用R中的部分匹配替换整个单词或单词

Question

我有一个包含数千个拼写错误的城市名称的数据框。我需要更正这些并且无法找到解决方案，尽管我已经广泛搜索了。我尝试了几种功能和方法

这是数据的微型样本:

citA <- data.frame("num" = c(1,2,3,4,5,6,7,8),
               "city" = c("BORNE","BOERNAE","BARNE","BOERNE",
                          "GALDEN","GELDON","GOELDEN","GOLDEN"))

   num    city
1   1   BORNE
2   2 BOERNAE
3   3   BARNE
4   4  BOERNE
5   5  GALDEN
6   6  GELDON
7   7 GOELDEN
8   8  GOLDEN

这些是我尝试过的一些函数，还尝试了更多，包括 str_replace 和 str_detect:

cit <- function(x){
  ifelse(x %in% grepl(c("BOR","BOE","BAR")),"BOERNE",
         ifelse(x %in% grepl(c("GAL","GEL","GOE")), "GOLDEN", "OTHER"))
}

或者

cit <- function(x){
  ifelse(x %in% c("BOR","BOE","BAR"),"BOERNE",
         ifelse(x %in% c("GAL","GEL","GOE"), "GOLDEN", "OTHER"))
}

运行代码:

`citA$city2 <- cit(citA$city)`

错误的结果:

  num    city city2
1   1  BOERNE OTHER
2   2 BOERNAE OTHER
3   3   BARNE OTHER
4   4  BOERNE OTHER
5   5  GALDEN OTHER
6   6  GELDON OTHER
7   7 GOELDEN OTHER
8   8  GOLDEN OTHER

还试过:

citA$city[grepl(c("BOR","BOE","BAR"),citA$city)] <- "BOERNE" 

但这会引发错误:

Warning message:
In grepl(c("BOR", "BOE", "BAR"), citA$city) :
  argument 'pattern' has length > 1 and only the first element will be used

您的想法会很有帮助!

Answer 1

如果你有很多这样的模式，你可以使用 dplyr 中的 case_when :

library(dplyr)
library(stringr)

citA %>%
  mutate(city2 = case_when(str_detect(city, 'BOR|BOE|BAR') ~ 'BOERNE', 
                           str_detect(city, 'GAL|GEL|GOE|GOL') ~ 'GOLDEN',
                           TRUE ~ 'OTHER'))

#  num    city  city2
#1   1   BORNE BOERNE
#2   2 BOERNAE BOERNE
#3   3   BARNE BOERNE
#4   4  BOERNE BOERNE
#5   5  GALDEN GOLDEN
#6   6  GELDON GOLDEN
#7   7 GOELDEN GOLDEN
#8   8  GOLDEN GOLDEN