R-正则表达式匹配模式，并且仅将模式存储在新列中？

Question

I've been searching for hours. This should be very easy but I don't see how :(

I have a dataframe called ds that contains a column structured like:

name
"Doe, Mr. John"
"Worth, Miss. Jane"

I want to extract the middle word and put it into a new column.

#This is how I'm doing it now
ds$title <- NA

mr  <- grep(", Mr. ", ds$name)
miss <- grep(", Miss. ", ds$name)

ds$title[mr] <- ", Mr. "
ds$title[miss] <- ", Miss. "

I'm trying to generalize this with regex so that it'll take any middle word matching the pattern of "comma space word period space"

This is my best guess but it only removes the pattern:

gsub(", .+\\.+ ", "", ds$name)

How do I keep the pattern and remove the rest?

Thank you!

Answer 1

You can use a capture group. Basically, you match the whole pattern, use a capture group to match the part you want to keep, and replace the whole match with the capture group:

# I often specify perl = TRUE, though it isn't necessary here
(ds$title <- gsub(".+(, .+\\.+ ).+", "\\1", ds$name, perl = TRUE))
#[1] ", Mr. "   ", Miss. "

The capture group is what's in the parentheses ( (, .+\\\\.+ ) ), and you refer back to it with \\\\1 . If you had a second capture group, you'd refer to it as \\\\2 .

Note that if you want to catch comma, space, word, period, space, then you could modify the capture group to (, .+\\\\. ) . You only need to match one period, not one or more.

---

A straightforward stringi alternative that does not use capture groups is stri_extract_first_regex (or in this case stri_extract_last_regex or stri_extract_all_regex work fine)

library(stringi)
ds$title <- stri_extract_first_regex(ds$name, ", .+\\. ")
#[1] ", Mr. "   ", Miss. "

and as thelatemail pointed out in a comment you can do a similar thing with base R, too, but it's a little harder to remember how to use the regmatches and regexpr functions:

regmatches(ds$name, regexpr(", .+\\. ", ds$name))
#[1] ", Mr. "   ", Miss. "

Answer 2

Matched capture groups are your BFF:

library(stringi)
library(purrr)

ds <- data.frame(name=c("Doe, Mr. John", "Worth, Miss. Jane"), stringsAsFactors=FALSE)

nonsp <- "[[:alnum:][:punct:]]+"
sp <- "[[:blank:]]+"

stri_match_all_regex(ds$name, nonsp %s+% sp %s+% "(" %s+% nonsp %s+% ")" %s+% sp %s+% nonsp) %>%
  map_chr(2)
## [1] "Mr."   "Miss."

For your "add column to a data frame" needs:

library(stringi)
library(dplyr)
library(purrr)

ds <- data.frame(name=c("Doe, Mr. John", "Worth, Miss. Jane"), stringsAsFactors=FALSE)

nonsp <- "[[:alnum:][:punct:]]+"
sp <- "[[:blank:]]+"

mutate(ds, title=stri_match_all_regex(ds$name, nonsp %s+% sp %s+% "(" %s+% nonsp %s+% ")" %s+% sp %s+% nonsp) %>% map_chr(2))
##                name title
## 1     Doe, Mr. John   Mr.
## 2 Worth, Miss. Jane Miss.