Regex extract thrid number

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I have some strings and I am trying to extract the third number that appears in it (in R). This is an example string, all of them have the same pattern:

string = "Speaks 2 times (1%) for a total of 34 words (1%)."

I have been able to get the first number with str_extract(string, "[0-9]+") But I have no idea on how to take only the third (the number of words). Any help would be gratly appreciated!

r regex
4 Answers

A better approach would be to target the number before the text words:

string <- "Speaks 2 times (1%) for a total of 34 words (1%)."
num <- sub("^.*\\b(\\d+) words\\b.*", "\\1", string)
num

[1] "34"
user_18632290

You can extract all matches and grab the 3rd item:

library(stringr)
string = "Speaks 2 times (1%) for a total of 34 words (1%)."
unlist(str_extract_all(string, "[0-9]+"))[3]
## => [1] "34"

Also, you can use sub:

sub("^(?:\\D+\\d+){2}\\D+(\\d+).*", "\\1", string)
## => [1] "34"

See the R demo. Details:

  • ^ - start of string
  • (?:\D+\d+){2} - two occurrences of any one or more non-digit chars and then one or more digits (note: replace {2} with {3} to extract the fourth number, or remove it to get the second number, adjust as you see fit)
  • \D+ - zero or more non-digit chars
  • (\d+) - Group 1 (\1): one or more digits
  • .* - any zero or more chars as many as possible.
user_38329700

We may use str_locate to find the location of the substring and extract

library(stringr)
ind <- str_locate_all(string, "\\d+")[[1]][3,]
str_sub(string, ind[1], ind[2])
[1] "34"

Or using base R with strsplit

setdiff(strsplit(string, "\\D+")[[1]], "")[3]
[1] "34"

data

string <-  "Speaks 2 times (1%) for a total of 34 words (1%)."
user_37322710

Here is an other way: First match all numbers then subset vector on position 3:

sapply(regmatches(string, gregexpr("[[:digit:]]+", string)), function(x) x[3])

[1] "34"
user_133216470
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