CODEWITHSUNDEEP
regexrpackagetext-mining
Thomas Jensen
Thomas Jensen

Reputation: 860

R Regular Expression Lookbehind

I have a vector filled with strings of the following format: <year1><year2><id1><id2>

the first entries of the vector looks like this:

199719982001
199719982002
199719982003
199719982003

For the first entry we have: year1 = 1997, year2 = 1998, id1 = 2, id2 = 001.

I want to write a regular expression that pulls out year1, id1, and the digits of id2 that are not zero. So for the first entry the regex should output: 199721.

I have tried doing this with the stringr package, and created the following regex:

"^\\d{4}|\\d{1}(?<=\\d{3}$)"

to pull out year1 and id1, however when using the lookbehind i get a "invalid regular expression" error. This is a bit puzzling to me, can R not handle lookaheads and lookbehinds?

Upvotes: 7

Views: 11054

Answers (3)

Wojciech Sobala
Wojciech Sobala

Reputation: 7561

You can use sub.

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

Upvotes: 1

mpiktas
mpiktas

Reputation: 11518

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.

Upvotes: 9

mpiktas
mpiktas

Reputation: 11518

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.

Upvotes: 9

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