如何在R中使用模糊匹配来加入数据?

时间:2019-07-05 19:19:52

标签: r fuzzy-search

我有一些主题和许可证数据,并且想创建一列来标记许可证是否适合列出的主题。另一个挑战是,有些老师教多个科目,用分号隔开,每个许可都有几个可接受的科目。

我认为我需要合并grep之类的东西,但是我不确定如何添加此功能,同时还要合并两个表中的数据。

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,其中包括所有许可以及每个许可下的可接受程序:

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")
)

我想做的是创建一列来标记主题/许可组合是否可接受:

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"))

在此先感谢您提供的任何帮助!

1 个答案:

答案 0 :(得分:4)

这里是tidyversefuzzyjoin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