如何遍历逻辑谓词列表？答案

【问题标题】：How can I iterate over a list of logical predicates?如何遍历逻辑谓词列表？
【发布时间】：2018-01-31 16:05:11
【问题描述】：

更新：以下问题可能是初学者在尝试编写包含一些dplyr 操作的自定义函数时不遵循的常见方法。在这些情况下，tt可能是时候学习非标准评估的概念和最佳实践了：http://dplyr.tidyverse.org/articles/programming.html

我正在尝试映射数据集以替换多个值。鉴于我想多次这样做，是否有可能（或一个好主意）创建一个包含指定逻辑谓词的列表，然后 purrr::map 它们？

例如，在starwars 数据集中，如果我想替换满足特定条件的skin_color 值，我可以这样做：

library(tidyverse)
#> -- Attaching packages -------------------------------------------------------------- tidyverse 1.2.1 --
#> v ggplot2 2.2.1.9000     v purrr   0.2.4     
#> v tibble  1.4.2          v dplyr   0.7.4     
#> v tidyr   0.7.2          v stringr 1.2.0     
#> v readr   1.1.1          v forcats 0.2.0
#> -- Conflicts ----------------------------------------------------------------- tidyverse_conflicts() --
#> x dplyr::filter() masks stats::filter()
#> x dplyr::lag()    masks stats::lag()

replaced_starwars <- starwars %>% 
  mutate(skin_color = replace(skin_color, (hair_color == "none" & eye_color == "black"), 
                              "SOMETHING_HERE"))

head(filter(replaced_starwars, hair_color == "none" & eye_color == "black"), 5)
#> # A tibble: 5 x 13
#>   name    height  mass hair_color skin_color   eye_color birth_year gender
#>   <chr>    <int> <dbl> <chr>      <chr>        <chr>          <dbl> <chr> 
#> 1 Nien N~    160  68.0 none       SOMETHING_H~ black           NA   male  
#> 2 Gasgano    122  NA   none       SOMETHING_H~ black           NA   male  
#> 3 Kit Fi~    196  87.0 none       SOMETHING_H~ black           NA   male  
#> 4 Plo Ko~    188  80.0 none       SOMETHING_H~ black           22.0 male  
#> 5 Lama Su    229  88.0 none       SOMETHING_H~ black           NA   male  
#> # ... with 5 more variables: homeworld <chr>, species <chr>, films <list>,
#> #   vehicles <list>, starships <list>

如何将逻辑谓词存储在列表中，然后用my_function 映射它们？

my_function <- function(my_data, lgc_predicates, replacement){
  out <- my_data %>% mutate(species = replace(species, lgc_predicates, replacement))
  return(out)
}

pred_list <- list('hair_color == "blond" & eye_color == "yellow"',
                  'hair_color == "none" & eye_color == "brown"',
                  'hair_color == "brown" & eye_color == "red"',
                  'hair_color == "none" & eye_color == "black"')

replaced_starwars <- map(starwars, 
                     my_function, 
                     lgc_predicates = pred_list, replacement = "SOMETHING_HERE")
#> Error in UseMethod("mutate_"): no applicable method for 'mutate_' applied to an object of class "character"

【问题讨论】：

您希望得到什么输出？一份清单就足够了吗？
在示例中，我希望有一个新的starwars（例如replaced_starwars）数据框，其中skin_color 值被替换为替换。
你的地图看起来很奇怪。听起来您想将一个函数映射到您的 pred_list 并将数据集 (starwars) 作为自定义函数的附加参数。您可能想要交换 my_function 中的参数和对 map 的调用

标签： r dplyr purrr

【解决方案1】：

在这种情况下，我个人喜欢dplyr::case_when

output <- starwars %>%
            mutate(skin_color = case_when(
                hair_color == "blond" & eye_color == "yellow" ~ "SOMETHING_HERE",
                hair_color == "none" & eye_color == "brown" ~ "SOMETHING_HERE",
                hair_color == "brown" & eye_color == "red" ~ "SOMETHING_HERE",
                hair_color == "none" & eye_color == "black" ~ "SOMETHING_HERE",
                TRUE ~ skin_color))

head(filter(output, hair_color == "none" & eye_color == "black"), 5)

# A tibble: 5 x 13
  # name  heig~  mass hair~ skin~ eye_~ birt~ gend~ home~ spec~ films vehi~ star~
  # <chr> <int> <dbl> <chr> <chr> <chr> <dbl> <chr> <chr> <chr> <lis> <lis> <lis>
# 1 Nien~   160  68.0 none  SOME~ black  NA   male  Sull~ Sull~ <chr~ <chr~ <chr~
# 2 Gasg~   122  NA   none  SOME~ black  NA   male  Troi~ Xexto <chr~ <chr~ <chr~
# 3 Kit ~   196  87.0 none  SOME~ black  NA   male  Glee~ Naut~ <chr~ <chr~ <chr~
# 4 Plo ~   188  80.0 none  SOME~ black  22.0 male  Dorin Kel ~ <chr~ <chr~ <chr~
# 5 Lama~   229  88.0 none  SOME~ black  NA   male  Kami~ Kami~ <chr~ <chr~ <chr~

你甚至可以传递一个参数列表，但你必须取消引用列表和列名：

pred_list <- list(!! hair_color == "blond" & !! eye_color == "yellow" ~ "SOMETHING_HERE",
                  !! hair_color == "none" & !! eye_color == "brown" ~ "SOMETHING_HERE",
                  !! hair_color == "brown" & !! eye_color == "red" ~ "SOMETHING_HERE",
                  !! hair_color == "none" & !! eye_color == "black" ~ "SOMETHING_HERE",
                  TRUE ~ !! skin_color)

output <- starwars %>%
        mutate(skin_color = case_when(!!! pred_list))

【讨论】：