R Regular Expression Lookbehind

Question

I have a vector filled with strings of the following format:

the first entries of the vector looks like

Answer 1

You will need to use gregexpr from the base package. This works:

> s <- "199719982001"
> gregexpr("^\\d{4}|\\d{1}(?<=\\d{3}$)",s,perl=TRUE)
[[1]]
[1]  1 12
attr(,"match.length")
[1] 4 1
attr(,"useBytes")
[1] TRUE

Note the perl=TRUE setting. For more details look into ?regex.

Judging from the output your regular expression does not catch id1 though.

Answer 2

Since this is fixed format, why not use substr? year1 is extracted using substr(s,1,4), id1 is extracted using substr(s,9,9) and the id2 as as.numeric(substr(s,10,13)). In the last case I used as.numeric to get rid of the zeroes.

Answer 3

You can use sub.

sub("^(.{4}).{4}(.{1}).*([1-9]{1,3})$","\\1\\2\\3",s)