聚类分析 练习题

d = matrix(data=c(0,0,0,0,0,4,0,0,0,0,6,9,0,0,0,1,7,10,0,0,6,3,5,8,0),nrow=5,ncol=5,byrow = TRUE);d
x = scale(d) # 数据标准化
D = dist(x);D # 默认为 Euclidean 距离

# 最短距离法
hc1 = hclust(D, method='single')
data.frame(hc1$merge,hc1$height)
plot(hc1)

# 最长距离法
hc2=hclust(D,"complete");hc2
data.frame(hc2$merge,hc2$height)
plot(hc2)

# 类平均法
hc3=hclust(D,"average");hc3
data.frame(hc3$merge,hc3$height)
plot(hc3)

library(openxlsx)
d7.2 = read.xlsx("../Res/mvexer5.xlsx", "E7.2", rowNames=T);head(d7.2)

plot(d7.2,gap=0)

summary(d7.2)

Z = scale(d7.2);Z  # 数据标准化
boxplot(Z) # 箱型图
D = dist(Z)

hc = hclust(D, 'single') # 最短距离法
plot(hc);rect.hclust(hc, 2);cutree(hc, 2) # 分2类
plot(hc);rect.hclust(hc, 3);cutree(hc, 3) # 分3类
plot(hc);rect.hclust(hc, 4);cutree(hc, 4) # 分4类

hc2 = hclust(D, 'ward.D2') # Ward.D2 法
plot(hc2);rect.hclust(hc2, 2);cutree(hc2, 2) # 分2类
plot(hc2);rect.hclust(hc2, 3);cutree(hc2, 3) # 分3类
plot(hc2);rect.hclust(hc2, 4);cutree(hc2, 4) # 分4类

# kmeans 聚类法
kmeans(Z, 2)$cluster # 分2类

kmeans(Z, 3)$cluster # 分3类

kmeans(Z, 4)$cluster # 分4类

kMock<-function(n, svar=0.3) {
  set.seed(123) # 设定种子数
  x1 = matrix(rnorm(n*10, 0, svar), ncol = 10) # 均值为 0, 标准差为 svar 的 n*10 正态随机矩阵
  x2 = matrix(rnorm(n*10, 1, svar), ncol = 10) # 均值为 1, 标准差为 svar 的 n*10 正态随机矩阵
  X = rbind(x1, x2) # 形成 随机矩阵
  H = hclust(dist(X))
  plot(H);rect.hclust(H,2);
  cutree(H, 2) # 系统聚类结果
  
  km = kmeans(X, 2) # K-means 聚类
  #table(km)
  kc = km$cluster        # 分类结果
  #plot(X, pch = kc)
  plot(X, pch = kc, col = kc)
}

kMock(20)

kMock(50)

kMock(100)

kMock(1000)

kMock(10000)