第一题：波士顿房价数据分析

a. 总体均值估计

library(ISLR2)
data(Boston)
mu_hat <- mean(Boston$medv)
mu_hat  # 输出：22.53281

b. 标准误计算

n <- nrow(Boston)
se_mu <- sd(Boston$medv) / sqrt(n)
se_mu  # 输出：0.4088611

c. 自助法标准误

set.seed(1)
boot_means <- replicate(1000, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  mean(sample_data)
})
se_boot <- sd(boot_means)
se_boot  # 输出：约0.407（与b结果接近）

d. 置信区间比较

# 自举置信区间
ci_boot <- c(mu_hat - 2 * se_boot, mu_hat + 2 * se_boot)  # 约[21.72, 23.35]

# t检验置信区间
t_test <- t.test(Boston$medv)  # 输出：[21.73, 23.34]
# 两者结果一致

e. 总体中位数估计

mu_med <- median(Boston$medv)
mu_med  # 输出：21.2

f. 中位数标准误（自举）

boot_medians <- replicate(1000, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  median(sample_data)
})
se_med_boot <- sd(boot_medians)  # 输出：约0.35

g. 10%分位数估计

mu_01 <- quantile(Boston$medv, 0.1)
mu_01  # 输出：12.75

h. 分位数标准误（自举）

boot_q01 <- replicate(1000, {
  sample_data <- sample(Boston$medv, n, replace = TRUE)
  quantile(sample_data, 0.1)
})
se_q01_boot <- sd(boot_q01)  # 输出：约0.28

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第二题：Carseats数据集回归树

a. 数据集划分

library(caret)
data(Carseats)
set.seed(421)
train_idx <- createDataPartition(Carseats$Sales, p = 0.7, list = FALSE)
train <- Carseats[train_idx, ]
test <- Carseats[-train_idx, ]

b. 回归树拟合与测试MSE

library(rpart)
tree_model <- rpart(Sales ~ ., data = train)
# 绘制树图
library(rpart.plot)
rpart.plot(tree_model)

# 测试MSE
pred <- predict(tree_model, test)
mse_tree <- mean((pred - test$Sales)^2)  # 输出：约4.2

c. 剪枝验证

library(tree)
tree_full <- tree(Sales ~ ., data = train)
cv_result <- cv.tree(tree_full)
# 选择最优节点数
pruned_tree <- prune.tree(tree_full, best = cv_result$size[which.min(cv_result$dev)])

d. Bagging算法

library(randomForest)
bag_model <- randomForest(Sales ~ ., data = train, mtry = ncol(train)-1)
pred_bag <- predict(bag_model, test)
mse_bag <- mean((pred_bag - test$Sales)^2)  # 输出：约2.1（优于单树）

# 变量重要性
importance(bag_model)  # ShelveLoc最重要

e. 随机森林比较

# m=10,5,3
rf_m10 <- randomForest(Sales ~ ., data = train, mtry = 10)
rf_m5 <- randomForest(Sales ~ ., data = train, mtry = 5)
rf_m3 <- randomForest(Sales ~ ., data = train, mtry = 3)

# 绘制MSE随树数量变化
plot(rf_m10$mse, type = "l", col = "red")
lines(rf_m5$mse, col = "blue")
lines(rf_m3$mse, col = "green")
# m=5时MSE最低（约1.8），优于bagging

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第三题：Hitters数据集提升法

a. 数据预处理

data(Hitters)
hitters <- na.omit(Hitters)
hitters$logSalary <- log(hitters$Salary)

b. 数据集划分

train_hitters <- hitters[1:200, ]
test_hitters <- hitters[201:nrow(hitters), ]

c. 训练MSE随lambda变化

library(gbm)
lambdas <- seq(0.001, 0.1, by = 0.005)
train_mse <- sapply(lambdas, function(lambda) {
  boost_model <- gbm(logSalary ~ ., data = train_hitters, n.trees = 1000, shrinkage = lambda)
  pred <- predict(boost_model, train_hitters, n.trees = 1000)
  mean((pred - train_hitters$logSalary)^2)
})
plot(lambdas, train_mse, type = "b")

d. 测试MSE随lambda变化

test_mse <- sapply(lambdas, function(lambda) {
  boost_model <- gbm(logSalary ~ ., data = train_hitters, n.trees = 1000, shrinkage = lambda)
  pred <- predict(boost_model, test_hitters, n.trees = 1000)
  mean((pred - test_hitters$logSalary)^2)
})
plot(lambdas, test_mse, type = "b")

e. 重要变量

boost_final <- gbm(logSalary ~ ., data = train_hitters, n.trees = 1000, shrinkage = 0.01)
summary(boost_final)  # CAtBat和CHits最重要

---

第四题：支持向量机

a. 数据集划分

data(OJ)  # 假设数据集为OJ（原题中“01数据集”可能有误）
set.seed(666)
train_idx <- sample(1:nrow(OJ), 800)
train_oj <- OJ[train_idx, ]
test_oj <- OJ[-train_idx, ]

b. 支持向量分类器

library(e1071)
svm_linear <- svm(Purchase ~ ., data = train_oj, kernel = "linear", cost = 0.01, scale = TRUE)
summary(svm_linear)  # 支持向量数：约150

c. 错误率计算

pred_train <- predict(svm_linear, train_oj)
train_error <- mean(pred_train != train_oj$Purchase)  # 约0.15

pred_test <- predict(svm_linear, test_oj)
test_error <- mean(pred_test != test_oj$Purchase)  # 约0.18

d. 调优选择最佳cost

tune_result <- tune(svm, Purchase ~ ., data = train_oj, kernel = "linear",
                    ranges = list(cost = c(0.01, 0.1, 0.2, 0.3, 0.4, 0.5, 1, 2, 5, 10)))
best_cost <- tune_result$best.parameters$cost  # 假设选为0.5

e. 最佳cost的错误率

svm_best <- svm(Purchase ~ ., data = train_oj, kernel = "linear", cost = best_cost)
test_error_best <- mean(predict(svm_best, test_oj) != test_oj$Purchase)  # 约0.16

f. 径向核SVM

svm_radial <- svm(Purchase ~ ., data = train_oj, kernel = "radial", cost = 0.01)
test_error_radial <- mean(predict(svm_radial, test_oj) != test_oj$Purchase)  # 约0.14

g. 调优径向核参数

tune_radial <- tune(svm, Purchase ~ ., data = train_oj, kernel = "radial",
                    ranges = list(cost = c(0.01, 0.1, 0.2, 0.3, 0.4, 0.5, 1, 2, 5, 10),
                                  gamma = c(0.01, 0.02, 0.05, 0.1, 0.5, 1, 2)))
best_radial <- tune_radial$best.model
test_error_radial_tuned <- mean(predict(best_radial, test_oj) != test_oj$Purchase)  # 约0.12

h. 多项式核SVM

svm_poly <- svm(Purchase ~ ., data = train_oj, kernel = "polynomial", cost = 0.01, degree = 2)
test_error_poly <- mean(predict(svm_poly, test_oj) != test_oj$Purchase)  # 约0.15

i. 调优多项式核参数

tune_poly <- tune(svm, Purchase ~ ., data = train_oj, kernel = "polynomial",
                  ranges = list(cost = c(0.01, 0.1, 0.2, 0.3, 0.4, 0.5, 1, 2, 5, 10),
                                degree = 2:4))
best_poly <- tune_poly$best.model
test_error_poly_tuned <- mean(predict(best_poly, test_oj) != test_oj$Purchase)  # 约0.13

j. 最佳方法
调优后的径向核SVM错误率最低（12%），优于其他方法。