For my project I am looking into removing parts of text based on the pattern of special characters. I have a long .txt file that has the below structure:
mycharobj=c("---------Some text is here.---------More text is here - [3548]----- Even more text is here.-----------More text is here - [408]--------- Even more text is here again.")
String continues following the above pattern.
My target is to remove parts that start with - and end - [number], such as:
"-----------------------More text is here - [3548]"
"-----------More text is here - [408]"
I am planning to use the below to remove these parts with (will be looped in the future)
library(stringr)
library(qdapRegex)
temp=unlist(regmatches(mycharobj, gregexpr("[[:digit:]]+", mycharobj)))
mycharobj=rm_between(mycharobj, "-", paste(temp[1],"]", sep=""))
but for this to work, I need a regex expression that will remove the first occurrence of "-----------" in text until the first word or word character. If a string starts with text (word or word characters), it needs to ignore this and identify the first occurrence of "-----------" for my potential loop to work.
I was wondering if this can be done with regular expressions? Any help is appreciated. I have a very computationally demanding solution for this; split the string based on the special character "-" and then identify the parts of the text that I need through a set of conditionals. But due to the fact that it takes a lot more of the processing time, this solution is not very scalable for processing a large number of such .txt files.
You can use
gsub("-{9,}(?:(?!-{9}).)*?- \\[\\d+]", "", mycharobj, perl=TRUE)
See the regex demo.
Details:
-{9,} - nine or more - chars
(?:(?!-{9}).)*? - any one char, other than a line break char, zero or more but as few as possible occurrences, that does not start a nine hyphen char sequence
- \[ - a - [ string
\d+ - one or more digits
] - a ] char.
Related
I have a question similar to this one but instead of having two specific characters to look between, I want to get the text between a space and a specific character. In my example, I have this string:
myString <- "This is my string I scraped from the web. I want to remove all instances of a picture. picture-file.jpg. The text continues here. picture-file2.jpg"
but if I were to do something like this: str_remove_all(myString, " .*jpg) I end up with
[1] "This"
I know that what's happening is R is finding the first instance of a space and removing everything between that space and ".jpg" but I want it to be the first space immediately before ".jpg". My final result I hope for looks like this:
[1] "This is my string I scraped from the web. I want to remove all instances of a picture. the text continues here.
NOTE: I know that a solution may arise which does what I want, but ends up putting two periods next to each other. I do not mind a solution like that because later in my analysis I am removing punctuation.
You can use
str_remove_all(myString, "\\S*\\.jpg")
Or, if you also want to remove optional whitespace before the "word":
str_remove_all(myString, "\\s*\\S*\\.jpg")
Details:
\s* - zero or more whitespaces
\S* - zero or more non-whitespaces
\.jpg - .jpg substring.
To make it case insensitive, add (?i) at the pattern part: "(?i)\\s*\\S*\\.jpg".
If you need to make sure there is no word char after jpg, add a word boundary: "(?i)\\s*\\S*\\.jpg\\b"
I am trying to remove the decimal points in decimal numbers in R. Please note I want to keep the full stop of strings.
Example:
data= c("It's 6.00pm, and is late.")
I know that I have to use regex for this, but I am struggling. My desired output is:
"It's 6 00pm, and is late."
Thank you in advance.
Try this:
sub("(?<=\\d)\\.(?=\\d)", " ", data, perl = TRUE)
This solution uses lookbehind (?<=...) and lookahead (?=...)to assert that the period you wish to remove be enclosed by digits (thus avoiding matching the period at the sentence end). If you have several such cases within strings, then use gsubinstead of sub.
I suggest using a simple pattern to find the target text, then adding parenthesis to identify the parts of the matching text that you want to retain.
# Test data
data <- c("It's 6.00pm, and is late.")
The target pattern is a literal dot with a string of digits before and after it. \\d+ matches one or more digits and \\. matches a literal dot. Testing the pattern to see if it works:
grepl("\\d+\\.\\d+", data)
Result
TRUE
If we wanted too eliminate the whole thing we could do a simple replacement with an empty string. Testing if this targets the correct text:
sub("\\d+\\.\\d+", "", data)
Result
"It's pm, and is late."
Instead, to discard only a section of matched text we can identify the parts we want to keep, which is done by surrounding them with parenthesis. Once done we can refer to the captured text in the replacement. \\1 refers to the first chunk of text captured and \\2 refers to the second chunk of text, corresponding to the first and second sets of parenthesis
# pattern replacement
sub("(\\d+)\\.(\\d+)", "\\1\\2", data)
Result
[1] "It's 600pm, and is late."
This effectively removes the dot by omitting it from the replacement text.
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 would like to split strings like the following:
x <- "abc-1230-xyz-[def-ghu-jkl---]-[adsasa7asda12]-s-[klas-bst-asdas foo]"
by dash (-) on the condition that those dashes must not be contained inside a pair of []. The expected result would be
c("abc", "1230", "xyz", "[def-ghu-jkl---]", "[adsasa7asda12]", "s",
"[klas-bst-asdas foo]")
Notes:
There is no nesting of square brackets inside each other.
The square brackets can contain any characters / numbers / symbols except square brackets.
The other parts of the string are also variable so that we can only assume that we split by - whenever it's not inside [].
There's a similar question for python (How to split a string by commas positioned outside of parenthesis?) but I haven't yet been able to accurately adjust that to my scenario.
You could use look ahead to verify that there is no ] following sooner than a [:
-(?![^[]*\])
So in R:
strsplit(x, "-(?![^[]*\\])", perl=TRUE)
Explanation:
-: match the hyphen
(?! ): negative look ahead: if that part is found after the previously matched hyphen, it invalidates the match of the hyphen.
[^[]: match any character that is not a [
*: match any number of the previous
\]: match a literal ]. If this matches, it means we found a ] before finding a [. As all this happens in a negative look ahead, a match here means the hyphen is not a match. Note that a ] is a special character in regular expressions, so it must be escaped with a backslash (although it does work without escape, as the engine knows there is no matching [ preceding it -- but I prefer to be clear about it being a literal). And as backslashes have a special meaning in string literals (they also denote an escape), that backslash itself must be escaped again in this string, so it appears as \\].
Instead of splitting, extract the parts:
library(stringr)
str_extract_all(x, "(\\[[^\\[]*\\]|[^-])+")
I am not familiar with r language, but I believe it can do regex based search and replace. Instead of struggling with one single regex split function, I would go in 3 steps:
replace - in all [....] parts by a invisible char, like \x99
split by -
for each element in the above split result(array/list), replace \x99 back to -
For the first step, you can find the parts by \[[^]]
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.