7.2, 3. ],
       [7.4, 2.8],
       [7.9, 3.8],
       [6.4, 2.8],
       [6.3, 2.8],
       [6.1, 2.6],
       [7.7, 3. ],
       [6.3, 3.4],
       [6.4, 3.1],
       [6. , 3. ],
       [6.9, 3.1],
       [6.7, 3.1],
       [6.9, 3.1],
       [5.8, 2.7],
       [6.8, 3.2],
       [6.7, 3.3],
       [6.7, 3. ],
       [6.3, 2.5],
       [6.5, 3. ],
       [6.2, 3.4],
       [5.9, 3. ]])

iris.target

array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2])

plt.scatter(data[:,0],data[:,1],color="blue")
plt.show()

df51b623fa924a9d847079d330967d83.png

#真实的类别长成这样
plt.scatter(data[:,0],data[:,1],c=iris.target)
plt.show()

1fd045f957a64513a7f206a9f8970abc.png

1 定义和调用更新每个样本所属聚类，聚类中心更新，初始化聚类中心的参数

def update_centorids(X,idx,k):
    centorids=np.zeros((k,X.shape[1]))
    for i in range(k):
        index=np.where(idx==i)[0]
        centorids[i]=np.sum(X[index,:],axis=0)/len(index)
    return centorids
def initialize_centroid(X,k):
    np.random.seed(30)
    index_initial=[np.random.randint(1,X.shape[0]) for i in range(k)]
    centorids=np.zeros((k,X.shape[1]))
    print(index_initial)
    for i,j in enumerate(index_initial):
        centorids[i]=X[j,:]
    return centorids

2 定义Kmeans算法获得最终的聚类中心和样本所属聚类索引

def k_means(X,k,iter_num):
    centroids=np.zeros((k,X.shape[1]))
    #初始化聚类中心
    centroids=initialize_centroid(X,k)
    for i in range(iter_num):
        #每个样本找到所属聚类
        idx=find_closest(X,centroids)
        print(idx)
        #更新新的聚类中心
        centroids=update_centorids(X,idx,k)
        print(centroids)
    return centroids,idx

centroids,idx=k_means(data,3,30)

[38, 46, 141]
[1. 0. 0. 0. 1. 1. 0. 1. 0. 0. 1. 1. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 0. 0. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0.
 1. 1. 2. 2. 2. 0. 2. 1. 2. 0. 2. 0. 0. 2. 2. 2. 1. 2. 1. 2. 2. 0. 1. 2.
 2. 2. 2. 2. 2. 2. 2. 2. 0. 0. 2. 2. 1. 2. 2. 2. 1. 0. 0. 2. 2. 0. 1. 1.
 1. 2. 0. 1. 2. 2. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2.
 2. 1. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2.
 2. 2. 2. 2. 2. 2.]
[[4.9137931  2.78965517]
 [5.29772727 3.44772727]
 [6.50519481 2.93506494]]
[1. 0. 0. 0. 1. 1. 0. 1. 0. 0. 1. 1. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 0. 0. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0.
 1. 1. 2. 2. 2. 0. 2. 1. 2. 0. 2. 0. 0. 2. 2. 2. 1. 2. 1. 2. 2. 0. 1. 2.
 2. 2. 2. 2. 2. 2. 2. 0. 0. 0. 2. 2. 1. 2. 2. 2. 1. 0. 0. 2. 2. 0. 0. 1.
 1. 2. 0. 1. 2. 2. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2.
 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2.
 2. 2. 2. 2. 2. 2.]
[[5.0030303  2.77272727]
 [5.28333333 3.48095238]
 [6.52666667 2.94533333]]
[1. 0. 0. 0. 1. 1. 1. 1. 0. 0. 1. 1. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 0. 0. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0.
 1. 1. 2. 2. 2. 0. 2. 0. 2. 0. 2. 0. 0. 2. 2. 2. 0. 2. 1. 2. 2. 0. 2. 2.
 2. 2. 2. 2. 2. 2. 2. 0. 0. 0. 2. 2. 0. 2. 2. 2. 1. 0. 0. 2. 2. 0. 0. 1.
 0. 2. 0. 0. 2. 2. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2.
 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2.
 2. 2. 2. 2. 2. 2.]
[[5.0972973  2.77027027]
 [5.2027027  3.56756757]
 [6.51842105 2.94868421]]
[1. 0. 0. 0. 1. 1. 1. 1. 0. 0. 1. 1. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 0. 0. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 0. 0. 1. 1. 0. 1. 0.
 1. 1. 2. 2. 2. 0. 2. 0. 2. 0. 2. 0. 0. 2. 2. 2. 0. 2. 0. 0. 2. 0. 2. 2.
 2. 2. 2. 2. 2. 2. 2. 0. 0. 0. 0. 2. 0. 2. 2. 2. 0. 0. 0. 2. 0. 0. 0. 0.
 0. 2. 0. 0. 2. 0. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 2. 2.
 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 0. 2.
 2. 2. 2. 2. 2. 2.]
[[5.22391304 2.77608696]
 [5.16470588 3.61764706]
 [6.58       2.97      ]]
[1. 0. 1. 0. 1. 1. 1. 1. 0. 0. 1. 1. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 0. 1. 1. 1. 0. 1. 1.
 1. 1. 2. 2. 2. 0. 2. 0. 2. 0. 2. 0. 0. 2. 2. 2. 0. 2. 0. 0. 2. 0. 2. 2.
 2. 2. 2. 2. 2. 2. 2. 0. 0. 0. 0. 2. 0. 2. 2. 2. 0. 0. 0. 2. 0. 0. 0. 0.
 0. 2. 0. 0. 2. 0. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 2. 2.
 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 0. 2.
 2. 2. 2. 2. 2. 2.]
[[5.28333333 2.73571429]
 [5.10526316 3.57368421]
 [6.58       2.97      ]]
[1. 0. 1. 1. 1. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 0. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 0. 1. 1. 1. 0. 1. 1.
 1. 1. 2. 2. 2. 0. 2. 0. 2. 0. 2. 0. 0. 0. 0. 2. 0. 2. 0. 0. 2. 0. 2. 2.
 2. 2. 2. 2. 2. 2. 2. 0. 0. 0. 0. 2. 0. 2. 2. 2. 0. 0. 0. 2. 0. 0. 0. 0.
 0. 2. 0. 0. 2. 0. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 2. 0.
 2. 0. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 2. 0. 2.
 2. 2. 2. 2. 2. 0.]
[[5.445      2.6725    ]
 [5.04318182 3.50454545]
 [6.61818182 2.99242424]]
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 2. 0. 2. 0. 2. 2. 0. 0. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 0. 2. 2. 2. 0. 2.
 2. 2. 0. 2. 2. 0.]
[[5.77358491 2.69245283]
 [5.006      3.428     ]
 [6.81276596 3.07446809]]
[1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1. 1.
 1. 1. 2. 2. 2. 0. 2. 0. 2. 0. 2. 0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 0. 0. 0.
 0. 0. 2. 2. 2. 2. 0. 0. 0. 0. 0. 0. 0. 0. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 2. 0. 2. 2. 2. 2. 0. 2. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 2. 0.
 2. 0. 2. 0. 2. 2. 0. 0. 2. 2. 2. 2. 2. 0. 0. 2. 2. 2. 0. 2. 2. 2. 0. 2.
 2. 2. 0. 2. 2. 0.]
[[5.77358491 2.69245283]
 [5.006      3.428     ]
 [6.81276596 3.07446809]]

3 绘制各个聚类的图

#画图
import matplotlib.pyplot as plt
cluster1=data[np.where(idx==0)[0],:]
cluster2=data[np.where(idx==1)[0],:]
cluster3=data[np.where(idx==2)[0],:]
fig,axe=plt.subplots(figsize=(8,6))
axe.scatter(cluster1[:,0],cluster1[:,1],s=10,color="red",label="cluster1")
axe.scatter(cluster2[:,0],cluster2[:,1],s=10,color="yellow",label="cluster2")
axe.scatter(cluster3[:,0],cluster3[:,1],s=10,color="green",label="cluster2")
axe.scatter(centroids[:,0],centroids[:,1],s=30,marker="+",c="k")
plt.show()

01189883328f4b59adcb08a3b76d609c.png


4 定义评价函数–即任意一点所在聚类与聚类中心的距离平方和

定义一个评价函数

def metric_square(X,idx,centroids,k):
    lst_dist=[]
    for i in range(k):
        cluster=X[np.where(idx==i)[0],:]
        dist=np.sum(np.power(cluster-centroids[i,:],2))
        lst_dist.append(dist)
    return sum(lst_dist)

5 使用“肘部法则”选取k值

def selecte_K(X,iter_num):
    dist_arry=[]
    for k in range(1,10):
        centroids,idx=k_means(data,k,iter_num)
        dist_arry.append((k,metric_square(X,idx,centroids,k)))
    best_k=sorted(dist_arry,key=lambda item:item[1])[0][0]
    return dist_arry,best_k  
dist_arry,best_k=selecte_K(data,10)
y=[item[1] for item in dist_arry]
plt.plot(range(1,len(dist_arry)+1),y)

[38]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
 0. 0. 0. 0. 0. 0.]
 ……   
[1. 0. 0. 0. 1. 1. 0. 1. 0. 0. 1. 0. 0. 0. 1. 1. 1. 1. 1. 1. 1. 1. 0. 1.
 0. 0. 1. 1. 1. 0. 0. 1. 1. 1. 0. 0. 1. 1. 0. 1. 1. 7. 0. 1. 1. 0. 1. 0.
 1. 0. 8. 6. 8. 5. 4. 5. 6. 7. 4. 7. 7. 6. 5. 6. 5. 8. 5. 5. 4. 5. 6. 4.
 4. 4. 4. 8. 8. 8. 6. 5. 5. 5. 5. 5. 5. 6. 8. 4. 5. 5. 5. 6. 5. 7. 5. 5.
 5. 4. 7. 5. 6. 5. 8. 4. 4. 2. 7. 2. 4. 8. 8. 4. 8. 5. 5. 6. 4. 3. 2. 5.
 8. 5. 2. 4. 8. 8. 4. 6. 4. 2. 2. 3. 4. 4. 4. 2. 6. 6. 6. 8. 8. 8. 5. 8.
 8. 8. 4. 4. 6. 6.]
[[4.71428571 3.16190476]
 [5.24285714 3.66785714]
 [7.51428571 2.87142857]
 [7.8        3.8       ]
 [6.34545455 2.73636364]
 [5.68518519 2.66666667]
 [6.14375    3.14375   ]
 [4.94285714 2.38571429]
 [6.825      3.13      ]]

6 对任意样本来预测其所属的聚类

def predict_cluster(x,centroids):
    lst_dist=[]
    for i in range(centroids.shape[0]):
        dist=np.sum(np.power(x-centroids[i],2))
        lst_dist.append((i,dist))
    return sorted(lst_dist,key=lambda item:item[1])[0][0],lst_dist

for item in np.array([[1.0,2.0],[2.4,1.6]]):
    print(predict_cluster(item,centroids))

(1, [(0, 23.266603773584908), (1, 18.08722), (2, 34.94272974196466)])
(1, [(0, 12.574528301886792), (1, 10.132819999999999), (2, 21.646559529198715)])

7 试试SKLERAN

#导入包
from sklearn.cluster import KMeans
estimator = KMeans(n_clusters=3)
estimator.fit(data)
print(estimator.labels_)
print(estimator.predict(np.array([[1.0,2.0],[2.4,1.6]])))
estimator.inertia_
estimator.cluster_centers_

[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 0 2 0 2 0 2 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0
 2 2 2 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 2 2 2 2 0 2 2 2 2
 2 2 0 0 2 2 2 2 0 2 0 2 0 2 2 0 0 2 2 2 2 2 0 0 2 2 2 0 2 2 2 0 2 2 2 0 2
 2 0]
[1 1]
array([[5.77358491, 2.69245283],
   [5.006     , 3.428     ],
   [6.81276596, 3.07446809]])


文章标签：
数据挖掘
Python
机器学习/深度学习
数据可视化
算法
关键词：
人工智能平台 PAI python
python人工智能平台 PAI
Python机器学习
Python机器学习k-means聚类
人工智能平台 PAI k-means
Want595
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