Extracting numbers from vectors of strings
RegexRRegex Problem Overview
I have string like this:
years<-c("20 years old", "1 years old")
I would like to grep only the numeric number from this vector. Expected output is a vector:
c(20, 1)
How do I go about doing this?
Regex Solutions
Solution 1 - Regex
How about
# pattern is by finding a set of numbers in the start and capturing them
as.numeric(gsub("([0-9]+).*$", "\\1", years))
or
# pattern is to just remove _years_old
as.numeric(gsub(" years old", "", years))
or
# split by space, get the element in first index
as.numeric(sapply(strsplit(years, " "), "[[", 1))
Solution 2 - Regex
Update
Since extract_numeric is deprecated, we can use parse_number from readr package.
library(readr)
parse_number(years)
Here is another option with extract_numeric
library(tidyr)
extract_numeric(years)
#[1] 20 1
Solution 3 - Regex
I think that substitution is an indirect way of getting to the solution. If you want to retrieve all the numbers, I recommend gregexpr:
matches <- regmatches(years, gregexpr("[[:digit:]]+", years))
as.numeric(unlist(matches))
If you have multiple matches in a string, this will get all of them. If you're only interested in the first match, use regexpr instead of gregexpr and you can skip the unlist.
Solution 4 - Regex
Here's an alternative to Arun's first solution, with a simpler Perl-like regular expression:
as.numeric(gsub("[^\\d]+", "", years, perl=TRUE))
Solution 5 - Regex
Or simply:
as.numeric(gsub("\\D", "", years))
# [1] 20 1
Solution 6 - Regex
A stringr pipelined solution:
library(stringr)
years %>% str_match_all("[0-9]+") %>% unlist %>% as.numeric
Solution 7 - Regex
You could get rid of all the letters too:
as.numeric(gsub("[[:alpha:]]", "", years))
Likely this is less generalizable though.
Solution 8 - Regex
We can also use str_extract from stringr
years<-c("20 years old", "1 years old")
as.integer(stringr::str_extract(years, "\\d+"))
#[1] 20 1
If there are multiple numbers in the string and we want to extract all of them, we may use str_extract_all which unlike str_extract returns all the macthes.
years<-c("20 years old and 21", "1 years old")
stringr::str_extract(years, "\\d+")
#[1] "20" "1"
stringr::str_extract_all(years, "\\d+")
#[[1]]
#[1] "20" "21"
#[[2]]
#[1] "1"
Solution 9 - Regex
Extract numbers from any string at beginning position.
x <- gregexpr("^[0-9]+", years) # Numbers with any number of digits
x2 <- as.numeric(unlist(regmatches(years, x)))
Extract numbers from any string INDEPENDENT of position.
x <- gregexpr("[0-9]+", years) # Numbers with any number of digits
x2 <- as.numeric(unlist(regmatches(years, x)))
Solution 10 - Regex
Using the package unglue we can do :
# install.packages("unglue")
library(unglue)
years<-c("20 years old", "1 years old")
unglue_vec(years, "{x} years old", convert = TRUE)
#> [1] 20 1
Created on 2019-11-06 by the reprex package (v0.3.0)
More info: https://github.com/moodymudskipper/unglue/blob/master/README.md
Solution 11 - Regex
After the post from Gabor Grothendieck post at the r-help mailing list
years<-c("20 years old", "1 years old")
library(gsubfn)
pat <- "[-+.e0-9]*\\d"
sapply(years, function(x) strapply(x, pat, as.numeric)[[1]])