如何快速清理职业清单？

Question

我需要清理一个包含200多万个人及其职业条目的数据库。现在，我研究了一些独特的职业，我遇到了一些我需要清理的东西。例如，在职业列表中，对人力资源员工的描述有以下变化：

HR-employee
HR: employee
HR:Employee
HR - employee 
HR : employee
HR employee

现在，我需要找到一种方法来获取上面描述的所有条目，以获得“人力资源员工”的描述。对其他2000份工作说明也是如此。有没有一种简单的方法可以将所有这些“重复的职务说明”合并成一个职务说明？

Answer 1

假设"employee“的变体都在"HR”之前，如您的示例所示，您可以使用str_detect和ifelse

library(dplyr)
library(stringr)
df %>%
  mutate(job_clean = ifelse(str_detect(job, "(?i)\\bemployee\\b"),
                            "HR Employee",
                            job))
             job       job_clean
1     HR-employee     HR Employee
2    HR: employee     HR Employee
3 ZZ - Unemployed ZZ - Unemployed
4     HR:Employee     HR Employee
5  XY : smth else  XY : smth else
6   HR - employee     HR Employee
7   HR : employee     HR Employee
8     HR employee     HR Employee

请注意，(?i)用于使该匹配不区分大小写，而\\b是一个单词边界标记，用于断言只匹配确切的单词"employee“，而不是例如"employees”。

数据：

df <- data.frame(
  job = c("HR-employee", "HR: employee", "ZZ - Unemployed", "HR:Employee", "XY : smth else", "HR - employee", "HR : employee", "HR employee"))

编辑

如果不仅有"HR“类别ut也有”楼层“类别，那么您可以执行嵌套的操作：

df %>%
  mutate(job_clean = ifelse(str_detect(job, "(?i)hr.*\\bemployee\\b"),"HR Employee",
                            ifelse(str_detect(job, "(?i)floor.*\\bemployee\\b"),"Floor Employee", job)))
               job       job_clean
1      HR-employee     HR Employee
2     HR: employee     HR Employee
3  ZZ - Unemployed ZZ - Unemployed
4      HR:Employee     HR Employee
5   XY : smth else  XY : smth else
6    HR - employee     HR Employee
7    HR : employee     HR Employee
8      HR employee     HR Employee
9   Floor employee  Floor Employee
10  Floor-employee  Floor Employee

数据：

df <- data.frame(
  job = c("HR-employee", "HR: employee", "ZZ - Unemployed", "HR:Employee", "XY : smth else", "HR - employee", "HR : employee", "HR employee",
          "Floor employee", "Floor-employee"))