r-如何预测保持回归常数的回归结果

Question

大家好，基于工资数据集(工资是因变量(和下面创建的工作流程，我想了解以下内容：

1. 对于每个分段模型，age等于30的人的预测wage是多少
2. 考虑到灵活的pw6_wf_fit模型配置，特别是上面的六个断点：超过age的哪个(近似(值与wage的相关性最强

我尝试使用extract的版本，但到目前为止我不知道如何在R中应用它。对任何评论都很有用

我使用的代码如下：

if (!require("pacman")) install.packages("pacman")
# load (or install if pacman cannot find an existing installation) the relevant packages
pacman::p_load(
tidyverse, tidymodels, ISLR, patchwork,
rpart, rpart.plot, randomForest, gbm, kernlab, parsnip, skimr
)

data(Wage, package = "ISLR")
Wage %>% 
tibble::as_tibble() %>% 
skimr::skim()

lin_rec   <- recipe(wage ~ age, data = Wage)
# Specify as linear regression
lm_spec <- 
linear_reg() %>%
set_mode("regression") %>%
set_engine("lm")

plot_model <- function(wf_fit, data) {

predictions <- 
tibble::tibble(age = seq(min(data$age), max(data$age))) %>% 
dplyr::bind_cols(
predict(wf_fit, new_data = .),
predict(wf_fit, new_data = ., type = "conf_int")
)

p <- ggplot2::ggplot(aes(age, wage), data = data) +
geom_point(alpha = 0.05) +
geom_line(aes(y = .pred),
data = predictions, color = "darkgreen") +
geom_line(aes(y = .pred_lower),
data = predictions, linetype = "dashed", color = "blue") +
geom_line(aes(y = .pred_upper),
data = predictions, linetype = "dashed", color = "blue") +
scale_x_continuous(breaks = seq(20, 80, 5)) +
labs(title = substitute(wf_fit)) +
theme_classic()

return(p)

}

pw3_rec <- lin_rec %>% step_discretize(age, num_breaks = 3, min_unique = 5) 
pw4_rec <- lin_rec %>% step_discretize(age, num_breaks = 4, min_unique = 5) 
pw5_rec <- lin_rec %>% step_discretize(age, num_breaks = 5, min_unique = 5) 
pw6_rec <- lin_rec %>% step_discretize(age, num_breaks = 6, min_unique = 5)

pw3_wf_fit <- workflow(pw3_rec, lm_spec) %>% fit(data = Wage)
pw4_wf_fit <- workflow(pw4_rec, lm_spec) %>% fit(data = Wage)
pw5_wf_fit <- workflow(pw5_rec, lm_spec) %>% fit(data = Wage)
pw6_wf_fit <- workflow(pw6_rec, lm_spec) %>% fit(data = Wage)

(plot_model(pw3_wf_fit, Wage) + plot_model(pw4_wf_fit, Wage)) /
(plot_model(pw5_wf_fit, Wage) + plot_model(pw6_wf_fit, Wage)) 

Answer 1

第一个问题的答案非常简单：

map(list(pw3_wf_fit, pw4_wf_fit, pw5_wf_fit, pw6_wf_fit), 
~predict(.x, new_data=tibble(age=30))) %>% 
bind_rows()
# # A tibble: 4 × 1
#   .pred
#   <dbl>
# 1  99.3
# 2  94.2
# 3  92.3
# 4  89.5