Combine two grep functions using AND operator - r

Question

I want to combine the following commands using AND operator:
grep("^ab", strings, value = TRUE)
grep("ab$", strings, value = TRUE)
Here is an example for OR operator
http://r.789695.n4.nabble.com/grep-for-multiple-pattern-td4685244.html#a4685247
Would you please advise?

Answer 1

The search for an AND operator in regex (whether in R or elsewhere) can be a long and sad search. The boolean AND means that both of two statements have to be true. How would you apply that to regex? Consider the regex pattern "ab", in grep("ab", strings). Even this simple pattern has several requirements, ALL of which have to be true. It has to have an "a", AND it has to have a "b", AND the "b" has to follow the "a" directly.
strings <- c("abraham, not ahab", "no it was ahab",
"abraham was the one they left on ceti alpha V",
"You're talking about Sherlock Holmes", "He tasks me", "ab")
grep("ab", strings, value = TRUE)
# [1] "abraham, not ahab"
# [2] "no it was ahab"
# [3] "abraham was the one they left on ceti alpha V"
# [4] "You're talking about Sherlock Holmes"
# [5] "ab"
If what you'd like is to match strings that BOTH start with "ab" AND end with "ab", then #r2evans pattern will work for you: grep("^ab.*ab$", strings, value = TRUE) will show them to you. This means it starts with "ab", has zero or more other characters, and then ends with "ab".
grep("^ab.*ab$", strings, value = TRUE)
# [1] "abraham, not ahab"
# NOTICE THAT THIS DOESN'T MATCH "ab", despite "ab" being at the beginning
# AND the end
If what you'd like is to match all the strings that start with an "a" immediately followed by a "b", AND ALSO all those that end with an "a" immediately followed by a "b", then you actually want grep("(^ab)|(ab$)", strings, value = TRUE)
grep("^ab|ab$", strings, value = TRUE)
# [1] "abraham, not ahab"
# [2] "no it was ahab"
# [3] "abraham was the one they left on ceti alpha V"
# [4] "ab"
So what about that solitary "ab" case? What regex pattern would match that and only that?
grep("^ab$", strings, value = TRUE)
# [1] "ab"
In this case, we wanted all of the matches to BOTH start AND end with "ab", but it had to be the same "ab". Of course, we could combine this with the other "AND" version, and get all of the matches where ab was at the start and ab was at the end:
grep("^ab$|^ab.*ab$", strings, value = TRUE)
# [1] "abraham, not ahab" "ab"
..and one more thing:
We can use #r2evans comment to demonstrate a sort of DeMorgan's law with regex. Notice that the pattern with the | metacharacter produces the same thing that you would by subsetting the strings object with the logical vector produced by combining both individual regex matches with a boolean AND:
strings[grepl("^ab", strings) & grepl("ab$", strings)]
# [1] "abraham, not ahab" "ab"
Here grepl returns a logical vector, and we use it twice. The first is TRUE for every element of strings that matches "^ab", and the second for every element that matches "ab$". Combining those logical vectors with an & operator produces the same thing as a pattern with a | metacharacter.

Answer 2

You may use
grep("^ab(.*ab)?$", strings, value = TRUE)
The pattern matches a string that starts with ab and then has an optional substring ending with ab and then end of string should follow:
^ - start of string
ab - an ab substring
(.*ab)? - 1 or 0 repetitions (due to ? quantifier) of
.* - any 0+ chars, as many as possible
ab - an ab substring
$ - end of string.
See the regex demo.