为什么 tidymodels/recipes 中的 "id variable" 会起到预测作用？
Why does an "id variable" in tidymodels/recipes play a predictor role?
这与 是同一个问题，但即使有一个可接受的答案，OP 的最后评论指出 "id variable" 被用作预测变量的问题，正如查看 model$fit$variable.importance.
时所见
我有一个 "id variables" 的数据集，我想保留。
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 package (v0.3.0)
于 2020-01-27 创建
但即使我确实指定 label 是一个 id 变量，它也被用作预测变量。
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 作为 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 package (v0.3.0)
于 2020-01-27 创建
我可以尽量不提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 package (v0.3.0)
于 2020-01-27 创建
好的，模型可以用了，但是我丢失了我的 label。
我应该怎么做？
您 运行 遇到的主要 issue/conceptual 问题是，一旦您 juice() 配方，它 只是数据 ，即只是从字面上看是一个数据框。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，train_juiced 只是一个普通的小标题。如果您使用 fit() 在这个 tibble 上训练一个模型，它不会知道任何关于用于转换数据的方法。
tidymodels 框架 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 workflows.
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 package (v0.3.0)
于 2020-02-15 创建
模型中没有更多标签！
这与 model$fit$variable.importance.
我有一个 "id variables" 的数据集，我想保留。 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 package (v0.3.0)
于 2020-01-27 创建但即使我确实指定 label 是一个 id 变量，它也被用作预测变量。
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 作为 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 package (v0.3.0)
于 2020-01-27 创建我可以尽量不提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 package (v0.3.0)
于 2020-01-27 创建好的，模型可以用了，但是我丢失了我的 label。
我应该怎么做？
您 运行 遇到的主要 issue/conceptual 问题是，一旦您 juice() 配方，它 只是数据 ，即只是从字面上看是一个数据框。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，train_juiced 只是一个普通的小标题。如果您使用 fit() 在这个 tibble 上训练一个模型，它不会知道任何关于用于转换数据的方法。
tidymodels 框架 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 workflows.
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 package (v0.3.0)
于 2020-02-15 创建模型中没有更多标签！