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kmeans.py
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/
kmeans.py
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# -*- coding: utf-8 -*-
"""
Created on Tue Jun 5 00:05:04 2018
@author: Administrator
"""
from numpy import *
# 计算欧几里得距离
def distEclud(vecA, vecB):
return sqrt(sum(power(vecA - vecB, 2))) # 求两个向量之间的距离
# 构建聚簇中心，取k个(此例中为4)随机质心
def randCent(dataSet, k):
n = shape(dataSet)[1]
centroids = mat(zeros((k,n))) # 每个质心有n个坐标值，总共要k个质心
for j in range(n):
minJ = min(dataSet[:,j])
maxJ = max(dataSet[:,j])
rangeJ = float(maxJ - minJ)
centroids[:,j] = minJ + rangeJ * random.rand(k, 1)
return centroids
# k-means 聚类算法
def kMeans(dataSet, k):
m = shape(dataSet)[0] #电影个数
n = shape(dataSet)[1] #评价元素个数
division = mat(zeros((m,2))) # 用于存放该样本属于哪类及质心距离
# clusterAssment第一列存放该数据所属的中心点，第二列是该数据到中心点的距离
centroids = randCent(dataSet, k)
flag = True # 用来判断聚类是否已经收敛
while flag:
flag = False;
for i in range(m): # 把每一个数据点划分到离它最近的中心点
minDist = 1000000000;
minIndex = -1;
for j in range(k):
distJI = distEclud(centroids[j,:], dataSet[i,:])
if distJI < minDist:
minDist = distJI
minIndex = j # 如果第i个数据点到第j个中心点更近，则将i归属为j
if division[i,0] != minIndex:
flag = True # 如果分配发生变化，则需要继续迭代
division[i,0] = minIndex
division[i,1] = minDist # 并将第i个数据点的分配情况存入字典
#print (centroids)
for i in range(k): # 重新计算中心点
ptsInClust = dataSet[nonzero(division[:,0].A == i)[0]] # 去第一列等于cent的所有列
centroids[i,:] = mean(ptsInClust, axis = 0) # 算出这些数据的中心点"""
return division
def KTH(choosename,docs,docs2): #选中电影名称 选中及筛选后电影各项元素评分 选中及筛选后各项电影名称
dataMat = []
for i in range(len(docs)):
dataMat1 = []
for j in range(len(docs[i])):
if docs[i][j]==' ':
continue
docs[i][j] = float(float(docs[i][j])*100)
dataMat = mat(docs)
m = shape(dataMat)[0]
choosenum=[] #选择电影的编号
ansnum=[]
vis=[]
for i in range(m):
vis.append(0)
for i in range(m):
for j in range(len(choosename)):
if choosename[j]==docs2[i]:
choosenum.append(i)
vis[i]=1
break
for i in range(m):
ansnum.append([i,0])
for p in range(9):
Division = kMeans(dataMat,9)
for t in range(len(choosename)):
tmp = Division[choosenum[t],0]
ans=[] #编号 距离
for i in range(m):
if Division[i,0]==tmp:
ans.append([i,distEclud(dataMat[i,:],dataMat[choosenum[t],:])])
ans.sort(key=lambda x:x[1])
for i in range(len(ans)):
ansnum[ans[i][0]][1]+=1
ansnum.sort(key=lambda x:x[1],reverse=True)
templist = []
flag=0
for i in range(len(docs)):
if vis[ansnum[i][0]]==0:
templist.append(docs2[ansnum[i][0]])
flag+=1
if flag==4:
break
return templist