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我想在同一个数据集上评估几个（主要是）线性回归模型的性能。我想也许使用tidymodels包和workflowsets::workflow_set()可能的工作。我遵循了此处的示例，但我无法弄清楚如何从代码中实际获得合适的结果。

# Load packages
  library("tidyverse")
  library('workflowsets')
  library("parsnip")
  library("recipes")

# Data
  dat <- 
    structure(list(q = c(66.65, 75.58, 83.06, 91.28, 119.26, 133.14, 
    146.32, 153.39, 168.57, 182.36, 210.09, 188.19, 213.42, 296.95, 
    326.33, 358.63, 475.99, 475.99, 683.44, 683.44, 838.49, 1282.1, 
    1648.97, 1572.97, 2055.14, 2521.39, 2685.11, 2859.46, 3242.87, 
    6899.19, 6377.42, 7581.96, 9599.32), c = c(317.06, 283.99, 279.56, 
    283.99, 227.84, 227.84, 262.5, 242.64, 270.9, 266.67, 210.6, 
    235.12, 235.12, 210.6, 207.31, 227.84, 220.78, 194.67, 177.13, 
    207.31, 179.94, 177.13, 182.79, 139.89, 148.98, 144.36, 137.71, 
    158.66, 142.11, 142.11, 119.52, 110.48, 158.66), c_less_c_nought = c(300.06, 
    266.99, 262.56, 266.99, 210.84, 210.84, 245.5, 225.64, 253.9, 
    249.67, 193.6, 218.12, 218.12, 193.6, 190.31, 210.84, 203.78, 
    177.67, 160.13, 190.31, 162.94, 160.13, 165.79, 122.89, 131.98, 
    127.36, 120.71, 141.66, 125.11, 125.11, 102.52, 93.48, 141.66
    )), row.names = c(NA, -33L), class = c("tbl_df", "tbl", "data.frame"
    )) 

  # Recipes for models
  eq1_mod1_recipe <-
    recipes::recipe(c ~ q, data = dat) %>% 
    step_log(c, q, base = 10)
  
  eq2_mod2_a_recipe <- 
    recipes::recipe(c_less_c_nought ~ q, data = dat) %>% 
    step_log(c_less_c_nought, q, base = 10)  

  # Define model types
  lm_model <-
    parsnip::linear_reg() %>% 
    parsnip::set_engine("lm") %>% 
    parsnip::set_mode("regression")

  # Run the models?
  cq_models <-
    workflowsets::workflow_set(
      preproc = list(eq1m1 = eq1_mod1_recipe, e2m2a = eq2_mod2_a_recipe),
      models = list(lm = lm_model)
    )

看来这实际上并不适合模型本身。我需要什么/在哪里添加代码以适应线性模型？

或者，是否有更好但仍然“整洁”的方式来做到这一点？接受建议。

4

1 回答 1

1

工作流集仅适用于重新采样的数据，例如交叉验证或引导折叠。这是设计使然，因此人们不会尝试比较将单个时间拟合到数据集的性能指标。

library("tidymodels")

# Data
dat <- 
  structure(list(q = c(66.65, 75.58, 83.06, 91.28, 119.26, 133.14, 
                       146.32, 153.39, 168.57, 182.36, 210.09, 188.19, 213.42, 296.95, 
                       326.33, 358.63, 475.99, 475.99, 683.44, 683.44, 838.49, 1282.1, 
                       1648.97, 1572.97, 2055.14, 2521.39, 2685.11, 2859.46, 3242.87, 
                       6899.19, 6377.42, 7581.96, 9599.32), c = c(317.06, 283.99, 279.56, 
                                                                  283.99, 227.84, 227.84, 262.5, 242.64, 270.9, 266.67, 210.6, 
                                                                  235.12, 235.12, 210.6, 207.31, 227.84, 220.78, 194.67, 177.13, 
                                                                  207.31, 179.94, 177.13, 182.79, 139.89, 148.98, 144.36, 137.71, 
                                                                  158.66, 142.11, 142.11, 119.52, 110.48, 158.66), c_less_c_nought = c(300.06, 
                                                                                                                                       266.99, 262.56, 266.99, 210.84, 210.84, 245.5, 225.64, 253.9, 
                                                                                                                                       249.67, 193.6, 218.12, 218.12, 193.6, 190.31, 210.84, 203.78, 
                                                                                                                                       177.67, 160.13, 190.31, 162.94, 160.13, 165.79, 122.89, 131.98, 
                                                                                                                                       127.36, 120.71, 141.66, 125.11, 125.11, 102.52, 93.48, 141.66
                                                                  )), row.names = c(NA, -33L), class = c("tbl_df", "tbl", "data.frame"
                                                                  )) 


folds <- bootstraps(dat, times = 10)

eq1_mod1_recipe <-
  recipe(c ~ q, data = dat) %>% 
  step_log(c, q, base = 10)

eq2_mod2_a_recipe <- 
  recipe(c_less_c_nought ~ q, data = dat) %>% 
  step_log(c_less_c_nought, q, base = 10)  

lm_model <- linear_reg()

res <-
  workflow_set(
    preproc = list(eq1m1 = eq1_mod1_recipe, e2m2a = eq2_mod2_a_recipe),
    models = list(lm = lm_model)
  ) %>%
  workflow_map("fit_resamples", resamples = folds)


collect_metrics(res)
#> # A tibble: 4 × 9
#>   wflow_id .config         preproc model .metric .estimator   mean     n std_err
#>   <chr>    <chr>           <chr>   <chr> <chr>   <chr>       <dbl> <int>   <dbl>
#> 1 eq1m1_lm Preprocessor1_… recipe  line… rmse    standard   0.0454    10 0.00214
#> 2 eq1m1_lm Preprocessor1_… recipe  line… rsq     standard   0.857     10 0.0220 
#> 3 e2m2a_lm Preprocessor1_… recipe  line… rmse    standard   0.0502    10 0.00245
#> 4 e2m2a_lm Preprocessor1_… recipe  line… rsq     standard   0.856     10 0.0221

由reprex 包于 2022-02-23 创建(v2.0.1)

于 2022-02-23T22:50:54.110 回答