使用正则表达式提取R括号中的数字

Question

I am trying to extract the 22 from below:

"Feb22 19  (22) 100  (Weeklys) "

I have tried the below but with no luck. Any suggestions?

grep("\\(.*\\)", "Feb22 19  (22) 100  (Weeklys) ", value= TRUE

Answer 1

We can try using sub with a capture group:

x <- "Feb22 19  (22) 100  (Weeklys) "
sub(".*\\((\\d+)\\).*", "\\1", x)

[1] "22"

The above pattern can be explained as:

.*     consume anything, up until the last
\(     literal open parenthesis, which is then followed by
(\d+)  one or more digits (which are captured)
\)     followed by a closing parenthesis
.*     followed by anything

The replacement is \\\\1 , which is the number captured in the pattern. Note that should the input not contain a number in parentheses, the above call to sub would actually return the original input string. If you don't like this behavior, then you will have to do more work.

Answer 2

We can also use:

    string<-"Feb22 19 (22) 100 (Weeklys) "
    unlist(stringr::str_extract_all(string,"\\d{1,}(?=\\))"))
    #[1] "22"

I was recently advised to use simplify although I find unlist 's output better.

Using stringr::str_extract_all(string,"\\\\d{1,}(?=\\\\))",simplify=TRUE)

    [,1]
[1,] "22"