Question

我使用sklearn使用Kmeans进行了聚类。虽然它有一种打印质心的方法，但我觉得scikit-learn没有找到簇长度的方法（或者到目前为止我还没有看到它）。是否有一种巧妙的方法来获取每个集群的集群长度或与集群相关的许多点？我目前有这个相当繁琐的代码来做这个，我发现长度为1的簇，需要通过测量点之间的欧几里德距离来添加其他点，并且必须更新标签

import numpy as np
from clustering.clusternew import Kmeans_clu
from evolution.generate import reproduction
from mapping.somnew import mapping, no_of_neurons, neuron_weights_init
from population_creation.population import pop_create
from New_SOL import newsol


data = genfromtxt('iris.csv', delimiter=',', skip_header=0, usecols=range(0, 4)) ##Read the input data
actual_label = genfromtxt('iris.csv', delimiter=',', dtype=str,skip_header=0, usecols=(4))
chromosome = int(input("Enter the number of chromosomes: "))  #Input the population size
max_gen = int(input("Enter the maximum number of generation: "))  #Input the maximum number of generation

for i in range(0, chromosome):
    cluster = 3#random.randint(2, max_cluster)  ##Randomly selects cluster number from 2 to root(poplation)
    K.insert(i, cluster)  ##Store the number of clusters in clu
    print('value of K is ',K)
    u, label,z1,A1= Kmeans_clu(cluster, data)
    #print("centers and labels : ", u, label)
    lab.insert(i, label)  ##Store the labels in lab
    center.insert(i, u) 
    new_center = pop_create(max_cluster, features, cluster, u)
    population.insert(i, new_center)
    print("VAlue of population in main\n" ,population)

newsol(max_gen,population,data)

对于newsol方法，我们从上面的方法生成代码传递新的种群，然后再次对种群进行K-Means

def ClusterIndicesComp(clustNum, labels_array):   #list comprehension for accessing the features in iris data set 
    return np.array([i for i, x in enumerate(labels_array) if x == clustNum])

def newsol(max_gen,population,data):
    #print('VAlue of NewSol Population is',population)
    for i in range(max_gen):
        cluster1=5
        u,label,t,l=Kmeans_clu(cluster1, population)
        A1.insert(i,t)
        plab.insert(i,label)
        pcenter.insert(i,u)
        k2=Counter(l.labels_)  #Count number of elements in each cluster
        k1=[t for (t, v) in k2.items() if v == 1] #element whose length is one will be fetched 
        t1= np.array(k1) #Iterating through the cluster which have one point associated with them 
        for b in range(len(t1)):
            print("Value in NEW_SOL is of 1 length cluster\n",t1[b])
            plot1=data[ClusterIndicesComp(t1[b], l.labels_)]
            print("Values are in sol of plot1",plot1)
            for q in range(cluster1):
                plot2=data[ClusterIndicesComp(q, l.labels_)]
                print("VAlue of plot2 is for \n",q,plot2)
                for i in range(len(plot2)):#To get one element at a time from plot2
                    plotk=plot2[i]
                    if([t for (t, v) in k2.items() if v >2]):#checking if the cluster have more than 2 points than only the distance will be calculated 
                        S=np.linalg.norm(np.array(plot1) - np.array(plotk))
                        print("Distance between plot1 and plotk is",plot1,plotk,np.linalg.norm(np.array(plot1) - np.array(plotk)))#euclidian distance is calculated 
                    else:
                        print("NO distance between them\n")

我做过的Kmeans是

from sklearn.cluster import KMeans
import numpy as np 

def Kmeans_clu(K, data):

    kmeans = KMeans(n_clusters=K, init='random', max_iter=1, n_init=1).fit(data) ##Apply k-means clustering
    labels = kmeans.labels_
    clu_centres = kmeans.cluster_centers_
    z={i: np.where(kmeans.labels_ == i)[0] for i in range(kmeans.n_clusters)} #getting cluster for each label 

    return clu_centres, labels ,z,kmeans

Answer 1

要获取每个群集中的实例数，您可以尝试使用Counter：

from collections import Counter, defaultdict
print(Counter(estimator.labels_))

结果：

Counter({0: 62, 1: 50, 2: 38})

其中群集0有62个实例，群集1有50个实例，群集2有38个实例

并且可能存储每个群集的实例索引，您可以使用defaultdict：

clusters_indices = defaultdict(list)
for index, c  in enumerate(estimator.labels_):
    clusters_indices[c].append(index)

现在，要查找集群0中的实例索引，请调用：

print(clusters_indices[0])

结果：

[50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 
 71, 72, 73, 74, 75, 76, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92,
 93, 94, 95, 96, 97, 98, 99, 101, 106, 113, 114, 119, 121, 123, 126, 127, 133, 138, 142, 146, 149]

在KMeans聚类之后查找聚类的长度（与聚类关联的点数）（scikit learn）

1 个答案: