Question

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

例如：

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

我尝试过：

  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 = ""))

但是存在向量化问题。

Answer 1

可以吗？

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

我用\\w（一个文字字符：https://www.regular-expressions.info/shorthand.html

）替换了正则表达式

我正在使用\\1来表示将字符串替换为在(\\w{2})中捕获的第一个捕获区域，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)

基于str中匹配正则表达式的条件str_replace？

2 个答案: