R extract substring from end of pattern until first occurance of character

Question

Struggling for hours to get this match and replace in R gsub to work and still no success. I'm trying to match the pattern "Reason:" in a string, and extact everything AFTER this pattern and until the first occurance of a dot (.) For instance:

Offer Disposition. MSISDN: 7183067962. Offer: . Disposition: DECLINED. Reason: Not interested. ChannelID: CARE.

would return "Not interested"

Answer 1

Here's a solution:

s <- "Offer Disposition. MSISDN: 7183067962. Offer: . Disposition: DECLINED. Reason: Not interested. ChannelID: CARE."

sub(".*Reason: (.*?)\\..*", "\\1", s)
# [1] "Not interested"

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Update (to address comments):

If you also have strings that do not match the pattern, I recommend using regexpr instead of sub:

s2 <- c("no match example",
        "Offer Disposition. MSISDN: 7183067962. Offer: . Disposition: DECLINED. Reason: Not interested. ChannelID: CARE.")

match <- regexpr("(?<=Reason: ).*?(?=\\.)", s2, perl = TRUE)
ifelse(match == -1, NA, regmatches(s2, match))
# [1] NA                                "Not interested. ChannelID: CARE"

For you second example, you can use the following regex:

s3 <- "Delete Payment Arrangement of type Proof of Payment for BAN : 907295267 on date 02/01/2014, from reason PAERR."

# a)
sub(".*type (.*?) for.*", "\\1", s3)
# [1] "Proof of Payment"

# b)
match <- regexpr("(?<=type ).*?(?= for)", s3, perl = TRUE)
ifelse(match == -1, NA, regmatches(s3, match))
# [1] "Proof of Payment"

Answer 2

Lots of different ways (as you can see from the submissions). I personally like to use stringr functions.

library(stringr)

rec <- "Offer Disposition. MSISDN: 7183067962. Offer: . Disposition: DECLINED. Reason: Not interested. ChannelID: CARE."
str_match(rec, "Reason: ([a-zA-Z0-9\ ]+)\\.")[2]
## [1] "Not interested"