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模拟:Fig. 13.5

R Notebook 模拟:Fig. 13.5
作者 szcf-weiya
时间 2018-02-04

问题重述

对于Easy Problem,

对于Difficult Problem,

本模拟实验比较了k最近邻和k均值应用在上述两个问题的分类误差率随相应参数变化的曲线,结果表明与书中图13.5完美契合。

生成数据

## generate dataset X
genX <- function(N, p = 10){
  sapply(1:p, function(i) runif(N))
}
## generate dataset Y
genY <- function(X, id = 1){
  if (id == 1){
    apply(X, 1, function(x) ifelse(x[1] < 0.5, 0, 1) )
  }
  else if (id == 2){
    apply(X, 1, function(x) ifelse(prod(x[1:3]-0.5) > 0, 1, 0) )
  }
  else{
    cat("WARNING! Incorrect id for problems.")
  }
}

参数设定

n.train = 100
n.test = 1000
n.realization = 10

k最近邻

library(class)
seq.knn = seq(1, 71, by = 7)
n.knn = length(seq.knn)
err = array(NA, c(n.knn, n.realization))

编写在两个问题中k最近邻的模拟函数knn.sim()

knn.sim <- function(id){
  for (j in 1:n.realization){
    X.train = genX(n.train)
    Y.train = genY(X.train, id)
    X.test = genX(n.test)
    Y.test = genY(X.test, id)
    for (i in 1:n.knn){
      Y.pred = knn(X.train, X.test, factor(Y.train), k = seq.knn[i])
      err[i, j] = sum(Y.pred!=Y.test)/n.test
    }
  }
  return(err)
}

编写整个模拟实验中通用的绘图函数myplot()

myplot <- function(seq.knn, err, main, xlab, ylim){
  ## calculate the mean and std of misclassification error
  err.mean = apply(err, 1, mean)
  err.std = apply(err, 1, sd)

  n.knn = length(seq.knn)
  ## plot
  plot(seq.knn, err.mean,
       main = main,
       ylab = "Misclassification Error",
       xlab = xlab,
       col = "blue", type = "l",
       ylim = ylim)
  for(i in 1:n.knn){
    lines(c(seq.knn[i], seq.knn[i]),
          c(err.mean[i] - err.std[i], err.mean[i] + err.std[i]),
          col = "blue", pch = 3)
    lines(c(seq.knn[i]-0.2, seq.knn[i]+0.2),
          c(err.mean[i] - err.std[i], err.mean[i] - err.std[i]),
          col = "blue", pch = 3)
    lines(c(seq.knn[i]-0.2, seq.knn[i]+0.2),
          c(err.mean[i] + err.std[i], err.mean[i] + err.std[i]),
          col = "blue", pch = 3)
  }
}

则Easy Problem的分类误差随邻居数变化的曲线为

err = knn.sim(1)
myplot(seq.knn, err, "Nearest Neighbors / Easy", "Number of Neighbors", c(0.1, 0.5))

对于Difficult Problem,我们有

err = knn.sim(2)
myplot(seq.knn, err, "Nearest Neighbors / Difficult", "Number of Neighbors", c(0.4, 0.6))

k-means聚类

seq.kmeans = c(1, 2, 3, 5, seq(6, 31, by = 4))
n.kmeans = length(seq.kmeans)
err = array(NA, c(n.kmeans, n.realization))

编写两个问题中k-means的模拟函数kmeans.sim()

## predict function for k-means
predict.kmeans <- function(cl0, cl1, newpoint){
  center0 = cl0$centers
  center1 = cl1$centers
  res0 = apply(newpoint, 1, function(x) {
    min(apply(center0, 1, function(xx) sum((xx-x)^2)))
  })
  res1 = apply(newpoint, 1, function(x) {
    min(apply(center1, 1, function(xx) sum((xx-x)^2)))
  })
  ifelse(res0 < res1, 0, 1)
}
kmeans.sim <- function(id){
  for (j in 1:n.realization){
    X.train = genX(n.train)
    Y.train = genY(X.train, id)
    X.test = genX(n.test)
    Y.test = genY(X.test, id)
    for (i in 1:n.kmeans){
      cl0 = kmeans(X.train[Y.train==0, ], seq.kmeans[i])
      cl1 = kmeans(X.train[Y.train==1, ], seq.kmeans[i])
      Y.pred = predict.kmeans(cl0, cl1, X.test)
      err[i, j] = sum(Y.pred!=Y.test)/n.test
    }
  }
  return(err)
}

则Easy Problem的分类误差随原型个数变化的曲线为

err = kmeans.sim(1)
myplot(seq.kmeans, err, "K-means / Easy", "Number of Prototypes per Class", c(0.1, 0.5))

而Difficult Problem对应的图象为

err = kmeans.sim(2)
myplot(seq.kmeans, err, "K-means / Difficult", "Number of Prototypes per Class", c(0.4, 0.6))

待解决的问题

  • 为什么贝叶斯误差率为0.
  • 加入LVQ

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