Applying a function to a backreference within gsub in R

Question

I\'m new to R and am stuck with backreferencing that doesn\'t seem to work. In:

gsub(\"\\\\((\\\\d+)\\\\)\", f(\"\\\\1\"), string)

It corre

Answer 1

Here's a way by tweaking a bit stringr::str_replace(), in the replace argument, just use a lambda formula as the replace argument, and reference the captured group not by ""\\1" but by ..1, so your gsub("\\((\\d+)\\)", f("\\1"), string) will become str_replace2(string, "\\((\\d+)\\)", ~f(..1)), or just str_replace2(string, "\\((\\d+)\\)", f) in this simple case :

str_replace2 <- function(string, pattern, replacement, type.convert = TRUE){
  if(inherits(replacement, "formula"))
    replacement <- rlang::as_function(replacement)
  if(is.function(replacement)){
    grps_mat <- stringr::str_match(string, pattern)[,-1, drop = FALSE]
    grps_list <- lapply(seq_len(ncol(grps_mat)), function(i) grps_mat[,i])
    if(type.convert) {
      grps_list <- type.convert(grps_list, as.is = TRUE) 
      replacement <- rlang::exec(replacement, !!! grps_list)
      replacement <- as.character(replacement)
    } else {
      replacement <- rlang::exec(replacement, !!! grps_list)
    }
  }
  stringr::str_replace(string, pattern, replacement)
}

str_replace2(
  "foo (4)",
  "\\((\\d+)\\)", 
  sqrt)
#> [1] "foo 2"

str_replace2(
  "foo (4) (5)",
  "\\((\\d+)\\) \\((\\d+)\\)", 
  ~ sprintf("(%s)", ..1 * ..2))
#> [1] "foo (20)"

^{Created on 2020-01-24 by the reprex package (v0.3.0)}