I'm trying to extract the status -- in this case the word "Active" from this pattern:
Status\nActive\nHometown\
Using this regex: https://regex101.com/r/xegX00/1, but I cannot get it to work in R using str_extract. It does seem weird to have dual escapes, but I've tried every possible combination here and cannot get this to work. Any help appreciated!
mutate(status=str_extract(df, "(?<=Status\\\\n)(.*?)(?=\\\\)"))
You can use sub in base R -
x <- "Status\nActive\nHometown\n"
sub('.*Status\n(.*?)\n.*', '\\1', x)
#[1] "Active"
If you want to use stringr, here is a suggestion with str_match which avoids using lookahead regex
stringr::str_match(x, 'Status\n(.*)\n')[, 2]
#[1] "Active"
Your regex fails because you tested it against a wrong text.
"Status\nActive\nHometown" is a string literal that denotes (defines, represents) the following plain text:
Status
Active
Hometown
In regular expression testers, you need to test against plain text!
To match a newline, you can use "\n" (i.e. a line feed char, an LF char), or "\\n", a regex escape that also matches a line feed char.
You can use
library(stringr)
x <- "Status\nActive\nHometown\n"
stringr::str_extract(x, "(?<=Status\\n).*") ## => [1] "Active"
## or
stringr::str_extract(x, "(?<=Status\n).*") ## => [1] "Active"
See the R demo online and a correct regex test.
Note you do not need an \n at the end of the pattern, as in an ICU regex flavor (used in R stringr regex methods), the . pattern matches any chars other than line break chars, so it is OK to just use .* to match the whole line.
So I am trying to split my string based on all the punctuations and space wherever they occur in the string (hence the + sign) except for on "#" & "/" because I don't want it to split #n/a which it does. I did search a lot on this problem but can't get to the solution. Any suggestions?
t<-"[[:punct:][:space:]]+"
bh <- tolower(strsplit(as.character(a), t)[[1]])
I have also tried storing the following to t but it also gives error
t<-"[!"\$%&'()*+,\-.:;<=>?#\[\\\]^_`{|}~\\ ]+"
Error: unexpected input in "t<-"[!"\"
One alternate is to substitute #n/a but I want to know how to do it without having to do that.
You may use a PCRE regex with a lookahead that will restrict the bracket expression pattern:
t <- "(?:(?![#/])[[:punct:][:space:]])+"
bh <- tolower(strsplit(as.character(a), t, perl=TRUE)[[1]])
The (?:(?![#/])[[:punct:][:space:]])+ pattern matches 1 or more repetitions of any punctuation or whitespace that is not # and / chars.
See the regex demo.
If you want to spell out the symbols you want to match inside a bracket expression you may fix your other pattern like
t <- "[][!\"$%&'()*+,.:;<=>?#\\\\^_`{|}~ -]+"
Note that ] must be right after the opening [, [ inside the expression does not need to be escaped, - can be put unescaped at the end, a \ should be defined with 4 backslashes. $ does not have to be escaped.
I'm trying to parse some of my chess pgn data but I'm having some trouble capturing characters just inside one bracket.
testString <- '[Event \"?\"]\n[Site \"http://www.chessmaniac.com play free chess\"]\n[Date \"2018.08.25\"]\n[Round \"-\"]\n[White \"NothingFancy 1497\"]\n[Black \"JR Smith 1985\"]\n[Result \"1-0\"]\n\n1.'
#Attempt to just get who white is, which is inside a bracket [White xxx]
findWhite <- regexpr('\\[White.*\\]', tempString)
regmatches(tempString, findWhite)
The stringr package seems to do what I want, but I'm curious what is different about the use of the same regular expression. I'm fine using stringr, but I like to also know how to do this in base R.
library(stringr)
str_extract(tempString, '\\[White.*\\]')
If you need the whole match starting with [White and ending with ] you may use
regmatches(testString, regexpr("\\[White\\s*[^][]*]", testString))
[1] "[White \"NothingFancy 1497\"]"
If you only need the substring inside double quotes:
regmatches(testString, regexpr("\\[White\\s*\\K[^][]*", testString, perl=TRUE))
[1] "\"NothingFancy 1497\""
See the regex demo.
To strip the double quotes, you may use something like
regmatches(testString, regexpr('\\[White\\s*"\\K.*(?="])', testString, perl=TRUE))
[1] "NothingFancy 1497"
See another regex demo and an online R demo.
Details
\\[ - a [ char
White - a literal substring
\\s* - 0+ whitespaces
\\K - match reset operator discarding the text matched so far
[^][]* - 0+ chars other than [ and ]
.* (in the other version) - matches any 0+ chars other than line break chars, as many as possible
(?="]) - a positive lookahead that matches a position inside a string that is immediately followed with "].
At least one way to do it in base R is to use sub and only keep the part that you want.
sub(".*\\[White\\s(*.*?)\\].*", "\\1", testString)
[1] "\"NothingFancy 1497\""
Simple regex question. I have a string on the following format:
this is a [sample] string with [some] special words. [another one]
What is the regular expression to extract the words within the square brackets, ie.
sample
some
another one
Note: In my use case, brackets cannot be nested.
You can use the following regex globally:
\[(.*?)\]
Explanation:
\[ : [ is a meta char and needs to be escaped if you want to match it literally.
(.*?) : match everything in a non-greedy way and capture it.
\] : ] is a meta char and needs to be escaped if you want to match it literally.
(?<=\[).+?(?=\])
Will capture content without brackets
(?<=\[) - positive lookbehind for [
.*? - non greedy match for the content
(?=\]) - positive lookahead for ]
EDIT: for nested brackets the below regex should work:
(\[(?:\[??[^\[]*?\]))
This should work out ok:
\[([^]]+)\]
Can brackets be nested?
If not: \[([^]]+)\] matches one item, including square brackets. Backreference \1 will contain the item to be match. If your regex flavor supports lookaround, use
(?<=\[)[^]]+(?=\])
This will only match the item inside brackets.
To match a substring between the first [ and last ], you may use
\[.*\] # Including open/close brackets
\[(.*)\] # Excluding open/close brackets (using a capturing group)
(?<=\[).*(?=\]) # Excluding open/close brackets (using lookarounds)
See a regex demo and a regex demo #2.
Use the following expressions to match strings between the closest square brackets:
Including the brackets:
\[[^][]*] - PCRE, Python re/regex, .NET, Golang, POSIX (grep, sed, bash)
\[[^\][]*] - ECMAScript (JavaScript, C++ std::regex, VBA RegExp)
\[[^\]\[]*] - Java, ICU regex
\[[^\]\[]*\] - Onigmo (Ruby, requires escaping of brackets everywhere)
Excluding the brackets:
(?<=\[)[^][]*(?=]) - PCRE, Python re/regex, .NET (C#, etc.), JGSoft Software
\[([^][]*)] - Bash, Golang - capture the contents between the square brackets with a pair of unescaped parentheses, also see below
\[([^\][]*)] - JavaScript, C++ std::regex, VBA RegExp
(?<=\[)[^\]\[]*(?=]) - Java regex, ICU (R stringr)
(?<=\[)[^\]\[]*(?=\]) - Onigmo (Ruby, requires escaping of brackets everywhere)
NOTE: * matches 0 or more characters, use + to match 1 or more to avoid empty string matches in the resulting list/array.
Whenever both lookaround support is available, the above solutions rely on them to exclude the leading/trailing open/close bracket. Otherwise, rely on capturing groups (links to most common solutions in some languages have been provided).
If you need to match nested parentheses, you may see the solutions in the Regular expression to match balanced parentheses thread and replace the round brackets with the square ones to get the necessary functionality. You should use capturing groups to access the contents with open/close bracket excluded:
\[((?:[^][]++|(?R))*)] - PHP PCRE
\[((?>[^][]+|(?<o>)\[|(?<-o>]))*)] - .NET demo
\[(?:[^\]\[]++|(\g<0>))*\] - Onigmo (Ruby) demo
If you do not want to include the brackets in the match, here's the regex: (?<=\[).*?(?=\])
Let's break it down
The . matches any character except for line terminators. The ?= is a positive lookahead. A positive lookahead finds a string when a certain string comes after it. The ?<= is a positive lookbehind. A positive lookbehind finds a string when a certain string precedes it. To quote this,
Look ahead positive (?=)
Find expression A where expression B follows:
A(?=B)
Look behind positive (?<=)
Find expression A where expression B
precedes:
(?<=B)A
The Alternative
If your regex engine does not support lookaheads and lookbehinds, then you can use the regex \[(.*?)\] to capture the innards of the brackets in a group and then you can manipulate the group as necessary.
How does this regex work?
The parentheses capture the characters in a group. The .*? gets all of the characters between the brackets (except for line terminators, unless you have the s flag enabled) in a way that is not greedy.
Just in case, you might have had unbalanced brackets, you can likely design some expression with recursion similar to,
\[(([^\]\[]+)|(?R))*+\]
which of course, it would relate to the language or RegEx engine that you might be using.
RegEx Demo 1
Other than that,
\[([^\]\[\r\n]*)\]
RegEx Demo 2
or,
(?<=\[)[^\]\[\r\n]*(?=\])
RegEx Demo 3
are good options to explore.
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
Test
const regex = /\[([^\]\[\r\n]*)\]/gm;
const str = `This is a [sample] string with [some] special words. [another one]
This is a [sample string with [some special words. [another one
This is a [sample[sample]] string with [[some][some]] special words. [[another one]]`;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
Source
Regular expression to match balanced parentheses
(?<=\[).*?(?=\]) works good as per explanation given above. Here's a Python example:
import re
str = "Pagination.go('formPagination_bottom',2,'Page',true,'1',null,'2013')"
re.search('(?<=\[).*?(?=\])', str).group()
"'formPagination_bottom',2,'Page',true,'1',null,'2013'"
The #Tim Pietzcker's answer here
(?<=\[)[^]]+(?=\])
is almost the one I've been looking for. But there is one issue that some legacy browsers can fail on positive lookbehind.
So I had to made my day by myself :). I manged to write this:
/([^[]+(?=]))/g
Maybe it will help someone.
console.log("this is a [sample] string with [some] special words. [another one]".match(/([^[]+(?=]))/g));
if you want fillter only small alphabet letter between square bracket a-z
(\[[a-z]*\])
if you want small and caps letter a-zA-Z
(\[[a-zA-Z]*\])
if you want small caps and number letter a-zA-Z0-9
(\[[a-zA-Z0-9]*\])
if you want everything between square bracket
if you want text , number and symbols
(\[.*\])
This code will extract the content between square brackets and parentheses
(?:(?<=\().+?(?=\))|(?<=\[).+?(?=\]))
(?: non capturing group
(?<=\().+?(?=\)) positive lookbehind and lookahead to extract the text between parentheses
| or
(?<=\[).+?(?=\]) positive lookbehind and lookahead to extract the text between square brackets
In R, try:
x <- 'foo[bar]baz'
str_replace(x, ".*?\\[(.*?)\\].*", "\\1")
[1] "bar"
([[][a-z \s]+[]])
Above should work given the following explaination
characters within square brackets[] defines characte class which means pattern should match atleast one charcater mentioned within square brackets
\s specifies a space
+ means atleast one of the character mentioned previously to +.
I needed including newlines and including the brackets
\[[\s\S]+\]
If someone wants to match and select a string containing one or more dots inside square brackets like "[fu.bar]" use the following:
(?<=\[)(\w+\.\w+.*?)(?=\])
Regex Tester
I have a requirement where I am working on a large data which is having double byte characters, in korean text. i want to look for a character and replace it. In order to display the korean text correctly in the browser I have changed the locale settings in R. But not sure if it gets updated for the code as well. below is my code to change locale to korean and the korean text gets visible properly in viewer, however in console it gives junk character on printing-
Sys.setlocale(category = "LC_ALL", locale = "korean")
My data is in a data.table format that contains a column with text in korean. example -
"광주광역시 동구 제봉로 49 (남동,(지하))"
I want to get rid of the 1st word which ends with "시" character. Then I want to get rid of the "(남동,(지하))" an the end. I was trying gsub, but it does not seem to be working.
New <- c("광주광역시 동구 제봉로 49 (남동,(지하))")
data <- as.data.table(New)
data[,New_trunc := gsub("\\b시", "", data$New)]
Please let me know where I am going wrong. Since I want to search the end of word, I am using \\b and since I want to replace any word ending with "시" character I am giving it as \\b시.....is this not the way to give? How to take care of () at the end of the sentence.
What would be a good source to refer to for regular expressions.
Is a utf-8 setting needed for the script as well?How to do that?
Since you need to match the letter you have at the end of the word, you need to place \b (word boundary) after the letter, so as to require a transition from a letter to a non-letter (or end of string) after that letter. A PCRE pattern that will handle this is
"\\s*\\b\\p{L}*시\\b"
Details
\\s* - zero or more whitespaces
\\b - a leading word boundary
\\p{L}* - zero or more letters
시 - your specific letter
\\b - end of the word
The second issue is that you need to remove a set of nested parentheses at the end of the string. You need again to rely on the PCRE regex (perl=TRUE) that can handle recursion with the help of a subroutine call.
> sub("\\s*(\\((?:[^()]++|(?1))*\\))$", "", New, perl=TRUE)
[1] "광주광역시 동구 제봉로 49"
Details:
\\s* - zero or more whitespaces
(\\((?:[^()]++|(?1))*\\)) - Group 1 (will be recursed) matching
\\( - a literal (
(?:[^()]++|(?1))* - zero or more occurrences of
[^()]++ - 1 or more chars other than ( and ) (possessively)
| - or
(?1) - a subroutine call that repeats the whole Group 1 subpattern
\\) - a literal )
$ - end of string.
Now, if you need to combine both, you would see that R PCRE-powered gsub does not handle Unicode chars in the pattern so easily. You must tell it to use Unicode mode with (*UCP) PCRE verb.
> gsub("(*UCP)\\b\\p{L}*시\\b|\\s*(\\((?:[^()]++|(?1))*\\))$", "", New, perl=TRUE)
[1] " 동구 제봉로 49"
Or using trimws to get rid of the leading/trailing whitespace:
> trimws(gsub("(*UCP)\\b\\p{L}*시\\b|(\\((?:[^()]++|(?1))*\\))$", "", New, perl=TRUE))
[1] "동구 제봉로 49"
See more details about the verb at PCRE Man page.