如何在 R 中使用模糊匹配连接数据？

Question

我有一些主题和许可证数据，并想创建一个列来标记许可证是否适合列出的主题。额外的挑战是一些教师教授多个科目，用分号分隔，并且每个许可证都有几个可接受的科目。

我认为我需要合并诸如 grep 之类的东西，但我不太确定如何添加此功能，同时还要加入来自两个表的数据。

示例代码

以下是我的数据框的摘录：

df1 <- data.frame(Subject = c("Spanish Language Arts; I teach all subjects for my students", 
"Math; Science", "Mathematics; ELA", "ELA", "Science;Math;English Language Arts", 
"Spanish Language Arts; I teach all subjects for my students",
 "Math", "Science;Social Studies;Mathematics;English Language Arts", "ELA", 
"English Language Arts"), 
Licensure = c("Content Area - Early Childhood (preK-Grade 3)", 
"Core Subjects (Grades EC-6) 1770", "Mathematics (Grades 7-12) 1706", 
"English Language Arts and Reading (Grades 7-12) 1709", "Core Subjects (Grades EC-6) 1770", 
"English Language Arts and Reading (Grades 7-12) 1709", 
"English Language Arts and Reading (Grades 7-12) 1709", 
"Content Area - Elementary Education (Grades 1-6)", 
"Mathematics (Grades 7-12) 1706", "Content Area - Elementary Education (Grades 1-6)"))

这是我创建的列表，其中包括所有许可证以及每个许可证下的可接受程序：

lic.subject_index <- list(
  "Content Area - Early Childhood (preK-Grade 3)" = c("I teach all subjects for my students", "Math", "Mathematics", "ELA", "English Language Arts", "Language Arts"),
  "Content Area - Elementary Education (Grades 1-6)" = c("I teach all subjects for my students", "Math", "Mathematics", "ELA", "English Language Arts", "Language Arts"),
  "Core Subjects (Grades EC-6) 1770" = c("I teach all subjects for my students", "Math", "Mathematics", "ELA", "English Language Arts", "Language Arts"),
  "English Language Arts and Reading (Grades 7-12) 1709" = c("ELA", "English Language Arts", "Language Arts"),
  "Mathematics (Grades 7-12) 1706" = c("Math", "Mathematics")
)

我希望能够创建一个列来标记主题/许可证组合是否可接受：

ideal.df <- data.frame(Subject = c("Spanish Language Arts; I teach all subjects for my students", 
"Math; Science", "Mathematics; ELA", "ELA", "Science;Math;English Language Arts", 
"Spanish Language Arts; I teach all subjects for my students", "Math", 
"Science;Social Studies;Mathematics;English Language Arts", "ELA", "English Language Arts"), 
Licensure = c("Content Area - Early Childhood (preK-Grade 3)", "Core Subjects (Grades EC-6) 1770", 
"Mathematics (Grades 7-12) 1706", "English Language Arts and Reading (Grades 7-12) 1709", 
"Core Subjects (Grades EC-6) 1770", "English Language Arts and Reading (Grades 7-12) 1709", 
"English Language Arts and Reading (Grades 7-12) 1709", "Content Area - Elementary Education (Grades 1-6)", 
"Mathematics (Grades 7-12) 1706", "Content Area - Elementary Education (Grades 1-6)"), 
flag = c("True", "True", "True", "True", "True", "False", "False", "True", "False", "True"))

提前感谢您提供的任何帮助！

Answer 1

这是tidyverse 和fuzzyjoin 的选项

library(fuzzyjoin)
library(tidyverse)
out <- df1 %>%
       rownames_to_column('rn') %>% 
       separate_rows(Subject, sep = ';') %>% 
       stringdist_left_join(
         enframe(lic.subject_index, name = 'Licensure', value = 'Subject') %>% 
              unnest) %>% 
       group_by(rn = as.integer(rn)) %>%
       summarise(ind = any(!is.na(Licensure.y))) %>%
       ungroup %>% 
       pull(ind) %>% 
       mutate(df1, flag = .)
out$flag
#[1]  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE  TRUE FALSE  TRUE

-检查 OP 的理想输出

as.logical(ideal.df$flag)
#[1]  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE  TRUE FALSE  TRUE