This question already has answers here:
Regex - Split String on Comma, Skip Anything Between Balanced Parentheses
(2 answers)
Closed 1 year ago.
I have the following string:
Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews
I want to replace the semicolons that are not in parenthesis to commas. There can be any number of brackets and any number of semicolons within the brackets and the result should look like this:
Almonds , Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt), Cashews
This is my current code:
x<- Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews
gsub(";(?![^(]*\\))",",",x,perl=TRUE)
[1] "Almonds , Roasted Peanuts (Peanuts, Canola Oil (Antioxidants (319; 320)); Salt), Cashews "
The problem I am facing is if there's a nested () inside a bigger bracket, the regex I have will replace the semicolon to comma.
Can I please get some help on regex that will solve the problem? Thank you in advance.
The pattern ;(?![^(]*\)) means matching a semicolon, and assert that what is to the right is not a ) without a ( in between.
That assertion will be true for a nested opening parenthesis, and will still match the ;
You could use a recursive pattern to match nested parenthesis to match what you don't want to change, and then use a SKIP FAIL approach.
Then you can match the semicolons and replace them with a comma.
[^;]*(\((?>[^()]+|(?1))*\))(*SKIP)(*F)|;
In parts, the pattern matches
[^;]* Match 0+ times any char except ;
( Capture group 1
\( Match the opening (
(?> Atomic group
[^()]+ Match 1+ times any char except ( and )
| Or
(?1) Recurse the whole first sub pattern (group 1)
)* Close the atomic group and optionally repeat
\) Match the closing )
) Close group 1
(*SKIP)(*F) Skip what is matched
| Or
; Match a semicolon
See a regex demo and an R demo.
x <- c("Almonds ; Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt); Cashews",
"Peanuts (32.5%); Macadamia Nuts (14%; PPPG(AHA)); Hazelnuts (9%); nuts(98%)")
gsub("[^;]*(\\((?>[^()]+|(?1))*\\))(*SKIP)(*F)|;",",",x,perl=TRUE)
Output
[1] "Almonds , Roasted Peanuts (Peanuts; Canola Oil (Antioxidants (319; 320)); Salt), Cashews"
[2] "Peanuts (32.5%), Macadamia Nuts (14%; PPPG(AHA)), Hazelnuts (9%), nuts(98%)"
Related
I have several strings with open and unclosed parenthesis. I managed to remove the opening parenthesis (if there is no closing one), but I do not manage to remove the closing parenthesis if there is no opening one. I want to leave those with matching parenthesis alone
string1 = "This (is solved"
string2 = "This is (fine)"
string3 = "This is the problem)"
This is what I was able to remove the first Problem case with (Opening parenthesis but no opening)
str_remove(data, "[(](?!.*[)])")
But I cannot seem to turn it around. The following grabs all closing parenthesis, but not the one without an oping.
"(?!.*[(])[)]"
Any ideas are appreciated!
If you do not need to handle nested paired (balanced) parentheses, you can use
gsub("(\\([^()]*\\))|[()]", "\\1", string)
See the regex demo. Details:
(\([^()]*\)) - Group 1 (\1 refers to this group value): (, then zero or more chars other than ( and ), and then a ) char
| - or
[()] - a ( or ) char.
See the R demo:
x <- c("This (is solved", "This is (fine)", "This is the problem)")
gsub("(\\([^()]*\\))|[()]", "\\1", x)
# => [1] "This is solved" "This is (fine)" "This is the problem"
If the parentheses can be nested, you can use
gsub("(\\((?:[^()]++|(?1))*\\))|[()]", "\\1", string, perl=TRUE)
See this regex demo. Details:
(\((?:[^()]++|(?1))*\)) - Group 1:
\( - a ( char
(?:[^()\n]++|(?1))* - zero or more sequences of either one or more chars other than ( and ), or the whole Group 1 pattern that is recursed
\) - a ) char
|[()] - or a ( / ) char.
My string pattern is as follows:
1233 fox street, omaha NE ,69131-7233
Jeffrey Jones, 666 Church Street, Omaha NE ,69131-72339
Betty Davis, LLC, 334 Aloha Blvd., Fort Collins CO ,84444-00333
,1233 Decker street, omaha NE ,69131-7233
I need to separate the above string into four variables: name, address, city_state, zipcode.
Since the pattern has three to four commas, I am starting at the right to separate the field into multiple fields.
rubular.com says the pattern ("(,\\d.........)$"))) or the pattern ",\d.........$" will match the zipcode at the end of the string.
regex101.com, finds neither of the above patterns comes up with a match.
When I try to separate with:
#need to load pkg:tidyr for the `separate`
function
library(tidyr)
separate(street_add, c("street_add2", "zip", sep= ("(,\d.........)$")))
or with:
separate(street_add, c("street_add2", "zip", sep= (",\d.........$")))
In both scenarios, R splits at the first comma in the string.
How do I split the string into segments?
Thank you.
Use
sep=",(?=[^,]*$)"
See regex proof.
EXPLANATION
--------------------------------------------------------------------------------
, ','
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
[^,]* any character except: ',' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
I want to match every cases of "-", but not these ones:
[\d]-[A-Z]
[A-Z]-[\d]
I tried this pattern: ((?<![A-Z])-(?![0-9]))|((?<![0-9])-(?![A-Z])) but some results are incorrect like: "RUA VF-32 N"
Can anyone help me?
A simple approach is to use grep with your current logic and inverting the result, and then run another grep to only keep those items that have a hyphen in them:
x <- c("QUADRA 120 - ASA BRANCA","FAZENDA LAGE -RODOVIA RIO VERDE","C-15","99-B","A-A")
grep("-", grep("[A-Z]-\\d|\\d-[A-Z]", x, invert=TRUE, value=TRUE), value=TRUE, fixed=TRUE)
# => [1] "QUADRA 120 - ASA BRANCA" "FAZENDA LAGE -RODOVIA RIO VERDE"
# [3] "A-A"
Here, [A-Z]-\\d|\\d-[A-Z] matches a hyphen either in between an uppercase ASCII etter or a digit or betweena digit and an ASCII uppercase letter. If there is a match, the result is inverted due to invert=TRUE.
See the R demo.
To only match - in all contexts other than in between a letter and a digit, you may use the PCRE regex based on SKIP-FAIL technique like
> grep("(?:\\d-[A-Z]|[A-Z]-\\d)(*SKIP)(*F)|-", x, perl=TRUE)
[1] 1 2
See this regex demo
Details
(?:\d-[A-Z]|[A-Z]-\d) - a non-capturing group that matches either a digit, - and then uppercase ASCII letter, or an uppercase ASCII letter, - and a digit
(*SKIP)(*F) - omit the current match and proceed looking for the next match at the end of the "failed" match
| - or
- - a hyphen.
For a text field, I would like to expose those that contain invalid characters. The list of invalid characters is unknown; I only know the list of accepted ones.
For example for French language, the accepted list is
A-z, 1-9, [punc::], space, àéèçè, hyphen, etc.
The list of invalid charactersis unknown, yet I want anything unusual to resurface, for example, I would want
This is an 2-piece à-la-carte dessert to pass when
'Ã this Øs an apple' pumps up as an anomalie
The 'not contain' notion in R does not behave as I would like, for example
grep("[^(abc)]",c("abcdef", "defabc", "apple") )
(those that does not contain 'abc') match all three while
grep("(abc)",c("abcdef", "defabc", "apple") )
behaves correctly and match only the first two. Am I missing something
How can we do that in R ? Also, how can we put hypen together in the list of accepted characters ?
[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+
The above regex matches any of the following (one or more times). Note that the parameter ignore.case=T used in the code below allows the following to also match uppercase variants of the letters.
a-z Any lowercase ASCII letter
1-9 Any digit in the range from 1 to 9 (excludes 0)
[:punct:] Any punctuation character
The space character
àâæçéèêëîïôœùûüÿ Any valid French character with a diacritic mark
- The hyphen character
See code in use here
x <- c("This is an 2-piece à-la-carte dessert", "Ã this Øs an apple")
gsub("[a-z1-9[:punct:] àâæçéèêëîïôœùûüÿ-]+", "", x, ignore.case=T)
The code above replaces all valid characters with nothing. The result is all invalid characters that exist in the string. The following is the output:
[1] "" "ÃØ"
If by "expose the invalid characters" you mean delete the "accepted" ones, then a regex character class should be helpful. From the ?regex help page we can see that a hyphen is already part of the punctuation character vector;
[:punct:]
Punctuation characters:
! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { | } ~
So the code could be:
x <- 'Ã this Øs an apple'
gsub("[A-z1-9[:punct:] àéèçè]+", "", x)
#[1] "ÃØ"
Note that regex has a predefined, locale-specific "[:alpha:]" named character class that would probably be both safer and more compact than the expression "[A-zàéèçè]" especially since the post from ctwheels suggests that you missed a few. The ?regex page indicates that "[0-9A-Za-z]" might be both locale- and encoding-specific.
If by "expose" you instead meant "identify the postion within the string" then you could use the negation operator "^" within the character class formalism and apply gregexpr:
gregexpr("[^A-z1-9[:punct:] àéèçè]+", x)
[[1]]
[1] 1 8
attr(,"match.length")
[1] 1 1
How do I match the year such that it is general for the following examples.
a <- '"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}'
b <- 'Þegar það gerist (1998/I) (TV)'
I have tried the following, but did not have the biggest success.
gsub('.+\\(([0-9]+.+\\)).?$', '\\1', a)
What I thought it did was to go until it finds a (, then it would make a group of numbers, then any character until it meets a ). And if there are several matches, I want to extract the first group.
Any suggestions to where I go wrong? I have been doing this in R.
You could use
library(stringr)
strings <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)')
years <- str_match(strings, "\\((\\d+(?: B\\.C\\.)?)")[,2]
years
# [1] "1953" "1998"
The expression here is
\( # (
(\d+ # capture 1+ digits
(?: B\.C\.)? # B.C. eventually
)
Note that backslashes need to be escaped in R.
Your pattern contains .+ parts that match 1 or more chars as many as possible, and at best your pattern could grab last 4 digit chunks from the incoming strings.
You may use
^.*?\((\d{4})(?:/[^)]*)?\).*
Replace with \1 to only keep the 4 digit number. See the regex demo.
Details
^ - start of string
.*? - any 0+ chars as few as possible
\( - a (
(\d{4}) - Group 1: four digits
(?: - start of an optional non-capturing group
/ - a /
[^)]* - any 0+ chars other than )
)? - end of the group
\) - a ) (OPTIONAL, MAY BE OMITTED)
.* - the rest of the string.
See the R demo:
a <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)', 'Johannes Passion, BWV. 245 (1725 Version) (1996) (V)')
sub("^.*?\\((\\d{4})(?:/[^)]*)?\\).*", "\\1", a)
# => [1] "1953" "1998" "1996"
Another base R solution is to match the 4 digits after (:
regmatches(a, regexpr("\\(\\K\\d{4}(?=(?:/[^)]*)?\\))", a, perl=TRUE))
# => [1] "1953" "1998" "1996"
The \(\K\d{4} pattern matches ( and then drops it due to \K match reset operator and then a (?=(?:/[^)]*)?\\)) lookahead ensures there is an optional / + 0+ chars other than ) and then a ). Note that regexpr extracts the first match only.