如何在我的dataframe中创建一个列来映射到R中的列表？

Question

我有一些许可数据，并试图在我的dataframe中创建一个列，该列根据某人注册的程序告诉我所列出的许可是否可以接受。

为了做到这一点，我创建了一个列表，因为其中一些许可对于多个程序是可以接受的。理想情况下，我想的是，我可以以某种方式使用这个列表作为参考，以查看程序是否在许可名称下列出。我也尝试过case_when，但是不断地出错。我也希望有一个列表，我可以使用作为一种地图，因为许可证名称可能会每年变化。

示例代码

以下是我的数据摘要：

df1 <- data.frame(Program = c("Elementary Education", "Elementary Education", "Secondary Math", "Secondary Math", "Secondary ELA", "Secondary ELA"), Licensure = c("Content Area - Elementary Education (Grades 1-6)", "Content Area - Secondary Math (Grades 7-12)", "Content Area - Secondary Math (Grades 7-12)", "Mathematics (Grades 7-12) 1706", "Content Area - Secondary ELA (Grades 7-12)", "Content Area - Early Childhood (preK-Grade 3)"))

下面是我创建的列表，其中包含了所有许可，每个许可下面都有可接受的程序：

license_index <- list(
  "Content Area - Early Childhood (preK-Grade 3)" = "Elementary Education",
  "Content Area - Elementary Education (Grades 1-6)" = "Elementary Education",
  "Content Area - Middle Grades ELA (Grades 4-9)" = c("Elementary Education", "Secondary ELA"),
  "Content Area - Middle Grades Math (Grades 4-9)" = c("Elementary Education", "Secondary Math"),
  "Content Area - Middle School Mathematics (Grades 4-8)" = "Elementary Education",
  "Content Area - Secondary ELA (Grades 7-12)" = "Secondary ELA",
  "Content Area - Secondary Math (Grades 7-12)" = "Secondary Math",
  "Content Area - Secondary English (Grades 7-12)" = "Secondary ELA",
  "English Language Arts and Reading (Grades 4-8) 864" = "Elementary Education",
  "Core Subjects (Grades EC-6) 1770" = "Elementary Education",
  "English Language Arts and Reading (Grades 7-12) 1709" = "Secondary ELA",
  "Mathematics (Grades 4-8) 866" = "Elementary Education",
  "Mathematics (Grades 7-12) 1706" = "Secondary Math"
)

理想情况下，作为最后一栏，我想要的是许可和程序是否匹配：

ideal.df <- data.frame(Program = c("Elementary Education", "Elementary Education", "Secondary Math", "Secondary Math", "Secondary ELA", "Secondary ELA"), Licensure = c("Content Area - Elementary Education (Grades 1-6)", "Content Area - Secondary Math (Grades 7-12)", "Content Area - Secondary Math (Grades 7-12)", "Mathematics (Grades 7-12) 1706", "Content Area - Secondary ELA (Grades 7-12)", "Content Area - Early Childhood (preK-Grade 3)"), match = c("Match", "No", "Match", "Match", "Match", "No"))

我想我需要变异函数，也许我需要使用purrr映射函数，但我对tidyverse不太熟悉，我真的很感激帮助！提前感谢！

Answer 1

下面是使用tidyverse的一种方法，我们将命名的list转换为带有enframe的两列data.frame，用原始数据集将right_join转换为right_join，并通过将列“match”与“Program”进行比较来创建match

library(tidyverse)
enframe(license_index, name = "Licensure", value = "match") %>%
    unnest %>% 
    right_join(df1) %>% 
    mutate(match = match == Program) %>%
    select(names(df1), everything())
# A tibble: 6 x 3
#  Program              Licensure                                        match
#  <fct>                <chr>                                            <lgl>
#1 Elementary Education Content Area - Elementary Education (Grades 1-6) TRUE 
#2 Elementary Education Content Area - Secondary Math (Grades 7-12)      FALSE
#3 Secondary Math       Content Area - Secondary Math (Grades 7-12)      TRUE 
#4 Secondary Math       Mathematics (Grades 7-12) 1706                   TRUE 
#5 Secondary ELA        Content Area - Secondary ELA (Grades 7-12)       TRUE 
#6 Secondary ELA        Content Area - Early Childhood (preK-Grade 3)    FALSE

或者我们可以使用说唱包，这对这个场景很有帮助。

library(rap)
df1 %>% 
   rap(match = ~ license_index[[as.character(Licensure)]] == Program) %>%
   unnest
#              Program                                        Licensure match
#1 Elementary Education Content Area - Elementary Education (Grades 1-6)  TRUE
#2 Elementary Education      Content Area - Secondary Math (Grades 7-12) FALSE
#3       Secondary Math      Content Area - Secondary Math (Grades 7-12)  TRUE
#4       Secondary Math                   Mathematics (Grades 7-12) 1706  TRUE
#5        Secondary ELA       Content Area - Secondary ELA (Grades 7-12)  TRUE
#6        Secondary ELA    Content Area - Early Childhood (preK-Grade 3) FALSE

Answer 2

试试这个：

x<-stack(license_index)
x$values[match(df1$Licensure,x$ind)]==df1$Program
#[1]  TRUE FALSE  TRUE  TRUE  TRUE FALSE

如果需要，可以使用TRUE和No映射上述值的Match和FALSE值。

Answer 3

我真的帮不了你，但是这个基-R解决方案应该有效：

df1$match <-sapply(1:nrow(df1), function(i){
  ifelse(license_index[[which(names(license_index) == df1$Licensure[i])]] == df1$Program[i],'Match','No')})