str_replace: replacement depending on wildcard value [A-Z]

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I have a number of strings containing the pattern "of" followed by an uppercase letter without spaces (in regex: "of[A-Z]"). I want to add spaces, e.g. "PrinceofWales" should become "Prince of Wales" etc.). However, I couldn't find how to add the value of [A-Z] that was matched into the replacement value:

library(tidyverse) 

str_replace("PrinceofWales", "of[A-Z]", " of [A-Z]")
# Gives: Prince of [A-Z]ales
# Expected: Prince of Wales

str_replace("DukeofEdinburgh", "of[A-Z]", " of [A-Z]")
# Gives: Duke of [A-Z]dinburgh
# Expected: Duke of Edinburgh

Can someone enlighten me? :)

stringr tidyverse r
1 Answers

It needs to be captured as a group (([A-Z])) and replace with the backreference (\\1) of the captured group i.e. regex interpretation is in the pattern and not in the replacement

stringr::str_replace("PrinceofWales", "of([A-Z])", " of \\1")
[1] "Prince of Wales"

According to ?str_replace

replacement - A character vector of replacements. Should be either length one, or the same length as string or pattern. References of the form \1, \2, etc will be replaced with the contents of the respective matched group (created by ()).

Or another option is a regex lookaround

stringr::str_replace("PrinceofWales", "of(?=[A-Z])", " of ")
[1] "Prince of Wales"
user_37322710
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