CODEWITHSUNDEEP
rregexescaping
Ju Ko
Ju Ko

Reputation: 508

Match multiple strings containing escape characters in R

I have a vector of text strings containing smilies and a dictionary containing only the smilies.

A <- c("This :/ :/ :) ^^","is :/ ^^", "weird^^ :)")
B <- c(":)",":/","^^")

I would like to extract all matches of smilies for each text string including duplicates, so my output should look like this:

[[1]]
[1] ":/" ":/" ":)" "^^"

[[2]]
[1] ":/" "^^"

[[3]]
[1] "^^" ":)"

This is what I tried so far:

# does not return duplicates
sapply(A, function(x) B[str_detect(x, fixed(B))], USE.NAMES = FALSE)

[[1]]
[1] ":)" ":/" "^^"

[[2]]
[1] ":/" "^^"

[[3]]
[1] ":)" "^^"

# Only returns first instance
str_extract_all(A,fixed(B))

[[1]]
[1] ":)"

[[2]]
[1] ":/"

[[3]]
[1] "^^"

# returns error because of unescaped characters
rm_default(A,pattern=B,fixed=TRUE,extract=TRUE)
Error in stringi::stri_extract_all_regex(text.var, pattern) : 
  Incorrectly nested parentheses in regexp pattern. (U_REGEX_MISMATCHED_PAREN)
In addition: Warning messages:
1: In if (substring(pattern, 1, 4) == "@rm_") { :
  the condition has length > 1 and only the first element will be used
2: In if (substring(pattern, 1, 1) == "@") { :
  the condition has length > 1 and only the first element will be used

Any help is much appreciated.

Upvotes: 2

Views: 279

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626804

You may build a regex dynamically using the items in your B list by first sorting the items by length in a descending order (so that if you have :)) and :) the first could be extracted - that is a requirement for an unanchored NFA expression where the first alternative in an alternation group "wins", see [the Remember That The Regex Engine Is Eager section), and escape each item. Then just call regmatches / stringr::str_extract_all:

regex.escape <- function(string) {
  gsub("([][{}()+*^${|\\\\?.])", "\\\\\\1", string)
}

sort.by.length.desc <- function (v) v[order( -nchar(v)) ] 

A <- c("This :/ :/ :) ^^","is :/ ^^", "weird^^ :)")
B <- c(":)",":/","^^")

B <- sort.by.length.desc(B)
pattern <- paste(regex.escape(B), collapse="|")
regmatches(A, gregexpr(pattern, A))

See the R demo online.

In this case, the pattern will be :\)|:/|\^\^ and the output will be

[[1]]
[1] ":/" ":/" ":)" "^^"

[[2]]
[1] ":/" "^^"

[[3]]
[1] "^^" ":)"

Upvotes: 2

akrun
akrun

Reputation: 887118

One option is to do strsplit and then extract the elements that are contained in 'B'

lapply(strsplit(A, "[A-Za-z ]"), function(x) x[x %in% B])
#[[1]]
#[1] ":/" ":/" ":)" "^^"

#[[2]]
#[1] ":/" "^^"

#[[3]]
#[1] "^^" ":)"

Upvotes: 1

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