Find the consecutively matched pattern index

Question

I would like to find all the index of matched pattern in some string. For Example, I have a string x <- "1110001101", I want to match it with "11", the result should be c(1, 2, 7), however, I just can't get 2...

• Method 1: Use gregexpr

  x
  [1] "1110001101"
  
  gregexpr(pattern = "11", x)
  [[1]]
  [1] 1 7 # Why isn't there a 2?
  attr(,"match.length")
  [1] 2 2
  attr(,"useBytes")
  [1] TRUE
  

• Method 2: Use str_locate_all from package stringr

  library(stringr)
  str_locate_all(pattern = "11", x)
  [[1]]
       start end
  [1,]     1   2
  [2,]     7   8 # Why still isn't there a 2?
  

Did I lose some subtle arguments for these functions? Thanks for your suggestions!

Answer 1

We can use a regex lookaround i.e. a positive regex lookahead to match a character followed by two 1's to give the positions of the start of the match with gregexpr

as.integer(gregexpr("(?=11)", x, perl = TRUE)[[1]])
#[1] 1 2 7

---

Or with str_locate either a regex lookbehind (in that case subtract 1)

stringr::str_locate_all(x, "(?<=11)")[[1]][,2]-1
#[1] 1 2 7

Or a regex lookahead

stringr::str_locate_all(x, "(?=11)")[[1]][,1]
#[1] 1 2 7

---

The difference between this approach and the OP's is that with the OP's approach, once the match is made, it skips that part and looks for the next match. This can be better explained if we look at another string

x1 <- "11110001101"
str_locate_all(pattern = "11", x1)
#[[1]]
#      start end
#[1,]     1   2
#[2,]     3   4
#[3,]     8   9

With regex lookaround, there will be 4 matches

as.integer(gregexpr("(?=11)", x1, perl = TRUE)[[1]])
#[1] 1 2 3 8