How to remove all line breaks (enter symbols) from the string?
my_string <- "foo\nbar\rbaz\r\nquux"
I've tried gsub("\n", "", my_string), but it doesn't work, because new line and line break aren't equal.
You need to strip \r and \n to remove carriage returns and new lines.
x <- "foo\nbar\rbaz\r\nquux"
gsub("[\r\n]", "", x)
## [1] "foobarbazquux"
Or
library(stringr)
str_replace_all(x, "[\r\n]" , "")
## [1] "foobarbazquux"
I just wanted to note here that if you want to insert spaces where you found newlines the best option is to use the following:
gsub("\r?\n|\r", " ", x)
which will insert only one space regardless whether the text contains \r\n, \n or \r.
Have had success with:
gsub("\\\n", "", x)
With stringr::str_remove_all
library(stringr)
str_remove_all(my_string, "[\r\n]")
# [1] "foobarbazquux"
Related
I am trying to clean some garbage out of some text. While doing this, I am assuming that any word that has a letter (any letter) repeated three or more times is garbage - and I want to remove it.
I've come up with this:
gsub(pattern = "[a-zA-Z]\\1\\1", replacement = "", string)
in which string is the character vector, but this doesn't work. Everything else I've tried might find the pattern, but it just removes the pattern, leaving a mess. I'm trying to remove the whole word with the pattern in it.
Any ideas?
You need
gsub("\\s*[[:alpha:]]*([[:alpha:]])\\1{2}[[:alpha:]]*", "", string)
gsub("\\s*\\p{L}*(\\p{L})\\1{2}\\p{L}*", "", string, perl=TRUE)
stringr::str_replace_all(string, "\\s*\\p{L}*(\\p{L})\\1{2}\\p{L}*", "")
See an R demo:
string <- "This is a baaaad unnnnecessary short word"
gsub("\\s*[[:alpha:]]*([[:alpha:]])\\1{2}[[:alpha:]]*", "", string)
gsub("\\s*\\p{L}*(\\p{L})\\1{2}\\p{L}*", "", string, perl=TRUE)
library(stringr)
str_replace_all(string, "\\s*\\p{L}*(\\p{L})\\1{2}\\p{L}*", "")
All yielding [1] "This is a short word".
See the regex demo. Regex details:
\s* - zero or more whitespaces
\p{L}* / [[:alpha:]]* - zero or more letters
(\p{L}) - Capturing group 1: any single letter
\1{2} - two occurrences of the same value as in Group 1
\p{L}* / [[:alpha:]]* - zero or more letters.
You need to assign a "capture group" to the [.] class by wrapping it in parens, since the \\1 needs something to reference:
gsub("([a-zA-Z])\\1\\1", "", "aabbbccdddee")
# [1] "aaccee"
Updated on OP comment:
Try this:
gsub("([A-Z]&|[a-z])\\1{2, }", "", "AAA")
[1] "AAA"
gsub("([A-Z]&|[a-z])\\1{2, }", "", "aabbbccdddee")
[1] "aaccee"
r2evans example with different regex:
gsub("(\\w)\\1{2, }", "", "aabbbccdddee")
[1] "aaccee"
I would like to remove this character
c("
I use this
df <- gsub("c/(/"", " ", df$text)
But I receive this error:
Error: unexpected string constant in "inliwc <- gsub("c/(/"", ""
What can I do?
You need to escape the round brackets as well as the quotes which can be done as :
temp <- 'this is ac(" string'
gsub("c\\(\"", " ", temp)
#OR use single quotes in gsub
#gsub('c\\("', " ", temp)
#[1] "this is a string"
A faster way would be to use fixed = TRUE
gsub('c("', " ", temp, fixed = TRUE)
You can also use sub if there is a single occurrence of the pattern in the string.
The opening round bracket is a regex metacharacter; in R, its literal use needs to be escaped using \\:
text <- "c("
text <- gsub("c\\(", "", text)
We can also use sub
sub('c[()]"', '', temp)
#[1] "this is a string"
data
temp <- 'this is ac(" string'
I want to ignore the spaces and underscores in the beginning of a string in R.
I can write something like
txt <- gsub("^\\s+", "", txt)
txt <- gsub("^\\_+", "", txt)
But I think there could be an elegant solution
txt <- " 9PM 8-Oct-2014_0.335kwh "
txt <- gsub("^[\\s+|\\_+]", "", txt)
txt
The output should be "9PM 8-Oct-2014_0.335kwh ". But my code gives " 9PM 8-Oct-2014_0.335kwh ".
How can I fix it?
You could bundle the \s and the underscore only in a character class and use quantifier to repeat that 1+ times.
^[\s_]+
Regex demo
For example:
txt <- gsub("^[\\s_]+", "", txt, perl=TRUE)
Or as #Tim Biegeleisen points out in the comment, if only the first occurrence is being replaced you could use sub instead:
txt <- sub("[\\s_]+", "", txt, perl=TRUE)
Or using a POSIX character class
txt <- sub("[[:space:]_]+", "", txt)
More info about perl=TRUE and regular expressions used in R
R demo
The stringr packages offers some task specific functions with helpful names. In your original question you say you would like to remove whitespace and underscores from the start of your string, but in a comment you imply that you also wish to remove the same characters from the end of the same string. To that end, I'll include a few different options.
Given string s <- " \t_blah_ ", which contains whitespace (spaces and tabs) and underscores:
library(stringr)
# Remove whitespace and underscores at the start.
str_remove(s, "[\\s_]+")
# [1] "blah_ "
# Remove whitespace and underscores at the start and end.
str_remove_all(s, "[\\s_]+")
# [1] "blah"
In case you're looking to remove whitespace only – there are, after all, no underscores at the start or end of your example string – there are a couple of stringr functions that will help you keep things simple:
# `str_trim` trims whitespace (\s and \t) from either or both sides.
str_trim(s, side = "left")
# [1] "_blah_ "
str_trim(s, side = "right")
# [1] " \t_blah_"
str_trim(s, side = "both") # This is the default.
# [1] "_blah_"
# `str_squish` reduces repeated whitespace anywhere in string.
s <- " \t_blah blah_ "
str_squish(s)
# "_blah blah_"
The same pattern [\\s_]+ will also work in base R's sub or gsub, with some minor modifications, if that's your jam (see Thefourthbird`s answer).
You can use stringr as:
txt <- " 9PM 8-Oct-2014_0.335kwh "
library(stringr)
str_trim(txt)
[1] "9PM 8-Oct-2014_0.335kwh"
Or the trimws in Base R
trimws(txt)
[1] "9PM 8-Oct-2014_0.335kwh"
Is there any effective way to remove punctuation in text but keeping hyphenated expressions, such as "accident-prone"?
I used the following function to clean my text
clean.text = function(x)
{
# remove rt
x = gsub("rt ", "", x)
# remove at
x = gsub("#\\w+", "", x)
x = gsub("[[:punct:]]", "", x)
x = gsub("[[:digit:]]", "", x)
# remove http
x = gsub("http\\w+", "", x)
x = gsub("[ |\t]{2,}", "", x)
x = gsub("^ ", "", x)
x = gsub(" $", "", x)
x = str_replace_all(x, "[^[:alnum:][:space:]'-]", " ")
#return(x)
}
and apply it on hyphenated expressions that returned
my_text <- "accident-prone"
new_text <- clean.text(text)
new_text
[1] "accidentprone"
while my desired output is
"accident-prone"
I have referenced this thread but didn't find it worked on my situation. There must be some regex things that I haven't figured out. It will be really appreciated if someone could enlighten me on this.
Putting my two cents in, you could use (*SKIP)(*FAIL) with perl = TRUE and remove any non-word characters:
data <- c("my-test of #$%^&*", "accident-prone")
(gsub("(?<![^\\w])[- ](?=\\w)(*SKIP)(*FAIL)|\\W+", "", data, perl = TRUE))
Resulting in
[1] "my-test of" "accident-prone"
See a demo on regex101.com.
Here the idea is to match what you want to keep
(?<![^\\w])[- ](?=\\w)
# a whitespace or a dash between two word characters
# or at the very beginning of the string
let these fail with (*SKIP)(*FAIL) and put what you want to be removed on the right side of the alternation, in this case
\W+
effectively removing any non-word-characters not between word characters.
You'd need to provide more examples for testing though.
The :punct: set of characters includes the dash and you are removing them. You could make an alternate character class that omits the dash. You do need to pay special attention to the square-brackets placements and escape the double quote and the backslash:
(test <- gsub("[]!\"#$%&'()*+,./:;<=>?#[\\^_`{|}~]", "", "my-test of #$%^&*") )
[1] "my-test of "
The ?regex (help page) advises against using ranges. I investigated whether there might be any simplification using my local ASCII sequence of punctuation, but it quickly became obvious that was not the way to go for other reasons. There were 5 separate ranges, and the "]" was in the middle of one of them so there would have been 7 ranges to handle in addition to the "]" which needs to come first.
I want to replace a white space with ONE backslash and a whitespace like this:
"foo bar" --> "foo\ bar"
I found how to replace with multiple backslashes but wasn't able to adapt it to a single backslash.
I tried this so far:
x <- "foo bar"
gsub(" ", "\\ ", x)
# [1] "foo bar"
gsub(" ", "\\\ ", x)
# [1] "foo bar"
gsub(" ", "\\\\ ", x)
# [1] "foo\\ bar"
However, all the outcomes do not satisfy my needs. I need the replacement to dynamically create file paths which contain folders with names like
/some/path/foo bar/foobar.txt.
To use them for shell commands in system() white spaces have to be exited with a \ to
/some/path/foo\ bar/foobar.txt.
Do you know how to solve this one?
Your problem is a confusion between the content of a string and its representation. When you print out a string in the ordinary way in R you will never see a single backslash (unless it's denoting a special character, e.g. print("y\n"). If you use cat() instead, you'll see only a single backslash.
x <- "foo bar"
y <- gsub(" ", "\\\\ ", x)
print(y)
## [1] "foo\\ bar"
cat(y,"\n") ## string followed by a newline
## foo\ bar
There are 8 characters in the string; 6 letters, one space, and the backslash.
nchar(y) ## 8
For comparison, consider \n (newline character).
z <- gsub(" ", "\n ", x)
print(z)
## [1] "foo\n bar"
cat(z,"\n")
## foo
## bar
nchar(z) ## 8
If you're constructing file paths, it might be easier to use forward slashes instead - forward slashes work as file separators in R on all operating systems (even Windows). Or check out file.path(). (Without knowing exactly what you're trying to do, I can't say more.)
To replace a space with one backslash and a space, you do not even need to use regular expression, use your gsub(" ", "\\ ", x) first attempt with fixed=TRUE:
> x <- "foo bar"
> res <- gsub(" ", "\\ ", x, fixed=TRUE)
> cat(res, "\n")
foo\ bar
See an online R demo
The cat function displays the "real", literal backslashes.