Question

namez <- c("foo2003", "bar2340", "naught45")
patternz <- "03"
grepl("[patternz]$",namez)

It does not work. What should I substitute [patternz] with, so the regular expression will match the contents of the patternz variable?

[edit] Notice that I want to match the string “03”, not the digits “0” and “3” separately.

Answer 1

Must admit to struggling to see what the problem is here. For the example stated nothing more than

R> namez <- c("foo2003", "bar2340", "naught45")
R> patternz <- "03"
R> grepl(patternz, namez)
[1]  TRUE FALSE FALSE

is required as patternz is a character vector and the aim is not to match 0 & 3 but to match the literal "03"

If you need this to match only at the end of the strings, then we do need to add "$" either by hand:

R> patternz2 <- "03$"
R> grepl(patternz2, namez)
[1]  TRUE FALSE FALSE

or via a paste0() operation

R> grepl(paste0(patternz, "$"), namez)
[1]  TRUE FALSE FALSE

The issue is to use patternz as the actual regexp and base R functions handle this perfectly.