Question

我需要通过自举执行knn回归，并针对不同的K值进行迭代

说我有2个数据框，训练和测试

train <- read.csv("train.csv")
test <- read.csv("test.csv")

还有一个函数knn如下：

knn <- function(train_data, train_label, test_data, K){

  len_train <- nrow(train_data)
  len_test <- nrow(test_data)


  test_label <- rep(0, len_test)

  k_means <- function(training_pt){

    distances <- as.matrix(dist(rbind(training_pt, train_data)))[1, (1+1):(1+len_train)]
    data.frame(y = train_label) %>%
    # train_label %>%
      mutate(pt_dist = distances) %>%
      arrange(pt_dist) %>%
      select(y) %>%
      slice(1:K) %>% pull() %>% mean()
  }

  predictions <- apply(test_data, 1, k_means)
  return(predictions)

}

其中train_data采用带有预测变量列的数据帧， train_label是火车值的向量，而test_data是一个数据列，其列与train_data相似

此函数返回test_data每行的预测测试标签

现在，我编写了一个函数来生成引导受限样本：

gen_boot_sample <- function(df, sample_size = 25){
  df %>% sample_n(sample_size, replace = T)
}

我设法写了一些东西，将knn函数应用于固定值K的生成的自举样本。

但是我一直在努力遍历K

想法是生成一个数据帧，其中包含每个K值的每个引导捆绑样本（例如20个样本）的误差值

test_label <- test_data %>%
  select_at(.vars = vars(contains("y"))) %>%
  pull()

rerun(5, gen_boot_sample(train_data)) %>%
      map( ~ knn( 
      train_data = .x %>%
        select_at(.vars = vars(contains("x"))),
      train_label = .x %>%
        select_at(.vars = vars(contains("y"))) %>%
        pull(),
      test_data = test_data %>%
        select_at(.vars = vars(contains("x"))),
      K = 5
         )
      ) %>%
      map(~sum(. - test_label)^2)

我检查了答案 purrr map equivalent of nested for loop 但由于我的knn函数如何接受参数而感到困惑

编辑：添加部分数据

train_data <- structure(list(x1 = c(1973.5, 1967.5, 1970.5, 1978, 1964, 1962, 
1980, 1961.5, 1976.5, 1979.5), y = c(6.57, 1.83, 3.69, 11.88, 
0.92, 0.72, 16.2, 0.92, 8.28, 14.85)), row.names = c(28L, 16L, 
22L, 37L, 9L, 5L, 41L, 4L, 34L, 40L), class = "data.frame")

test_data <- structure(list(x1 = c(1978.75, 1962.75, 1974.25, 1975.75, 1963.75, 
1972.75, 1968.25, 1980.75, 1979.25, 1970.75), y = c(8.91, 0.6, 
6.39, 6.12, 0.77, 4.41, 2.07, 11.61, 12.96, 3.6)), row.names = c(38L, 
6L, 29L, 32L, 8L, 26L, 17L, 42L, 39L, 22L), class = "data.frame")

Answer 1

我们可以使用嵌套在map中的另一个循环来运行不同的“ K”值

library(tidyverse)
rerun(5, gen_boot_sample(train_data)) %>%
      map(~ {
         # create the subset datasets
         train_data <- .x %>%
                           select_at(vars(contains('x')))
         train_label <- .x %>%
                          select_at(.vars = vars(contains("y"))) %>% 
                          pull()
         test_data <- test_data %>% 
                         select_at(.vars = vars(contains("x")))
        # loop over different values for 'K'
        map_dbl(1:10, ~ {
               #apply the knn function
               out <- knn(train_data, train_label, test_data, K = .x)
               sum(out - test_label)^2}
             )
      })

使用purrr嵌套地图功能

1 个答案: