基于变异中匹配正则表达式的条件str_replace？

Question

对于“区”列中与regex("[:alpha:]{2}AL") 匹配的任何条目，我想将“AL”替换为“01”。

例如：

df <- tibble(district = c("NY14", "MT01", "MTAL", "PA10", "KS02", "NDAL", "ND01", "AL02", "AL01"))

我试过了：

  df %>% mutate(district=replace(district, 
                          str_detect(district, regex("[:alpha:]{2}AL")), 
                          str_replace(district,"AL","01"))) 

和

  df %>% mutate(district=replace(district, 
                          str_detect(district, regex("[:alpha:]{2}AL")), 
                          paste(str_sub(district, start = 1, end = 2),"01",sep = "")) 

但存在矢量化问题。

Answer 1

这样好吗？

str_replace_all(string=df$district,
                pattern="(\\w{2})AL",
                replacement="\\101")

我将正则表达式替换为\\w，一个单词字符：https://www.regular-expressions.info/shorthand.html

我使用\\1 表示将字符串替换为第一个捕获的区域，该区域在(\\w{2}) 中捕获，因此保留前2 个字母，然后添加01

Answer 2

您可以将replace 更改为ifelse

ifelse( str_detect(df$district, regex("[:alpha:]{2}AL")), 
         str_replace(df$district,"AL","01"),df$district)