CODEWITHSUNDEEP
regexr
user1471980
user1471980

Reputation: 10626

Extracting numbers from vectors of strings

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?

Upvotes: 171

Views: 293531

Answers (13)

Dan Adams
Dan Adams

Reputation: 5204

Slight variation on some other very good answers:

years <- c("20 years old", "1 years old")

as.numeric(gsub("[^0-9]", "", years))
#> [1] 20  1

Created on 2023-07-24 with reprex v2.0.2

Here we use ^ at the beginning of the regex to negate the pattern.

Upvotes: 0

terraviva
terraviva

Reputation: 41

I am interested in this question as it applies to extracting values from the base::summary() function. Another option you might want to consider to extract values from a table is to build a function that takes any entry of your summary() table and transforms it into a useful number. For example if you get:

(s <- summary(dataset))

sv_final_num_beneficiarios  sv_pfam_rec        sv_area_transf    
Min.   :    1.0            Min.   :0.0000036   Min.   :0.000004  
1st Qu.:   67.5            1st Qu.:0.0286363   1st Qu.:0.010107  
Median :  200.0            Median :0.0710803   Median :0.021865  
Mean   :  454.6            Mean   :0.1140274   Mean   :0.034802  
3rd Qu.:  515.8            3rd Qu.:0.1527177   3rd Qu.:0.044234  
Max.   :17516.0            Max.   :0.8217923   Max.   :0.360924

you might want to extract that 1st Qu for sv_pfam_rec and for that read the 2nd row of the 2nd col. In order to get the formatted single value I made a function

s_extract <- function(summary_entry){
separate(as_tibble(summary_entry),
sep = ":",
col = value, 
remove = FALSE, 
into = c("bad", "good"))[[3]] %>% 
as.numeric() 
}

You just have to feed a summary entry, for example summary_entry = s[3,3] to obtain the Median of sv_area_transf.

It is worth nothing that given that this function is based on separate() it makes it easier to navigate certain cases in which the name of the variable also contains numbers

Upvotes: 0

moodymudskipper
moodymudskipper

Reputation: 47300

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

Upvotes: 6

Ronak Shah
Ronak Shah

Reputation: 388817

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"

Upvotes: 31

akrun
akrun

Reputation: 886948

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

Upvotes: 126

sbaniwal
sbaniwal

Reputation: 337

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)))

Upvotes: 8

Joe
Joe

Reputation: 8601

A stringr pipelined solution:

library(stringr)
years %>% str_match_all("[0-9]+") %>% unlist %>% as.numeric

Upvotes: 26

989
989

Reputation: 12937

Or simply:

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

Upvotes: 62

juanbretti
juanbretti

Reputation: 438

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]])

Upvotes: 5

sebastian-c
sebastian-c

Reputation: 15395

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.

Upvotes: 75

Andrew
Andrew

Reputation: 38619

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

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

Upvotes: 41

Tyler Rinker
Tyler Rinker

Reputation: 109844

You could get rid of all the letters too: 

as.numeric(gsub("[[:alpha:]]", "", years))

Likely this is less generalizable though.

Upvotes: 18

Arun
Arun

Reputation: 118779

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))

Upvotes: 130

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