聚类
在本练习中，我们将实现K-means聚类
K-means 聚类
我们将实施和应用K-means到一个简单的二维数据集，以获得一些直观的工作原理。 K-means是一个迭代的，无监督的聚类算法，将类似的实例组合成簇。 该算法通过猜测每个簇的初始聚类中心开始，然后重复将实例分配给最近的簇，并重新计算该簇的聚类中心。 我们要实现的第一部分是找到数据中每个实例最接近的聚类中心的函数。
1 准备数据
无监督学习中，数据是不带任何标签的
import numpy as np import pandas as pd import matplotlib.pyplot as plt
import pandas as pd import numpy as np data=pd.read_csv("data/ex7data2.csv") data.head()
| X1 | X2 | |
| 0 | 1.842080 | 4.607572 |
| 1 | 5.658583 | 4.799964 |
| 2 | 6.352579 | 3.290854 |
| 3 | 2.904017 | 4.612204 |
| 4 | 3.231979 | 4.939894 |
import seaborn as sb plt.figure(figsize=(4,6)) sb.lmplot(x="X1",y="X2",data=data,fit_reg=False) plt.show()
<Figure size 288x432 with 0 Axes>
1 初始化聚类中心
2 所有样本点聚类（计算每个样本点与聚类中心的距离，选择最小距离的聚类中心所在的聚类作为该样本所属聚类）
3 重新计算聚类中心（每个聚类中所有样本的均值）
4 迭代执行2-3直到聚类中心不再变化（iter_num）
2 给定聚类中心，计算每个点属于哪个聚类，定义函数实现
#给定聚类中心，如何求每个样本所属的聚类 def find_closest(X,centroids): #样本数量 m=X.shape[0] idx=np.zeros(m) k=centroids.shape[0] #遍历所有样本 for i in range(m): distance=10000 #遍历所有聚类中心 for j in range(k): #计算样本与聚类中心的距离 dist=np.sum(np.power(X[i,:]-centroids[j,:],2)) if dist<distance: distance=dist idx[i]=j return idx
#测试一下 centorids=np.arange(1,5).reshape(2,2) centorids
array([[1, 2], [3, 4]])
idx=find_closest(data.values,centorids)
idx
array([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., 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., 0., 0., 1., 0., 0., 0., 0., 0., 0., 1., 0., 0., 0., 1., 0., 0., 1., 0., 0., 0., 0., 0., 0., 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1., 0., 1., 0., 0., 0., 0., 0., 0., 0., 1., 0., 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 0., 0., 0., 0., 0., 0., 0., 0., 1., 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.])
3 根据已有的数据的标记，来重新更新聚类中心，定义相应的函数
index=np.where(idx==0)[0] index
array([ 56, 100, 101, 103, 104, 105, 106, 107, 108, 110, 111, 112, 114, 115, 117, 118, 119, 120, 121, 122, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 140, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 162, 164, 165, 166, 167, 168, 169, 170, 172, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 188, 189, 190, 191, 192, 193, 194, 195, 197, 198, 199], dtype=int64)
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
#测试函数 update_centorids(data.values,idx,2)
array([[2.8195125 , 0.99467112], [4.03762952, 3.8009101 ]])
4 初始化聚类中心，定义相应的函数
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
#测试一下该函数 initialize_centroid(data.values,3)
[294, 141, 252] array([[6.48212628, 2.5508514 ], [3.7875723 , 1.45442904], [6.01017978, 2.72401338]])
data.values[294]
array([6.48212628, 2.5508514 ])
5 定义K-means算法
#设计K-means算法 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.values,3,10)
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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. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.] [[6.03366736 3.00052511] [1.95399466 5.02557006] [3.04367119 1.01541041]] [1. 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. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.] [[6.03366736 3.00052511] [1.95399466 5.02557006] [3.04367119 1.01541041]] [1. 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. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 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. 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. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.] [[6.03366736 3.00052511] [1.95399466 5.02557006] [3.04367119 1.01541041]]
centroids.shape[0]
3
6 绘制各个聚类的图
#画图 cluster1=data.values[np.where(idx==0)[0],:] cluster1 cluster2=data.values[np.where(idx==1)[0],:] cluster2 cluster3=data.values[np.where(idx==2)[0],:] cluster3
array([[3.20360621, 0.7222149 ], [3.06192918, 1.5719211 ], [4.01714917, 1.16070647], [1.40260822, 1.08726536], [4.08164951, 0.87200343], [3.15273081, 0.98155871], [3.45186351, 0.42784083], [3.85384314, 0.7920479 ], [1.57449255, 1.34811126], [4.72372078, 0.62044136], [2.87961084, 0.75413741], [0.96791348, 1.16166819], [1.53178107, 1.10054852], [4.13835915, 1.24780979], [3.16109021, 1.29422893], [2.95177039, 0.89583143], [3.27844295, 1.75043926], [2.1270185 , 0.95672042], [3.32648885, 1.28019066], [2.54371489, 0.95732716], [3.233947 , 1.08202324], [4.43152976, 0.54041 ], [3.56478625, 1.11764714], [4.25588482, 0.90643957], [4.05386581, 0.53291862], [3.08970176, 1.08814448], [2.84734459, 0.26759253], [3.63586049, 1.12160194], [1.95538864, 1.32156857], [2.88384005, 0.80454506], [3.48444387, 1.13551448], [3.49798412, 1.10046402], [2.45575934, 0.78904654], [3.2038001 , 1.02728075], [3.00677254, 0.62519128], [1.96547974, 1.2173076 ], [2.17989333, 1.30879831], [2.61207029, 0.99076856], [3.95549912, 0.83269299], [3.64846482, 1.62849697], [4.18450011, 0.45356203], [3.7875723 , 1.45442904], [3.30063655, 1.28107588], [3.02836363, 1.35635189], [3.18412176, 1.41410799], [4.16911897, 0.20581038], [3.24024211, 1.14876237], [3.91596068, 1.01225774], [2.96979716, 1.01210306], [1.12993856, 0.77085284], [2.71730799, 0.48697555], [3.1189017 , 0.69438336], [2.4051802 , 1.11778123], [2.95818429, 1.01887096], [1.65456309, 1.18631175], [2.39775807, 1.24721387], [2.28409305, 0.64865469], [2.79588724, 0.99526664], [3.41156277, 1.1596363 ], [3.50663521, 0.73878104], [3.93616029, 1.46202934], [3.90206657, 1.27778751], [2.61036396, 0.88027602], [4.37271861, 1.02914092], [3.08349136, 1.19632644], [2.1159935 , 0.7930365 ], [2.15653404, 0.40358861], [2.14491101, 1.13582399], [1.84935524, 1.02232644], [4.1590816 , 0.61720733], [2.76494499, 1.43148951], [3.90561153, 1.16575315], [2.54071672, 0.98392516], [4.27783068, 1.1801368 ], [3.31058167, 1.03124461], [2.15520661, 0.80696562], [3.71363659, 0.45813208], [3.54010186, 0.86446135], [1.60519991, 1.1098053 ], [1.75164337, 0.68853536], [3.12405123, 0.67821757], [2.37198785, 1.42789607], [2.53446019, 1.21562081], [3.6834465 , 1.22834538], [3.2670134 , 0.32056676], [3.94159139, 0.82577438], [3.2645514 , 1.3836869 ], [4.30471138, 1.10725995], [2.68499376, 0.35344943], [3.12635184, 1.2806893 ], [2.94294356, 1.02825076], [3.11876541, 1.33285459], [2.02358978, 0.44771614], [3.62202931, 1.28643763], [2.42865879, 0.86499285], [2.09517296, 1.14010491], [5.29239452, 0.36873298], [2.07291709, 1.16763851], [0.94623208, 0.24522253], [2.73911908, 1.10072284], [3.96162465, 2.72025046], [3.45928006, 2.68478445]])
centroids
array([[6.03366736, 3.00052511], [1.95399466, 5.02557006], [3.04367119, 1.01541041]])
import matplotlib.pyplot as plt fig,axe=plt.subplots(figsize=(6,9)) 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()
我们跳过的一个步骤是初始化聚类中心的过程。 这可以影响算法的收敛。 我们的任务是创建一个选择随机样本并将其用作初始聚类中心的函数。
7 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)
cluster=data.values[np.where(idx==0)[0],:]
centroids[0,:]
array([6.03366736, 3.00052511])
np.sum(np.power(cluster-centroids[0,:],2))
82.48594291556887
#测试该函数 metric_square(data.values,idx,centroids,3)
266.65851965491936
