regular expression in R-- new lines - regex

Question

I'm trying to using regular expression in R by using regexpr function. I have multiple conditions to match, therefore my regular expression is very long actually, for example "A\s+(\d+)|(\d+)\s+A". So I want to put each separate expression on different lines, like
"A\\s+(\\d+)|
(\\d+)\\s+A|"
But it's not working. The bracket tells R that I want to extract the digit number out. Can anyone give suggestions?

Answer 1

1) paste Try using paste:
paste("A\\s+(\\d+)",
"(\\d+)\\s+A",
sep = "|")
2) rex Another possibility is to use the rex package
library(rex)
rex(group("A", spaces, capture(digits)) %or%
group(capture(digits), spaces, "A"))
which gives:
(?:(?:A[[:space:]]+([[:digit:]]+))|(?:([[:digit:]]+)[[:space:]]+A))
3) rebus The rebus package is similar in intent:
library(rebus)
literal("A") %R% one_or_more(space()) %R% capture(one_or_more(ascii_digit())) %|%
capture(one_or_more(digit())) %R% one_or_more(space()) %R% literal("A")
which emits:
<regex> \QA\E[[:space:]]+([0-9]+)|([[:digit:]]+)[[:space:]]+\QA\E

Answer 2

If you want to break string literal up on to several lines in your script, one solution is to use paste0:
my_expr <- paste0('partone',
'parttwo',
'partthree')
Then you get the desired result:
> my_expr
[1] "partoneparttwopartthree"
You can't just break it up onto several lines in between quotes, b/c then the new line character is part of the expression.
If you are also trying to trouble shoot your regular expression, you'll need to post a sample of the data you are trying to work with and the desired result

Answer 3

Just use the x modifier with perl = TRUE in whatever function you're using. Place the x modifier ((?x)) at the beginning of the expression and white space is ignored. Additionally, comment charcters are ignored as well.
pat <- "(?x)\\\\ ## Grab a backslash followed by...
[a-zA-Z0-9]*cite[a-zA-Z0-9]* ## A word that contains ‘cite‘
(\\[([^]]+)\\]){0,2}\\** ## Look for 0-2 square brackets w/ content
\\{([a-zA-Z0-9 ,]+)\\}" ## Look for curly braces with viable bibkey
tex <- c(
"Many \\parencite*{Ted2005, Moe1999} say graphs \\textcite{Few2010}.",
"But \\authorcite{Ware2013} said perception good too.",
"Random words \\pcite[see][p. 22]{Get9999c}.",
"Still more \\citep[p. 22]{Foo1882c}?"
)
gsub(pat, "", tex, perl=TRUE)
## [1] "Many say graphs ." "But said perception good too."
## [3] "Random words ." "Still more ?"
A second approach...I maintain a package called regexr that attempts to enable maintainers of regular expressions libraries:
to write regular expressions in a way that is similar to the ways R code is written.
This may be overkill if you're aren't panning long term maintence of the expression but you could do the same thing with regexr by (no need for perl = TRUE). Note the minimal comments as the meaning is shared with sub expression names. The %:)% is a comment operator (commented code is happy code) but you need not use the leading names or comments, just construct.:
library(regexr)
pat2 <- construct(
backslash = "\\\\" %:)% "\\",
cite_command = "[a-zA-Z0-9]*cite[a-zA-Z0-9]*" %:)% "parencite",
square_brack = "(\\[([^]]+)\\]){0,2}\\**" %:)% "[e.g.][p. 12]",
bibkeys = "\\{([a-zA-Z0-9 ,]+)\\}" %:)% "{Rinker2014}"
)
gsub(pat2, "", tex)
## [1] "Many say graphs ." "But said perception good too."
## [3] "Random words ." "Still more ?"
The regexr frame work requires a bit of upfront time but the "code" is much easier to maintain, more modular, and easier to understand by others without learning a new "language". This is one approach of many and I tend to use a combination of standard regex, regexr and rebus (which works within the regexr framework). So for example we can grab any of the sub expressions from pat2 with the subs function as follows:
subs(pat2)
## $backslash
## [1] "\\\\"
##
## $cite_command
## [1] "[a-zA-Z0-9]*cite[a-zA-Z0-9]*"
##
## $square_brack
## [1] "(\\[([^]]+)\\]){0,2}\\**"
##
## $bibkeys
## [1] "\\{([a-zA-Z0-9 ,]+)\\}"
I also included simple way to test the main and sub expressions for perl validity as follows:
test(pat2)
## $regex
## [1] TRUE
##
## $subexpressions
## backslash cite_command square_brack bibkeys
## TRUE TRUE TRUE TRUE