食谱本身没有自动执行此操作的方法,但我认为您可以创建一个为您处理此问题的函数,如下所示:
library(recipes)
#> Loading required package: dplyr
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
#>
#> Attaching package: 'recipes'
#> The following object is masked from 'package:stats':
#>
#> step
data(crickets, package = "modeldata")
levels_more_than <- function(vec, num = 2) {
n_distinct(levels(vec)) > num
}
recipe(~ ., data = crickets) %>%
step_dummy(species, one_hot = !! levels_more_than(crickets$species)) %>%
prep() %>%
bake(new_data = NULL)
#> # A tibble: 31 × 3
#> temp rate species_O..niveus
#> <dbl> <dbl> <dbl>
#> 1 20.8 67.9 0
#> 2 20.8 65.1 0
#> 3 24 77.3 0
#> 4 24 78.7 0
#> 5 24 79.4 0
#> 6 24 80.4 0
#> 7 26.2 85.8 0
#> 8 26.2 86.6 0
#> 9 26.2 87.5 0
#> 10 26.2 89.1 0
#> # … with 21 more rows
recipe(~ ., data = iris) %>%
step_dummy(Species, one_hot = !! levels_more_than(iris$Species)) %>%
prep() %>%
bake(new_data = NULL)
#> # A tibble: 150 × 7
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species_setosa
#> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 5.1 3.5 1.4 0.2 1
#> 2 4.9 3 1.4 0.2 1
#> 3 4.7 3.2 1.3 0.2 1
#> 4 4.6 3.1 1.5 0.2 1
#> 5 5 3.6 1.4 0.2 1
#> 6 5.4 3.9 1.7 0.4 1
#> 7 4.6 3.4 1.4 0.3 1
#> 8 5 3.4 1.5 0.2 1
#> 9 4.4 2.9 1.4 0.2 1
#> 10 4.9 3.1 1.5 0.1 1
#> # … with 140 more rows, and 2 more variables: Species_versicolor <dbl>,
#> # Species_virginica <dbl>
由reprex 包于 2022-02-23 创建(v2.0.1)
以下是在 recipes 中使用非标准选择器的一些技巧。