I would like to use str_extract in the stringr package to extract the numbers from strings in the form XX nights etcetc.
I'm currently doing this:
library(stringr)
str_extract("17 nights$5 Days", "(\\d)+ nights")
but that returns
"17 nights"
instead of 17.
How can I extract just the number? I thought specifying the extract group with parentheses would work, but it doesn't.
You can use the look ahead regular express (?=)
library(stringr)
str_extract("17 nights$5 Days", "(\\d)+(?= nights)")
(\d) - a digit
(\d)+ - one or more digits
(?= nights) - that comes in front of " nights"
The look behind (?<=) can also come in handy.
A good reference cheatsheet is from Rstudio's website: https://raw.githubusercontent.com/rstudio/cheatsheets/main/regex.pdf
(\\d+) or \\d+
In base R, we can use sub to extract number which comes before "nights"
as.integer(sub("(\\d+)\\s+nights.*", "\\1","17 nights$5 Days"))
#[1] 17
Or if the number is always the first number in the string we can use readr::parse_number
readr::parse_number("17 nights$5 Days")
#[1] 17
as.integer(sub("(\\d+)\\s+nights.*", "\\1","17 nights$5 Days")) (even with the as.integer) is about 3X faster than the equivalent stringr
If you want to specify a specific group for return, use str_replace(). The pattern you want to capture is wrapped in (), then in the replacement argument you refer to that group as "\\1" as it is capture group number one.
I added the ^ to indicate you want numbers only at the beginning of the string.
library(stringer)
str_replace(string = "17 nights$5 Days",
pattern = "(^\\d+).*",
replacement = "\\1")
giving:
[1] "17"
You can use stringr::str_match which returns all of the matched groups as a matrix then select the correct column.
library(stringr)
str_match("17 nights$5 Days", "(\\d+?) nights")[[2]]
Using rebus. If the string always start with a number:
library(stringr)
library(rebus)
pattern = START %R% one_or_more(DGT)
str_extract("17 nights$5 Days", pattern)
#> [1] "17"
Created on 2021-05-30 by the reprex package (v2.0.0)
