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.
Related
I'm just getting to know the language R, previously worked with python. The challenge is to replace the last character of each word in the string with *.
How it should look: example text in string, and result work: exampl* tex* i* strin*
My code:
library(tidyverse)
library(stringr)
string_example = readline("Enter our text:")
string_example = unlist(strsplit(string_example, ' '))
string_example
result = str_replace(string_example, pattern = "*\b", replacement = "*")
result
I get an error:
> result = str_replace(string_example, pattern = "*\b", replacement = "*")
Error in stri_replace_first_regex(string, pattern, fix_replacement(replacement), :
Syntax error in regex pattern. (U_REGEX_RULE_SYNTAX, context=``)
Help solve the task
Oh, I noticed an error, the pattern should be .\b. this is how the code is executed, but there is no replacement in the string
If you mean words consisting of letters only, you can use
string_example <- "example text in string"
library(stringr)
str_replace_all(string_example, "\\p{L}\\b", "*")
## => [1] "exampl* tex* i* strin*"
See the R demo and the regex demo.
Details:
\p{L} - a Unicode category (propery) class matching any Unicode letter
\b - a word boundary, in this case, it makes sure there is no other word character immediately on the right. It will fails the match if the letter matched with \p{L} is immediately followed with a letter, digit or _ (these are all word chars). If you want to limit this to a letter check, replace \b with (?!\p{L}).
Note the backslashes are doubled because in regular string literals backslashes are used to form string escape sequences, and thus need escaping themselves to introduce literal backslashes in string literals.
Some more things to consider
If you do not want to change one-letter words, add a non-word boundary at the start, "\\B\\p{L}\\b"
If you want to avoid matching letters that are followed with - + another letter (i.e. some compound words), you can add a lookahead check: "\\p{L}\\b(?!-)".
You may combine the lookarounds and (non-)word boundaries as you need.
I'm trying to remove the numbers at the beginning of a row inside quotation marks.
> g<-"My name is Paul.\nI like playing football.\n\"55012\" And that's all."
> cat(g)
My name is Paul.
I like playing football.
"55012" And that's all.
> gsub("[\r\n]\"+[[:digit:]][^[[:space:]]]*"," ",g)
[1] "My name is Paul.\nI like playing football. 012\" And that's all."
This should work, but I don't know why only \n"55 is being replaced and not the entire number.
You closed the bracket expression with a couple of redundant [...]. [^[[:space:]]] is a sequence of [^[[:space:]] and ] patterns and matches any char other than [ and whitespace and then a ] char.
However, even that is not enough to fully fix the issue.
You may use
gsub("(^|\n)\"+[0-9]+\"+\\s*","\\1", g)
See the R demo
Pattern details
(^|\n) - start of string or a newline captured in Group 1 (referred to with \1 from the replacement pattern)
\"+ - one or more double quotes
[0-9]+ - 1+ digits
\"+ - one or more double quotes
\s* - 0+ whitespaces.
See the regex demo
I have the following string:
x = "marchTextIWantToDisplayWithSpacesmarch"
I would like to delete the 'march' portion at the beginning of the string and then add a space before each uppercase letter in the remainder to yield the following result:
"Text I Want To Display With Spacesmarch"
To insert whitepace, I used gsub("([a-z]?)([A-Z])", "\\1 \\2", x, perl= T) but I have no clue how to modify the pattern so that the first 'march' is excluded from the returned string. I'm trying to get better at this so any help would be greatly appreciated.
An option would be to capture the upper case letter as a group ((...)) and in the replacement create a space followed by the backreference (\\1) of the captured group
gsub("([A-Z])", " \\1", x)
#[1] "march Text I Want To Display With Spacesmarch"
If we need to remove the 'march'
sub("\\b[a-z]\\w+\\s+", "", gsub("([A-Z])", " \\1", x))
[#1] "Text I Want To Display With Spacesmarch"
data
x <- "marchTextIWantToDisplayWithSpacesmarch"
No, you can't achieve your replacement using single gsub because in one of your requirement, you want to remove all lowercase letters starting from the beginning, and your second requirement is to introduce a space before every capital letter except the first capital letter of the resultant string after removing all lowercase letters from the beginning of text.
Doing it in single gsub call would have been possible in cases where somehow we can re-use some of the existing characters to make the conditional replace which can't be the case here. So in first step, you can use ^[a-z]+ regex to get rid of all lowercase letters only from the beginning of string,
sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch")
leaving you with this,
[1] "TextIWantToDisplayWithSpacesmarch"
And next step you can use this (?<!^)(?=[A-Z]) regex to insert a space before every capital letter except the first one as you might not want an extra space before your sentence. But you can combine both and write them as this,
gsub('(?<!^)(?=[A-Z])', ' ', sub('^[a-z]+', '', "marchTextIWantToDisplayWithSpacesmarch"), perl=TRUE)
which will give you your desired string,
[1] "Text I Want To Display With Spacesmarch"
Edit:
Explanation of (?<!^)(?=[A-Z]) pattern
First, let's just take (?=[A-Z]) pattern,
See the pink markers in this demo
As you can see, in the demo, every capital letter is preceded by a pink mark which is the place where a space will get inserted. But we don't want space to be inserted before the very first letter as that is not needed. Hence we need a condition in regex, which will not select the first capital letter which appears at the start of string. And for that, we need to use a negative look behind (?<!^) which means that Do not select the position which is preceded by start of string and hence this (?<!^) helps in discarding the upper case letter that is preceded by just start of string.
See this demo where the pink marker is gone from the very first uppercase letter
Hope this clarifies how every other capital letter is selected but not the very first. Let me know if you have any queries further.
You may use a single regex call to gsub coupled with trimws to trim the resulting string:
trimws(gsub("^\\p{Ll}+|(?<=.)(?=\\p{Lu})", " ", x, perl=TRUE))
## => [1] "Text I Want To Display With Spacesmarch"
It also supports all Unicode lowercase (\p{Ll}) and uppercase (\p{Lu}) letters.
See the R demo online and the regex demo.
Details
^\\p{Ll}+ - 1 or more lowercase letters at the string start
| - or
(?<=.)(?=\\p{Lu}) - any location between any char but linebreak chars and an uppercase letter.
Here is an altenative with a single call to gsubfn regex with some ifelse logic:
> gsubfn("^\\p{Ll}*(\\p{L})|(?<=.)(?=\\p{Lu})", function(n) ifelse(nchar(n)>0,n," "), x, perl=TRUE,backref=-1)
[1] "Text I Want To Display With Spacesmarch"
Here, the ^\\p{Ll}*(\\p{L}) part matches 0+ lowercase letters and captures the next uppercase into Group 1 that will be accessed by passing n argument to the anonymous function. If n length is non-zero, this alternative matched and the we need to replace with this value. Else, we replace with a space.
Since this is tagged perl, my 2 cents:
Can you chain together the substitutions inside sub() and gsub()? In newer perl versions an /r option can be added to the s/// substitution so the matched string can be returned "non-destructively" and then matched again. This allows hackish match/substitution/rematches without mastering advanced syntax, e.g.:
perl -E '
say "marchTextIWantToDisplayWithSpacesmarch" =~
s/\Amarch//r =~ s/([[:upper:]])/ $1/gr =~ s/\A\s//r;'
Output
Text I Want To Display With Spacesmarch
This seems to be what #pushpesh-kumar-rajwanshi and #akrun are doing by wrapping gsub inside sub() (and vice versa). In general I don't thinkperl = T captures the full magnificently advanced madness of perl regexps ;-) but gsub/sub must be fast operating on vectors, no?
I have a following string. I tried to remove all the strings before the last space but it seems I can't achieve it.
I tried to follow this post
Use gsub remove all string before first white space in R
str <- c("Veni vidi vici")
gsub("\\s*","\\1",str)
"Venividivici"
What I want to have is only "vici" string left after removing everything before the last space.
Your gsub("\\s*","\\1",str) code replaces each occurrence of 0 or more whitespaces with a reference to the capturing group #1 value (which is an empty string since you have not specified any capturing group in the pattern).
You want to match up to the last whitespace:
sub(".*\\s", "", str)
If you do not want to get a blank result in case your string has trailing whitespace, trim the string first:
sub(".*\\s", "", trimws(str))
Or, use a handy stri_extract_last_regex from stringi package with a simple \S+ pattern (matching 1 or more non-whitespace chars):
library(stringi)
stri_extract_last_regex(str, "\\S+")
# => [1] "vici"
Note that .* matches any 0+ chars as many as possible (since * is a greedy quantifier and . in a TRE pattern matches any char including line break chars), and grabs the whole string at first. Then, backtracking starts since the regex engine needs to match a whitespace with \s. Yielding character by character from the end of the string, the regex engine stumbles on the last whitespace and calls it a day returning the match that is removed afterwards.
See the R demo and a regex demo online:
str <- c("Veni vidi vici")
gsub(".*\\s", "", str)
## => [1] "vici"
Also, you may want to see how backtracking works in the regex debugger:
Those red arrows show backtracking steps.
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"