使用dplyr :: case_when进行整洁的评估编程

Question

我尝试编写一个包含dplyr :: case_when()函数的简单函数.我在https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html上阅读了dplyr文档的编程,但无法弄清楚它如何与case_when()函数一起工作.

我有以下数据:

data <- tibble(
   item_name = c("apple", "bmw", "bmw")
)

以下列表:

cat <- list(
   item_name == "apple" ~ "fruit",
   item_name == "bmw" ~ "car"
)

然后我想写一个函数,如:

category_fn <- function(df, ...){
   cat1 <- quos(...)
   df %>%
     mutate(category = case_when((!!!cat1)))
}

不幸的是category_fn(data,cat),在这种情况下会给出评估错误 我想获得与通过以下方式获得的输出相同的输出:

data %>% 
   mutate(category = case_when(item_name == "apple" ~ "fruit",
                               item_name == "bmw" ~ "car"))

这样做的方法是什么？

Answer 1

1)通列表使用let从wrapr包和data并cat从该工作没有以任何方式修改所述输入的问题.

library(dplyr)
library(wrapr)

category_fn <- function(data, List) {
  let(c(CATEGORY = toString(sapply(List, format))),
      data %>% mutate(category = case_when(CATEGORY)),
      subsMethod = "stringsubs",
      strict = FALSE)
}
category_fn(data, cat) # test

赠送:

# A tibble: 3 x 2
  item_name category
      <chr>    <chr>
1     apple    fruit
2       bmw      car
3       bmw      car

1a)中使用tidyeval/rlang和data和cat从所述的问题:

category_fn <- function(data, List) {
  cat_ <- lapply(List, function(x) do.call("substitute", list(x)))
  data %>% mutate(category = case_when(!!!cat_))
}
category_fn(data, cat)

给出与上述相同的结果.

2)单独传递列表组件如果您打算单独传递每个组件cat而不是cat它自己,那么这可以工作:

category_fn <- function(data, ...) eval.parent(substitute({
   data %>% mutate(category = case_when(...))
}))

category_fn(data, item_name == "apple" ~ "fruit",
                   item_name == "bmw" ~ "car") # test

赠送:

# A tibble: 3 x 2
  item_name category
      <chr>    <chr>
1     apple    fruit
2       bmw      car
3       bmw      car

2a)如果您更喜欢tidyeval/rlang,那么这种情况很简单:

library(dplyr)
library(rlang)

category_fn <- function(data, ...) {
   cat_ <- quos(...)
   data %>% mutate(category = case_when(!!!cat_))
}

category_fn(data, item_name == "apple" ~ "fruit",
                   item_name == "bmw" ~ "car") # test

Answer 2

首先引用列表中的每个元素:

cat <- list(
  quo(item_name == "apple" ~ "fruit"),
  quo(item_name == "bmw" ~ "car")
)

然后,您的函数不必引用cat对象本身.我还改变了使用"everything else"...参数来在调用中明确引用类别参数:

category_fn <- function(df, categories){
  df %>%
    mutate(category = case_when(!!!categories))
}

然后函数的输出符合预期:

category_fn(data, cat)
# A tibble: 3 x 2
  item_name category
      <chr>    <chr>
1     apple    fruit
2       bmw      car
3       bmw      car

为了完整起见,我注意到类别列表也适用于使用基本R quote()函数定义时的函数:

cat <- list(
  quote(item_name == "apple" ~ "fruit"),
  quote(item_name == "bmw" ~ "car")
)
> cat
[[1]]
item_name == "apple" ~ "fruit"

[[2]]
item_name == "bmw" ~ "car"

> category_fn(data, cat)
# A tibble: 3 x 2
  item_name category
      <chr>    <chr>
1     apple    fruit
2       bmw      car
3       bmw      car