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.
Related
I have looked at lots of posts here on SO with suggestions on REGEX patterns to grab texts from parentheses. However, from what I have looked into I cannot find a solution that works.
For example, I have had a look at the following:
R - Regular Expression to Extract Text Between Parentheses That Contain Keyword, Extract text in parentheses in R, regex to pickout some text between parenthesis [duplicate]
In the following order, here were the top answers solutions (with some amendments):
pattern1= '\\([^()]*[^()]*\\)'
pattern2= '(?<=\\()[^()]*(?=\\))'
pattern3= '.*\\((.*)\\).*'
all_patterns = c(pattern1, pattern2, pattern3)
I have used the following:
sapply(all_patterns , function(x)stringr::str_extract('I(data^2)', x))
\\([^()]*[^()]*\\) (?<=\\()[^()]*(?=\\)) .*\\((.*)\\).*
"(data^2)" "data^2" "I(data^2)"
None of these seem to only grab the characters within the brackets, so how can I just grab the characters inside brackets?
Expected output:
data
With str_extract, it would extract all those characters matched in the patterns. Instead, use a regex lookaround to match one or more characters that are not a ^ or the closing bracket ()) ([^\\^\\)]+) that succeeds an opening bracket ((?<=\\() - these are escaped (\\) as they are metacharacters
library(stringr)
str_extract('I(data^2)', '(?<=\\()[^\\^\\)]+')
# [1] "data"
Here is combinations of str_extract and str_remove
library(stringr)
str_extract(str_remove('I(data^2)', '.\\('), '\\w*')
[1] "data"
I'm trying to use stringr or R base calls to conditionally add a white-space for instances in a large vector where there is a numeric value then a special character - in this case a $ sign without a space. str_pad doesn't appear to allow for a reference vectors.
For example, for:
$6.88$7.34
I'd like to add a whitespace after the last number and before the next dollar sign:
$6.88 $7.34
Thanks!
If there is only one instance, then use sub to capture digit and the $ separately and in the replacement add the space between the backreferences of the captured group
sub("([0-9])([$])", "\\1 \\2", v1)
#[1] "$6.88 $7.34"
Or with a regex lookaround
gsub("(?<=[0-9])(?=[$])", " ", v1, perl = TRUE)
data
v1 <- "$6.88$7.34"
This will work if you are working with a vectored string:
mystring<-as.vector('$6.88$7.34 $8.34$4.31')
gsub("(?<=\\d)\\$", " $", mystring, perl=T)
[1] "$6.88 $7.34 $8.34 $4.31"
This includes cases where there is already space as well.
Regarding the question asked in the comments:
mystring2<-as.vector('Regular_Distribution_Type† Income Only" "Distribution_Rate 5.34%" "Distribution_Amount $0.0295" "Distribution_Frequency Monthly')
gsub("(?<=[[:alpha:]])\\s(?=[[:alpha:]]+)", "_", mystring2, perl=T)
[1] "Regular_Distribution_Type<U+2020> Income_Only\" \"Distribution_Rate 5.34%\" \"Distribution_Amount $0.0295\" \"Distribution_Frequency_Monthly"
Note that the \ appears due to nested quotes in the vector, should not make a difference. Also <U+2020> appears due to encoding the special character.
Explanation of regex:
(?<=[[:alpha:]]) This first part is a positive look-behind created by ?<=, this basically looks behind anything we are trying to match to make sure what we define in the look behind is there. In this case we are looking for [[:alpha:]] which matches a alphabetic character.
We then check for a blank space with \s, in R we have to use a double escape so \\s, this is what we are trying to match.
Finally we use (?=[[:alpha:]]+), which is a positive look-ahead defined by ?= that checks to make sure our match is followed by another letter as explained above.
The logic is to find a blank space between letters, and match the space, which then is replaced by gsub, with a _
See all the regex here
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\""
I'm trying to use a regex to replace the last instance of a phrase (and everything after that phrase, which could be any character):
stringi::stri_replace_last_regex("_AB:C-_ABCDEF_ABC:45_ABC:454:", "_ABC.*$", "CBA")
However, I can't seem to get the refex to function properly:
Input: "_AB:C-_ABCDEF_ABC:45_ABC:454:"
Actual output: "_AB:C-CBA"
Desired output: "_AB:C-_ABCDEF_ABC:45_CBA"
I have tried gsub() as well but that hasn't worked.
Any ideas where I'm going wrong?
One solution is:
sub("(.*)_ABC.*", "\\1_CBA", Input)
[1] "_AB:C-_ABCDEF_ABC:45_CBA"
Have a look at what stringi::stri_replace_last_regex does:
Replaces with the given replacement string last substring of the input that matches a regular expression
What does your _ABC.*$ pattern match inside _AB:C-_ABCDEF_ABC:45_ABC:454:? It matches the first _ABC (that is right after C-) and all the text after to the end of the line (.*$ grabs 0+ chars other than line break chars to the end of the line). Hence, you only have 1 match, and it is the last.
Solutions can be many:
1) Capturing all text before the last occurrence of the pattern and insert the captured value with a replacement backreference (this pattern does not have to be anchored at the end of the string with $):
sub("(.*)_ABC.*", "\\1_CBA","_AB:C-_ABCDEF_ABC:45_ABC:454:")
2) Using a tempered greedy token to make sure you only match any char that does not start your pattern up to the end of the string after matching it (this pattern must be anchored at the end of the string with $):
sub("(?s)_ABC(?:(?!_ABC).)*$", "_CBA","_AB:C-_ABCDEF_ABC:45_ABC:454:", perl=TRUE)
Note that this pattern will require perl=TRUE argument to be parsed with a PCRE engine with sub (or you may use stringr::str_replace that is ICU regex library powered and supports lookaheads)
3) A negative lookahead may be used to make sure your pattern does not appear anywhere to the right of your pattern (this pattern does not have to be anchored at the end of the string with $):
sub("(?s)_ABC(?!.*_ABC).*", "_CBA","_AB:C-_ABCDEF_ABC:45_ABC:454:", perl=TRUE)
See the R demo online, all these three lines of code returning _AB:C-_ABCDEF_ABC:45_CBA.
Note that (?s) in the PCRE patterns is necessary in case your strings may contain a newline (and . in a PCRE pattern does not match newline chars by default).
Arguably the safest thing to do is using a negative lookahead to find the last occurrence:
_ABC(?:(?!_ABC).)+$
Demo
gsub("_ABC(?:(?!_ABC).)+$", "_CBA","_AB:C-_ABCDEF_ABC:45_ABC:454:", perl=TRUE)
[1] "_AB:C-_ABCDEF_ABC:45_CBA"
Using gsub and back referencing
gsub("(.*)ABC.*$", "\\1CBA","_AB:C-_ABCDEF_ABC:45_ABC:454:")
[1] "_AB:C-_ABCDEF_ABC:45_CBA"
I am trying to replace instances in a string which begin with abc in a text I'm working with in R. The output text is highlighted in HTML over a couple of passes, so I need the replacement to ignore text inside HTML carets.
The following seems to work in Python but I'm not getting any hits on my regex in R. All help appreciated.
test <- 'abcdef abc<span abc>defabc abcdef</span> abc defabc'
gsub('\\babc\\(?![^<]*>\\)', 'xxx', test)
Expected output:
xxxdef xxx<span abc>defabc xxxdef</span> xxx defabc
Instead it is ignoring all instances of abc.
You need to remove unnecessary escapes and use perl=TRUE:
test <- 'abcdef abc<span abc>defabc abcdef</span> abc defabc'
gsub('\\babc(?![^<]*>)', 'xxx', test, perl=TRUE)
## => [1] "xxxdef xxx<span abc>defabc xxxdef</span> xxx defabc"
See the online R demo
When you escape (, it matches a literal ( symbol, so, in your pattern, \\(?![^<]*>\\) matches a ( 1 or 0 times, then !, then 0+ chars other than <, then > and a literal ). In my regex, (?![^<]*>) is a negative lookahead that fails the match if an abc is followed with any 0+ chars other than < and then a >.
Without perl=TRUE, R gsub uses the TRE regex flavor that does not support lookarounds (even lookaheads). Thus, you have to tell gsub via perl=TRUE that you want the PCRE engine to be used.
See the online PCRE regex demo.