ap5*_*p53 2 r r-recipes tidymodels
这与使用 step_naomit 进行预测并使用 tidymodels 保留 ID是相同的问题，但即使有一个可接受的答案，OP 的最后一条评论指出了“id 变量”被用作预测器的问题，正如在查看时所看到的model$fit$variable.importance.
我有一个我想保留的带有“id 变量”的数据集。我以为我可以通过 recipe() 规范来实现这一点。
library(tidymodels)
# label is an identifier variable I want to keep even though it's not
# a predictor
df <- tibble(label = 1:50,
x = rnorm(50, 0, 5),
f = factor(sample(c('a', 'b', 'c'), 50, replace = TRUE)),
y = factor(sample(c('Y', 'N'), 50, replace = TRUE)) )
df_split <- initial_split(df, prop = 0.70)
# Make up any recipe: just note I specify 'label' as "id variable"
rec <- recipe(training(df_split)) %>%
update_role(label, new_role = "id variable") %>%
update_role(y, new_role = "outcome") %>%
update_role(x, new_role = "predictor") %>%
update_role(f, new_role = "predictor") %>%
step_corr(all_numeric(), -all_outcomes()) %>%
step_dummy(all_predictors(),-all_numeric()) %>%
step_meanimpute(all_numeric(), -all_outcomes())
train_juiced <- prep(rec, training(df_split)) %>% juice()
logit_fit <- logistic_reg(mode = "classification") %>%
set_engine(engine = "glm") %>%
fit(y ~ ., data = train_juiced)
# Why is label a variable in the model ?
logit_fit[['fit']][['coefficients']]
#> (Intercept) label x f_b f_c
#> 1.03664140 -0.01405316 0.22357266 -1.80701531 -1.66285399
由reprex 包(v0.3.0)于 2020 年 1 月 27 日创建
但即使我确实指定label了一个 id 变量，它也被用作预测器。所以也许我可以在公式中使用我想要的特定术语，并专门添加label为 id 变量。
rec <- recipe(training(df_split), y ~ x + f) %>%
update_role(label, new_role = "id variable") %>%
step_corr(all_numeric(), -all_outcomes()) %>%
step_dummy(all_predictors(),-all_numeric()) %>%
step_meanimpute(all_numeric(), -all_outcomes())
#> Error in .f(.x[[i]], ...): object 'label' not found
由reprex 包(v0.3.0)于 2020 年 1 月 27 日创建
我可以试着不提 label
rec <- recipe(training(df_split), y ~ x + f) %>%
step_corr(all_numeric(), -all_outcomes()) %>%
step_dummy(all_predictors(),-all_numeric()) %>%
step_meanimpute(all_numeric(), -all_outcomes())
train_juiced <- prep(rec, training(df_split)) %>% juice()
logit_fit <- logistic_reg(mode = "classification") %>%
set_engine(engine = "glm") %>%
fit(y ~ ., data = train_juiced)
# Why is label a variable in the model ?
logit_fit[['fit']][['coefficients']]
#> (Intercept) x f_b f_c
#> -0.98950228 0.03734093 0.98945339 1.27014824
train_juiced
#> # A tibble: 35 x 4
#> x y f_b f_c
#> <dbl> <fct> <dbl> <dbl>
#> 1 -0.928 Y 1 0
#> 2 4.54 N 0 0
#> 3 -1.14 N 1 0
#> 4 -5.19 N 1 0
#> 5 -4.79 N 0 0
#> 6 -6.00 N 0 0
#> 7 3.83 N 0 1
#> 8 -8.66 Y 1 0
#> 9 -0.0849 Y 1 0
#> 10 -3.57 Y 0 1
#> # ... with 25 more rows
由reprex 包(v0.3.0)于 2020 年 1 月 27 日创建
好的，模型可以工作了，但是我丢失了label.
我该怎么做？
Jul*_*lge 10
您遇到的主要问题/概念问题是，一旦您juice()编写了配方，它就只是 data，即字面上的数据框。当您使用它来拟合模型时，模型无法知道某些变量具有特殊作用。
library(tidymodels)
# label is an identifier variable to keep even though it's not a predictor
df <- tibble(label = 1:50,
x = rnorm(50, 0, 5),
f = factor(sample(c('a', 'b', 'c'), 50, replace = TRUE)),
y = factor(sample(c('Y', 'N'), 50, replace = TRUE)) )
df_split <- initial_split(df, prop = 0.70)
rec <- recipe(y ~ ., training(df_split)) %>%
update_role(label, new_role = "id variable") %>%
step_corr(all_numeric(), -all_outcomes()) %>%
step_dummy(all_predictors(),-all_numeric()) %>%
step_meanimpute(all_numeric(), -all_outcomes()) %>%
prep()
train_juiced <- juice(rec)
train_juiced
#> # A tibble: 35 x 5
#> label x y f_b f_c
#> <int> <dbl> <fct> <dbl> <dbl>
#> 1 1 1.80 N 1 0
#> 2 3 1.45 N 0 0
#> 3 5 -5.00 N 0 0
#> 4 6 -4.15 N 1 0
#> 5 7 1.37 Y 0 1
#> 6 8 1.62 Y 0 1
#> 7 10 -1.77 Y 1 0
#> 8 11 -3.15 N 0 1
#> 9 12 -2.02 Y 0 1
#> 10 13 2.65 Y 0 1
#> # … with 25 more rows
请注意，这train_juiced实际上只是一个普通的小标题。如果您使用 对这个 tibble 训练模型fit()，它不会知道用于转换数据的配方的任何信息。
tidymodels 框架确实有一种方法可以使用配方中的角色信息来训练模型。可能最简单的方法是使用工作流。
logit_spec <- logistic_reg(mode = "classification") %>%
set_engine(engine = "glm")
wf <- workflow() %>%
add_model(logit_spec) %>%
add_recipe(rec)
logit_fit <- fit(wf, training(df_split))
# No more label in the model
logit_fit
#> ?? Workflow [trained] ??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
#> Preprocessor: Recipe
#> Model: logistic_reg()
#>
#> ?? Preprocessor ????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
#> 3 Recipe Steps
#>
#> ? step_corr()
#> ? step_dummy()
#> ? step_meanimpute()
#>
#> ?? Model ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
#>
#> Call: stats::glm(formula = formula, family = stats::binomial, data = data)
#>
#> Coefficients:
#> (Intercept) x f_b f_c
#> 0.42331 -0.04234 -0.04991 0.64728
#>
#> Degrees of Freedom: 34 Total (i.e. Null); 31 Residual
#> Null Deviance: 45
#> Residual Deviance: 44.41 AIC: 52.41
由reprex 包(v0.3.0)于 2020 年 2 月 15 日创建
模型中不再有标签！