如何按R中单列中的字符串标签对行值进行子集化？

Question

I have a column that I would like to subset its row value based on the first and last 'string' label in R. The level values are as followed:

[1] "60022 (Location; 9TH FLOOR; Snacks)"
[3] "60024 (Location; 9TH FLOOR; Lg Snacks)"
[5] "60027 (Location; 9TH FLOOR; Sml Snacks)"

I would like the to pull the # and the last string separated by the ';'. Is there a function or syntax in R to do this? So remove "Location; 9TH FLOOR" and just keep the last ; "" string.

I have tried this to pull just the first value but am unable to keep the "snacks" as well with this code:

#updated_df_2020$Machine <- sub("([A-Za-z]+).*", "\\1", updated_df_2020$Machine) 

End result for each row should be the number (60022 and then Snacks) like so:

[1] "60022 (Snacks)" 
[1] "60024 (Lg Snacks)" 
[1] "60027 (Sml Snacks)" 

Answer 1

If we need to remove the substring, capture the digits ( \\\\d+ ) at the start ( ^ ) of the string, and then capture the non white space ( \\\\S ) that succeeds the ; and zero or more space ( \\\\s* ) and other characters that follows ( .* ) till the ) at the end ( $ ) as second capture group. In the replacement, specify the backreference ( \\\\1 , \\\\2 ) of the captured group and modify it by adding the (

updated_df_2020$Machine <- sub("^(\\d+)\\b.*;\\s*\\b(\\S.*\\))$", 
        "\\1 (\\2", updated_df_2020$Machine)
updated_df_2020$Machine
#[1] "60022 (Snacks)"     "60024 (Lg Snacks)"  "60027 (Sml Snacks)"

If the start of the string is not a digit and still wants to get extract, replace ( (\\\\d+) ) with (\\\\w+)

data

updated_df_2020 <- data.frame(Machine = c("60022 (Location; 9TH FLOOR; Snacks)",
   "60024 (Location; 9TH FLOOR; Lg Snacks)", "60027 (Location; 9TH FLOOR; Sml Snacks)"),
   stringsAsFactors = FALSE)

Answer 2

You could do

> a <- c("60022 (Location; 9TH FLOOR; Snacks)", "60024 (Location; 9TH FLOOR; Snacks)", "60027 (Location; 9TH FLOOR; Snacks)")
> strs <- strsplit(a, split = " ")
> sapply(strs, function(s) paste(s[1], paste0("(", s[length(s)])))
#
# "60022 (Snacks)" "60024 (Snacks)" "60027 (Snacks)"
#

which is uglier, but i guess a bit easier to understand

Answer 3

We can extract the number at the begining and everything followed by colon afterwards using sub :

sub("(\\d+).*;(.*)", "\\1 (\\2", x)
#[1] "60022 ( Snacks)"     "60024 ( Lg Snacks)"  "60027 ( Sml Snacks)"

where x is

x <- c("60022 (Location; 9TH FLOOR; Snacks)", 
       "60024 (Location; 9TH FLOOR; Lg Snacks)",
       "60027 (Location; 9TH FLOOR; Sml Snacks)")