r - 在 dplyr select 中使用 perl=TRUE regex

Question

如何使用 perl = TRUE 选择列像正则表达式。

data.frame(baa=0,boo=0,boa=0,lol=0,bAa=0) %>% dplyr::select(matches("(?i)b(?!a)"))

Error in grep(needle, haystack, ...) : invalid regular expression '(?i)b(?!a)', reason 'Invalid regexp'

正则表达式确实有效。

grep("(?i)b(?!a)",c("baa","boo","boa","lol","bAa"),perl=T)

> [1] 2 3

有快捷功能/方式吗？

Answer 1

matches在 dplyr不支持perl = TRUE .但是，您可以创建自己的函数。在对源代码进行一些挖掘之后，这有效:

快捷方式:

library(dplyr)

#notice the 3 colons because grep_vars is not exported from dplyr
matches2 <- function (match, ignore.case = TRUE, vars = current_vars()) 
{
  dplyr:::grep_vars(match, vars, ignore.case = ignore.case, perl = TRUE)
}

data.frame(baa=0,boo=0,boa=0,lol=0,bAa=0) %>% select(matches2("(?i)b(?!a)"))
#boo boa
#1   0   0

或者更解释性的解决方案:

matches2 <- function (match, ignore.case = TRUE, vars = current_vars()) 
{
  grep_vars2(match, vars, ignore.case = ignore.case)
}

#this is pretty much my only change in the original dplyr:::grep_vars
#to make it accept perl.
grep_vars2 <- function (needle, haystack, ...) 
{
  grep(needle, haystack, perl = TRUE, ...)
}

 data.frame(baa=0,boo=0,boa=0,lol=0,bAa=0) %>% 
   select(matches2("(?i)b(?!a)"))
 #boo boa
 #1   0   0