2

如果variable_names包含名称我想更新那些成为 Not_predictors 的角色;而如果variable_names是 NA 我想在函数中跳过这一步。但是,在添加条件时,我得到了错误。请参阅下面的 repex 代码。

# Example Data
pred1 <- c(4,2,1)
pred2 <- c(4,2,1)
pred3 <- c(4,2,1)
pred4 <- c(4,2,1)
id_nr <- c(1, 2, 3)
y <- c(1,5,2)

data <- tibble::tibble(pred1, pred2, pred3, pred4, id_nr, y)

# Want to remove thes variables from predictor role if variable_names is not NA
variable_names <- c("pred1", "pred2")


# Without condition it works
recipe <- data %>%
  recipes::recipe(y ~ .) %>%
  recipes::update_role(id_nr, new_role = "id variable") %>%
  recipes::update_role(dplyr::all_of(variable_names), new_role = "Not_predictors") %>% 
  recipes::step_pca(., recipes::all_predictors(), num_comp = 2) %>% 
  recipes::prep()
recipe


# But when adding the condition I get error. 
recipe <- data %>%
  recipes::recipe(y ~ .) %>%
  recipes::update_role(id_nr, new_role = "id variable") %>%
  {
    if(!is.na(variable_names[1])){
      recipes::update_role(dplyr::all_of(variable_names), new_role = "Not_predictors") 
    } else {
      .
    }
  } %>% 
  recipes::step_pca(., recipes::all_predictors(), num_comp = 2) %>% 
  prep()
recipe

这是错误:

Error: $ operator is invalid for atomic vectors
In addition: Warning message:
No selectors were found 

任何帮助深表感谢。

4

1 回答 1

3

我从RStudio 社区收到了一个解决方案:

recipe <- data %>%
  recipes::recipe(y ~ .) %>%
  recipes::update_role(id_nr, new_role = "id variable") %>%
  (function(x){
    if(!is.na(variable_names[1])){
      x %>%  recipes::update_role(dplyr::all_of(variable_names), new_role = "Not_predictors") 
    } else {
      x
    }
  }) %>% 
  recipes::step_pca(., recipes::all_predictors(), num_comp = 2) %>% 
  recipes::prep()
recipe
于 2020-11-23T13:18:40.910 回答