如果需要计算所有聚类中每个其他点的欧几里得距离，则必须检查任何聚类是否只有一个点与之相关联，并且最小距离点将被添加到长度为一的聚类中（与该点相关联的点）。在需要更新标签和中心的集群值之后，我已经做了所有事情，但是有时在代码运行时，当它必须返回值时会抛出错误“'NoneType'对象不可迭代” 。我对此部分感到困惑，请提供帮助，在此先谢谢您。这是我的代码
import numpy as np
import matplotlib.pyplot as mlt
import pandas as pd
from clustering import Kmeans_clu
import random
from collections import Counter, defaultdict
from math import sqrt
import matplotlib.pyplot as plt
import random
def ClusterIndicesComp(clustNum, labels_array): # For extracting points from the label
return np.array([i for i, x in enumerate(labels_array) if x == clustNum])
def index(lst, obj, n): # To find the index of of Lable which needs to be change
count = 0
for index, item in enumerate(lst):
if item == obj:
count += 1
if count == n:
return index
raise ValueError('{} is not in list at least {} times'.format(obj, n))
def UpdateCentetr_Label(Index,lablelPlot1, plot2, lablePlot2, pop, plab, plot1, pcenter,cluster1,max_gen):
s = list(plab) #Storing Value of Label in list
print("Value of in Label before\n", s) # Before Updating the value of label
s1 = Counter(s)
print("Value of counter in before is \n", s1)
Index=Index+1
q = index(s, lablePlot2, Index) #Storing the index of label which need to be changed
s[q] = lablelPlot1 #Changing the label by labelPlot1
ulabel = np.array(s)
print("Value of an updated label is \n", ulabel)
print("------------------------------------------")
# print("Value of Pcenter in Updatedlabel\n",pcenter)
ncenter = plot2 + plot1 / 2 #Calculating the center of label which have one point associated
print("Value of new center to be updated\n", ncenter)
ucenter = np.array(pcenter)
#print("Value of ucenter is\n",ucenter)
ucenter[lablelPlot1] = ncenter #Changing the value of label with new center
print("Value of updated center is\n", ucenter)
ko = Counter(ulabel)
print("Value of counter after updated label is\n", ko)
LC1 = [t for (t, v) in ko.items() if v == 1] # Again checking if there is label which have point associated
t1 = np.array(LC1)
if (t1.size):
One_LengthCluster(cluster1, ulabel, ucenter, pop,max_gen) #To update the it again
else:
return ulabel,ucenter,cluster1
def ClusterPopulation(max_gen, population): # to cluster the population
pop = population
plab1 = []
pcenter = []
u=[]
cluster1 = int(input("Enter the cluster for new population\n"))
plab=[]
for i in range(0,max_gen):
if (i % cluster1 == 1): # Checking the condition of Kmeans Clustering
u, label, t, l = Kmeans_clu(cluster1, population) # Storing the values Center(u) anb label(u)
plab1.insert(i, label)
print("Plab is",plab)
pcenter.insert(i, u)
plab = np.array(plab1)
plab=plab[0]
#pcenter=np.array(pcenter)
pcenter=pcenter[0]
ulabel, ucenter, cluster1=One_LengthCluster(cluster1, plab, pcenter, pop,max_gen) #To check if any label has one point associated
return ulabel,ucenter,cluster1 #returning the value of ulabel, ucenter and cluster1
else:
print("Not need of clustering for this generation of\n", i)
def One_LengthCluster(cluster1, plab, pcenter, pop,max_gen):
indexes = []
Index=[] #Store the index of Point which have minimum euclidean distance
D=[] #Store the minimum euclidean distance
labelplot2=[] #Store the Label which have more than 2 points associated
Point2=[] #Store the Point which have minimum euclidean distance
z=[]
Smin=[]
I=[]
L=[]
LC = Counter(plab) #Counting number of points associated with label
print("VAlue of LAbel and Number Cluster Associated with them\n", LC)
LC1 = [t for (t, v) in LC.items() if v == 1]
t1 = np.array(LC1)
if (t1.size):# To check if any of the Label has one point associated if yes than calculate the distance with all the points
for b in range(len(t1)):
plot1 = pop[ClusterIndicesComp(t1[b], plab)] # Extracting the point in the label which have one point associated
print("Point of label one Length PLOT1 is\n", np.array(plot1), t1[b])
z1 = [t for (t, v) in LC.items() if v > 2] # To check distance with label which more than 3 points associated
z = np.array(z1) #Storing the value in the array
for d in range(len(z)):
print("Value of Label which have more than two cluster is\n", z[d])
plot2 = pop[ClusterIndicesComp(z[d], plab)] # Extracting the point in the label more than one point associated
print("Value of plot2 in one length cluster is\n", plot2)
for i in range(len(plot2)):
plotk = plot2[i] # To get one point at a time from plot2
S = np.linalg.norm(np.array(plot1) - np.array(plotk))
print("Distance between {} and {} is {}\n".format(plot1,plotk,S)) # euclidian distance is calculated
if (i == 0):
Smin = S
Sminant = S
indexes.append(i)
else:
if (S < Sminant):
Smin = S
Sminant = Smin
indexes = []
indexes.append(i)
elif (S == Sminant):
indexes = []
indexes.append(i)
#print('indexes:')
print("Index at which the minimum value is stored\n", indexes) # To find the index of Label with which euclidian distance is minimum
for i in range(len(indexes)):
Point2 = plot2[indexes[i]]
I = indexes[i]
L = z[d]
print("VAlues of Point{} which have min distance with plot1 is in Label {} and have Index {} and distance {}\n".format(Point2,L,I,Smin))
if(len(z)==1): #If Label which have more than 2 point associated is only one
D = Smin
Index = indexes[i]
labelplot2=z[d]
Point2=plot2[indexes[i]]
print("Here is the value\n", D, Index, labelplot2, Point2)
ulabel, ucenter, cluster1=UpdateCentetr_Label(Index,t1[b], Point2, z[d], pop, plab, plot1, pcenter,cluster1,max_gen) #After Finding Point now update center and label
return ulabel,ucenter,cluster1
elif (len(z) > 1): #If Label which have more than 2 point associated is more than one
D.append(Smin)
Index.append(I)
labelplot2.append(L)
#print("Value in list are------------\n", labelplot2)
print("Index value is\n",Index)
print("Label value is\n", labelplot2)
z=min(D) #Finding the minimum distance among all the labels
k=D.index(z) #Finding the index where minimum distance is stored in D
Index=Index[k]
labelplot2=labelplot2[k]
Point2 = pop[ClusterIndicesComp(labelplot2, plab)]
Point2=Point2[Index]
print("Value of minimum distance is\n",z,Index,labelplot2,k,Point2)
ulabel, ucenter, cluster1=UpdateCentetr_Label(Index,t1[b], Point2, labelplot2, pop, plab, plot1, pcenter,cluster1,max_gen) #After Finding Point now update center and label
return ulabel,ucenter,cluster1
D=[]
indexes=[]
else: #If no solution have one point associated in the label
print("no lenght 1 cluster\n")
return plab,pcenter,cluster1
population =np.random.rand(10,4) #Generating the random population
max_gen=10 #Giving value of max_gen
ulabel, ucenter, cluster1=ClusterPopulation(max_gen, population) #Taking back the values of ucenter ,ulabel and cluster1
print("Value of ulabel is\n",ulabel)
我的K均值代码是这个
from sklearn.cluster import KMeans
import numpy as np
def Kmeans_clu(K, data):
"""
:param K: Number of cluster
:param data:
:return:
"""
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)} #
return clu_centres, labels ,z,kmeans
TraceBack是
Traceback (most recent call last):
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/Population Clustering.py", line 184, in <module>
ulabel, ucenter, cluster1=ClusterPopulation(max_gen, population) #Taking back the values of ucenter ,ulabel and cluster1
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/Population Clustering.py", line 85, in ClusterPopulation
ulabel, ucenter, cluster1=One_LengthCluster(cluster1, plab, pcenter, pop,max_gen) #To check if any label has one point associated
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/Population Clustering.py", line 169, in One_LengthCluster
ulabel, ucenter, cluster1=UpdateCentetr_Label(Index,t1[b], Point2, labelplot2, pop, plab, plot1, pcenter,cluster1,max_gen) #After Finding Point now update center and label
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/Population Clustering.py", line 54, in UpdateCentetr_Label
One_LengthCluster(cluster1, ulabel, ucenter, pop,max_gen) #To update the it again
File "C:/Users/hp/AppData/Local/Programs/Python/Python36/Population Clustering.py", line 169, in One_LengthCluster
ulabel, ucenter, cluster1=UpdateCentetr_Label(Index,t1[b], Point2, labelplot2, pop, plab, plot1, pcenter,cluster1,max_gen) #After Finding Point now update center and label
TypeError: 'NoneType' object is not iterable
答案 0 :(得分：4)
我认为您只是在return函数中丢失了UpdateCentetr_Label语句，这迫使该函数返回实际上不是不可迭代的None：
def UpdateCentetr_Label(Index,lablelPlot1, plot2, lablePlot2, pop, plab, plot1, pcenter,cluster1,max_gen):
# other code here...
if (t1.size):
return One_LengthCluster(cluster1, ulabel, ucenter, pop,max_gen) #To update the it again
答案 1 :(得分：1)
您错过了函数中的简单return语句
def UpdateCentetr_Label(Index,lablelPlot1, plot2, lablePlot2, pop, plab, plot1, pcenter,cluster1,max_gen):
您只需要在将t1的大小检查后添加到One_length群集中即可
if (t1.size):
return One_LengthCluster(cluster1, ulabel, ucenter, pop,max_gen) #To update the it again