如何在workflow_set中使用dplyr风格选择器高效指定变量？

我希望用workflow_set遍历不同变量组合构建模型。官方文档提到预处理器有三类：标准R公式、预处理前的recipe对象、dplyr风格选择器，但只给了R公式的示例。我倾向用dplyr风格，方便后续自定义函数指定结果和预测变量，但没找到相关示例。目前用recipe %>% update_role %>% update_role的方式可行但太繁琐（示例2）；尝试add_variables的方式失败（示例3）。请问有没有更简洁的dplyr风格写法用于workflow_set？

示例1 R公式风格（官方示例） #

set.seed(123)
data <- data.frame(
  x1 = rnorm(100),
  x2 = rnorm(100),
  x3 = rnorm(100),
  x4 = rnorm(100),
  y = rnorm(100)
)

variables<-list(
  first = y ~ x1,
  second = y ~ x2,
  third = y ~ x1+x2,
  fourth = y ~ x3+x4
)

lm_model <- 
  linear_reg() %>% 
  set_engine("lm")

location_models <- workflow_set(preproc = variables, models = list(lm = lm_model))
location_models
location_models$fit[[4]]
extract_workflow(location_models, id = "third_lm")

location_models <-
  location_models %>%
  mutate(fit = map(info, \(x) fit(x$workflow[[1]], data)))
location_models$fit[[4]]

示例2 使用update_role的写法（可行但繁琐） #

variables<-list(
  fist = recipe(data)%>%update_role( y, new_role = "outcome")%>%update_role( x1, new_role = "predictor"),
  second = recipe(data)%>%update_role( y, new_role = "outcome")%>%update_role( x2, new_role = "predictor"),
  third = recipe(data)%>%update_role( y, new_role = "outcome")%>%update_role( c(x1,x2), new_role = "predictor"),
  forth = recipe(data)%>%update_role( y, new_role = "outcome")%>%update_role( c(x3,x4), new_role = "predictor")
)

示例3 尝试add_variables（失败） #

variables<-list(
  fist = add_variables(outcomes = y, predictors = x1),
  second = add_variables(outcomes = y, predictors = x2),
  third = add_variables(outcomes = y, predictors = c(x1,x2)),
  forth = add_variables(outcomes = y, predictors = c(x3,x4))
)

方法1：封装自定义函数生成Recipe #

通过封装函数减少重复代码，同时支持dplyr的选择语法（比如starts_with("x")这类批量选择器）：

# 定义生成指定变量角色的Recipe函数
create_var_recipe <- function(outcome_col, predictor_cols) {
  recipe(data) %>%
    update_role({{outcome_col}}, new_role = "outcome") %>%
    update_role({{predictor_cols}}, new_role = "predictor")
}

# 用dplyr风格定义变量组合
variables <- list(
  first = create_var_recipe(y, x1),
  second = create_var_recipe(y, x2),
  third = create_var_recipe(y, c(x1, x2)),
  fourth = create_var_recipe(y, starts_with("x3")) # 示例：用dplyr批量选择器
)

# 构建workflow_set（和示例1逻辑一致）
lm_model <- linear_reg() %>% set_engine("lm")
location_models <- workflow_set(preproc = variables, models = list(lm = lm_model))

方法2：直接用Workflow结合add_variables #

add_variables是workflow的步骤函数，需要包裹在workflow()中才能作为预处理器传入workflow_set：

# 用workflow封装add_variables步骤
variables <- list(
  first = workflow() %>% add_variables(outcomes = y, predictors = x1),
  second = workflow() %>% add_variables(outcomes = y, predictors = x2),
  third = workflow() %>% add_variables(outcomes = y, predictors = c(x1, x2)),
  fourth = workflow() %>% add_variables(outcomes = y, predictors = c(x3, x4))
)

# 构建workflow_set
lm_model <- linear_reg() %>% set_engine("lm")
location_models <- workflow_set(preproc = variables, models = list(lm = lm_model))

两种方法都能实现简洁的dplyr风格变量选择，其中自定义函数的方式更适合后续扩展预处理逻辑（比如添加特征工程步骤）。

内容的提问来源于stack exchange，提问作者Niall Marsay