Inverting a regex in R

You can use

vec <- c("19980213", "19980214", "19980215", "19980216","19980217", "iffi","geometry", "date_consid")
grep("^(\\d{8}|date_consid)$", vec, value=TRUE, invert=TRUE)
## => [1] "iffi"     "geometry"

See the R demo

The ^(\d{8}|date_consid)$ regex matches a string that only consists of any eight digits or that is equal to date_consid.

The value=TRUE makes grep return values rather than indices and invert=TRUE inverses the regex match result (returns those that do not match).

The pattern that you tried gives all the matches because the lookaheads are unanchored.

Using separate statements with or | will still match all strings.

You can change to logic to asserting from the start of the string, what is directly to the right is not either 8 digits or date_consid in a single check.

Using a positive lookahead, you have to add perl=T and add an anchor ^ to assert the start of the string and add an anchor $ to assert the end of the string after the lookahead.

 ^(?!\\d{8}$|date_consid$)

• ^ Start of string
• (?! Negative lookahead
  • \\d{8}$ Match 8 digits until end of string
  • | Or
  • date_consid$Match date_consid until end of string
• ) Close lookahead

For example

vec <- c("19980213", "19980214", "19980215", "19980216","19980217", "iffi","geometry", "date_consid")
grep("^(?!\\d{8}$|date_consid$)", vec, value=T, perl=T)

Output

[1] "iffi"     "geometry"