String replace with regex condition

Question

I have a pattern that I want to match and replace with an X. However, I only want the pattern to be replaced if the preceding character is either an A, B or not preceeded by any character (beginning of string).

I know how to replace patterns using the str_replace_all function but I don't know how I can add this additional condition. I use the following code:

library(stringr)

string <- "0000A0000B0000C0000D0000E0000A0000"
pattern <- c("XXXX")



replacement <- str_replace_all(string, pattern, paste0("XXXX"))

Result:

[1] "XXXXAXXXXBXXXXCXXXXDXXXXEXXXXAXXXX"

Desired result:

Replacement only when preceding charterer is A, B or no character:

[1] "XXXXAXXXXBXXXXC0000D0000E0000AXXXX"

Answer 1

You may use

gsub("(^|[AB])0000", "\\1XXXX", string)

See the regex demo

Details

• (^|[AB]) - Capturing group 1 (\1): start of string (^) or (|) A or B ([AB])
• 0000 - four zeros.

R demo:

string <- "0000A0000B0000C0000D0000E0000A0000"
pattern <- c("XXXX")
gsub("(^|[AB])0000", "\\1XXXX", string)
## -> [1] "XXXXAXXXXBXXXXC0000D0000E0000AXXXX"

Answer 2

Could you please try following. Using positive lookahead method here.

string <- "0000A0000B0000C0000D0000E0000A0000"
gsub(x = string, pattern = "(^|A|B)(?=0000)((?i)0000?)",
    replacement = "\\1xxxx", perl=TRUE)

Output will be as follows.

[1] "xxxxAxxxxBxxxxC0000D0000E0000Axxxx"

Answer 3

Thanks to Wiktor Stribiżew for the answer! It also works with the stringr package:

library(stringr)

string <- "0000A0000B0000C0000D0000E0000A0000"
pattern <- c("0000")

replace <- str_replace_all(string, paste0("(^|[AB])",pattern), "\\1XXXX")
replace

[1] "XXXXAXXXXBXXXXC0000D0000E0000AXXXX"