如何使用R中的模糊匹配连接数据？
如何使用R中的模糊匹配连接数据？
提问于 2019-07-05 19:19:52
我有一些主题和许可数据，并希望创建一个列，该列标记给定所列主题的许可是否是合适的。另外的挑战是，一些教师教授多门学科，用分号分隔，每个许可证都有几个可以接受的科目。
我认为我需要合并一些类似grep的内容，但我不太确定如何在加入两个表的数据的同时添加这个函数。
示例代码
以下是我的数据摘要：
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)"))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：
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")
)我想要做的是创建一个列，标记subject/license组合是否可以接受：
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"))谢谢您能提供的任何帮助！
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-07-05 19:46:31
下面是tidyverse和fuzzyjoin的一个选项
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$flag
#[1] TRUE TRUE TRUE TRUE TRUE FALSE FALSE TRUE FALSE TRUE-checking OP的理想输出
as.logical(ideal.df$flag)
#[1] TRUE TRUE TRUE TRUE TRUE FALSE FALSE TRUE FALSE TRUE页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56908276
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