Extracting numbers from vectors of strings

Question

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. Expe

Answer 1

Or simply:

as.numeric(gsub("\\D", "", years))
# [1] 20  1

Answer 2

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.

Answer 3

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

Answer 4

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))

Answer 5

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

Answer 6

Here's an alternative to Arun's first solution, with a simpler Perl-like regular expression:

as.numeric(gsub("[^\\d]+", "", years, perl=TRUE))