Use perl=TRUE regex in dplyr select

Question

How can I select cols using perl = TRUE like regex.

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'

regex is indeed valid.

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

> [1] 2 3

Is there a shortcut function/way?

Answer 1

matches in dplyr does not support perl = TRUE. However, you can make your own functions. After a bit of digging in the source code this works:

The fast way:

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

Or a more explanatory solution:

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

Answer 2

Another approach, although along the lines and probably more dangerous than LyzandeR's suggestion:

body(matches)[[grep("grep_vars", body(matches))]] <- substitute(grep_vars(match, vars, ignore.case = ignore.case, perl=T))

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

I would not use body(matches)[[3]] as any updates would cause this little patch create problems.