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在 tidymodels 中，我想创建一个基于配方和模型规范的工作流。当我不包含 step_pca(); 但是当我包含 step_pca() 作为设置时，我得到了错误。请参阅重复打击。

（如果我不使用工作流（），它可以工作；但是我失去了包括更新角色在内的功能）

x1 <- c(1, 6, 4, 2, 3, 4, 5, 7, 8, 2)
x2 <- c(1, 3, 4, 2, 3, 4, 5, 7, 8, 2)
id <- c(1:10)
y <- c(1, 4, 2, 5, 6, 2, 3, 6, 2, 4)
df1_train <- tibble(x1, x2, id, y)

# NA works with workflow
step_PCA_PREPROCESSING = NA
# Does not work with workflow
step_PCA_PREPROCESSING = 0.9

# My recipe
df1_train_recipe <- df1_train %>%
  recipes::recipe(y ~ .) %>%
  recipes::update_role(id, new_role = "id variable") %>%
  recipes::step_center(recipes::all_predictors()) %>%
  recipes::step_scale(recipes::all_predictors()) %>%
  # Optional step_pca
  {
    if (!is.na(step_PCA_PREPROCESSING)) {
      if (step_PCA_PREPROCESSING >= 1) {
        recipes::step_pca(., recipes::all_predictors(), num_comp = step_PCA_PREPROCESSING)
      } else if (step_PCA_PREPROCESSING < 1) {
        recipes::step_pca(., recipes::all_predictors(), threshold = step_PCA_PREPROCESSING)
      } else {
        .
      }
    } else {
      .
    }
  } %>%
  recipes::prep()

# Model specifications
model_spec <- parsnip::linear_reg() %>% 
  parsnip::set_engine("glmnet") 

# Create workflow (to know variable roles from recipes)
df1_workflow <- workflows::workflow() %>%
  workflows::add_recipe(df1_train_recipe) %>%
  workflows::add_model(model_spec) 

# Fit model
mod <-  parsnip::fit(df1_workflow, data = df1_train)

提前致谢

4

1 回答 1

1

我认为做你所说的最好的方法是使用设置为零的能力step_pca()，num_comp这意味着没有 PCA 分解。这对您的用例来说非常方便，因为threshold将覆盖num_comp.

注意：使用此参数将覆盖并重置给定的任何值num_comp。

library(tidymodels)

x1 <- c(1, 6, 4, 2, 3, 4, 5, 7, 8, 2)
x2 <- c(1, 3, 4, 2, 3, 4, 5, 7, 8, 2)
id <- c(1:10)
y <- c(1, 4, 2, 5, 6, 2, 3, 6, 2, 4)
df1_train <- tibble(x1, x2, id, y)

turn_off_pca <- 0
turn_on_pca  <- 1

rec1 <- recipe(y ~ ., data = df1_train) %>%
  update_role(id, new_role = "id variable") %>%
  step_center(all_predictors()) %>%
  step_scale(all_predictors()) %>%
  step_pca(all_predictors(), threshold = 0.9, num_comp = turn_off_pca)
  
rec2 <- recipe(y ~ ., data = df1_train) %>%
  update_role(id, new_role = "id variable") %>%
  step_center(all_predictors()) %>%
  step_scale(all_predictors()) %>%
  step_pca(all_predictors(), threshold = 0.9, num_comp = turn_on_pca)

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

workflow() %>%
  add_model(lm_spec) %>%
  add_recipe(rec1) %>%
  fit(df1_train)
#> ══ Workflow [trained] ══════════════════════════════════════════════════════════
#> Preprocessor: Recipe
#> Model: linear_reg()
#> 
#> ── Preprocessor ────────────────────────────────────────────────────────────────
#> 3 Recipe Steps
#> 
#> ● step_center()
#> ● step_scale()
#> ● step_pca()
#> 
#> ── Model ───────────────────────────────────────────────────────────────────────
#> 
#> Call:
#> stats::lm(formula = ..y ~ ., data = data)
#> 
#> Coefficients:
#> (Intercept)           x1           x2  
#>      3.5000       0.4607      -0.3459

workflow() %>%
  add_model(lm_spec) %>%
  add_recipe(rec2) %>%
  fit(df1_train)
#> ══ Workflow [trained] ══════════════════════════════════════════════════════════
#> Preprocessor: Recipe
#> Model: linear_reg()
#> 
#> ── Preprocessor ────────────────────────────────────────────────────────────────
#> 3 Recipe Steps
#> 
#> ● step_center()
#> ● step_scale()
#> ● step_pca()
#> 
#> ── Model ───────────────────────────────────────────────────────────────────────
#> 
#> Call:
#> stats::lm(formula = ..y ~ ., data = data)
#> 
#> Coefficients:
#> (Intercept)          PC1  
#>     3.50000      0.08116

由reprex 包于 2020-12-06 创建(v0.3.0.9001)

于 2020-12-07T04:22:12.170 回答