CODEWITHSUNDEEP
rregexgsubbackreference
Bratt Swan
Bratt Swan

Reputation: 1146

Backreference in R

I got really confused about the usage of backreferences

strings <- c("^ab", "ab", "abc", "abd", "abe", "ab 12")

gsub("(ab) 12", "\\1 34", strings)
[1] "^ab"   "ab"    "abc"   "abd"   "abe"   "ab 34"

gsub("(ab)12", "\\2 34", strings)
[1] "^ab"   "ab"    "abc"   "abd"   "abe"   "ab 12"

I know \1 refers to the first subpattern (reading from the left), \2 refers to the second subpattern, and so on. But I dont know what this subpattern means. Why \1 and \2 give different output

gsub("(ab)", "\\1 34", strings)
[1] "^ab 34"   "ab 34"    "ab 34c"   "ab 34d"   "ab 34e"   "ab 34 12"

Also, why I remove 12 after (ab) then it gives such result?

gsub("ab", "\\1 34", strings)
[1] "^ 34"   " 34"    " 34c"   " 34d"   " 34e"   " 34 12"

Furthermore, what if ab has no parenthesis? What does it indicate?

I really messed up with backreference and hope someone could explain the logic clearly

Upvotes: 7

Views: 5466

Answers (1)

akrun
akrun

Reputation: 887213

In the first and second case, there is a single capture group i.e. groups that are captured using (...), however in the first case replacement we use the backreference correctly i.e. the first capture group and in the second case, used \\2 which never existed.

To illustrate it

gsub("(ab)(d)", "\\1 34", strings)
#[1] "^ab"   "ab"    "abc"   "ab 34" "abe"   "ab 12"

here we are using two capture groups ((ab) and (d)), in the replacement we have first backreference (\\1) followed by a space followed by 34. So, in 'strings' this will match the 4th element i.e. "abd", get "ab" for the first backreference (\\1) followed by a space and 34.

Suppose, we do with the second backreference

gsub("(ab)(d)", "\\2 34", strings)
#[1] "^ab"   "ab"    "abc"   "d 34"  "abe"   "ab 12"

the first one is removed and we have "d" followed by space and 34.

Suppose, we are using a general case instead of specific characters

gsub("([a-z]+)\\s*(\\d+)", "\\1 34", strings)
#[1] "^ab"   "ab"    "abc"   "abd"   "abe"   "ab 34"
gsub("([a-z]+)\\s*(\\d+)", "\\2 34", strings)
#[1] "^ab"   "ab"    "abc"   "abd"   "abe"   "12 34"

Note how the values are changed in the last element by switching from first backreference to second. The pattern used is one or more lower case letters (inside the capture group (([a-z]+)) followed by zero or more space (\\s*) followed by one or more numbers in the second capture group ((\\d+)) (this matches only with the last element of 'strings'). In the replacement, we use the first and second backreference as showed above.

Upvotes: 14

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