L'objet 'NoneType' n'est pas itérable lors du renvoi des valeurs
Je dois vérifier si une grappe n'a qu'un seul point associé si besoin de calculer la distance euclidienne avec tous les autres points de la grappe et le point de distance minimale sera ajouté à la grappe de longueur un (à laquelle un point est associé). Après que l'ajout d'un point aux valeurs de cluster de label et centre doit être mis à jour, j'ai tout fait, mais parfois, lorsque le code est exécuté, l'erreur "L'objet 'NoneType' n'est pas itérable" est renvoyée lorsque les valeurs . Je suis coincé avec cette partie s'il vous plaît aider, merci d'avance.Voici mon code
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)
Mon code K-means est ceci
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 est
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
Je pense qu'il vous manque simplement une instruction return dans votre fonction UpdateCentetr_Label qui force la fonction à renvoyer None qui en fait n'est pas itérable
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
Vous avez manqué la déclaration simple return dans votre fonction
def UpdateCentetr_Label(Index,lablelPlot1, plot2, lablePlot2, pop, plab, plot1, pcenter,cluster1,max_gen):
Vous devez juste ajouter return après avoir vérifié la taille de t1 dans le cluster One_length
if (t1.size):
return One_LengthCluster(cluster1, ulabel, ucenter, pop,max_gen) #To update the it again