R Extract number from string

Question

I have been trying to get this right. What I want to do is extract a year from a string. The string looks like this for example:

Toy Story (1995)

Or it could look like this

Twelve Monkeys (a.k.a. 12 Monkeys) (1995)

To extract the numbers, I currently use

year = gsub("(?<=\\()[^()]*(?=\\))(*SKIP)(*F)|.", "", x, perl=T)

Now, this would work in most cases, where the first one is used, but in the list the second one is also used.

[1] 1995
[2] a.k.a. 12 Monkeys1995

So obviously I do not want the string but only the year, how do I get this?

Have a look at the `gsubfn` suggestion [here](http://stackoverflow.com/questions/12735503/extract-numbers-between-brackets-within-a-string) - `library(gsubfn); strapplyc(x, "[(](\\d+)[)]", simplify = TRUE)` — Steven Beaupré, Oct 23 '16 at 12:38

akrun · Accepted Answer · 2016-10-23T12:58:00.683

4

We can use

library(stringr)
as.numeric(str_extract(x, "(?<=\\()[0-9]+(?=\\))"))
#[1] 1995 1995

x <-  c("Toy Story (1995)", "Twelve Monkeys (a.k.a. 12 Monkeys) (1995)")

edited Oct 23 '16 at 12:58

answered Oct 23 '16 at 12:25

akrun

1

as number, not string, lol.... `as.numeric(str_extract(x, "(?<=\\()[0-9]+(?=\\))"))` very precise @akrun. good one. – Manoj Kumar Oct 23 '16 at 12:56
@ManojKumar Yes, I forgot to wrap with `as.numeric` – akrun Oct 23 '16 at 12:57

score 2 · Answer 2 · answered Oct 23 '16 at 12:51

2

stringi::stri_match_last_regex(x, "\\(([[:digit:]]+)\\)")[,2]

Escaping the parens is still a pain, but it's a far more readable regex IMO.

answered Oct 23 '16 at 12:51

hrbrmstr

score 0 · Answer 3 · answered Oct 23 '16 at 13:20

If the years are always located at the end of each string circled by parentheses, you could do this in base R:

as.numeric(gsub("\\(|\\)", "", substr(x, nchar(x)-5,nchar(x))))
#[1] 1995 1995

Use trimws(x) beforehand in case there might be any head or tail spaces.

3 Answers3