I have a df = desc with a variable "value" that holds long text and would like to remove every word in that variable that ends with ".htm" . I looked for a long time around here and regex expressions and cannot find a solution.
Can anyone help? Thank you so much!
I tried things like:
library(stringr)
desc <- str_replace_all(desc$value, "\*.htm*$", "")
But I get:
Error: '\*' is an unrecognized escape in character string starting ""\*"
This regex:
Will Catch all that ends with .htm
Will not catch instances with .html
Is not dependent on being in the beginning / end of a string.
strings <- c("random text shouldbematched.htm notremoved.html matched.htm random stuff")
gsub("\\w+\\.htm\\b", "", strings)
Output:
[1] "random text notremoved.html random stuff"
I am not sure what exactly you would like to accomplish, but I guess one of those is what you are looking for:
words <- c("apple", "test.htm", "friend.html", "remove.htm")
# just replace the ".htm" from every string
str_replace_all(words, ".htm", "")
# exclude all words that contains .htm anywhere
words[!grepl(pattern = ".htm", words)]
# exlude all words that END with .htm
words[substr(words, nchar(words)-3, nchar(words)) != ".htm"]
I am not sure if you can use * to tell R to consider any value inside a string, so I would first remove it. Also, in your code you are setting a change in your variable "value" to replace the entire df.
So I would suggest the following:
desc$value <- str_replace(desc$value, ".htm", "")
By doing so, you are telling R to remove all .htm that you have in the desc$value variable alone. I hope it works!
Let's assume you have, as you say, a variable "value" that holds long text and you want to remove every word that ends in .html. Based on these assumptions you can use str_remove all:
The main point here is to wrap the pattern into word boundary markers \\b:
library(stringr)
str_remove_all(value, "\\b\\w+\\.html\\b")
[1] "apple and test2.html01" "the word must etc. and as well" "we want to remove .htm"
Data:
value <- c("apple test.html and test2.html01",
"the word friend.html must etc. and x.html as well",
"we want to remove .htm")
To achieve what you want just do:
desc$value <- str_replace(desc$value, ".*\\.htm$", "")
You are trying to escape the star and it is useless. You get an error because \* does not exist in R strings. You just have \n, \t etc...
\. does not exist either in R strings. But \\ exists and it produces a single \ in the resulting string used for the regular expression. Therefore, when you escape something in a R regexp you have to escape it twice:
In my regexp: .* means any chars and \\. means a real dot. I have to escape it twice because \ needs to be escape first from the R string.
Related
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 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 looked around both here and elsewhere, I found many similar questions but none which exactly answer mine. I need to clean up naming conventions, specifically replace/remove certain words and phrases from a specific column/variable, not the entire dataset. I am migrating from SPSS to R, I have an example of the code to do this in SPSS below, but I am not sure how to do it in R.
EG:
"Acadia Parish" --> "Acadia" (removes Parish and space before Parish)
"Fifth District" --> "Fifth" (removes District and space before District)
SPSS syntax:
COMPUTE county=REPLACE(county,' Parish','').
There are only a few instances of this issue in the column with 32,000 cases, and what needs replacing/removing varies and the cases can repeat (there are dozens of instances of a phrase containing 'Parish'), meaning it's much faster to code what needs to be removed/replaced, it's not as simple or clean as a regular expression to remove all spaces, all characters after a specific word or character, all special characters, etc. And it must include leading spaces.
I have looked at the replace() gsub() and other similar commands in R, but they all involve creating vectors, or it seems like they do. What I'd like is syntax that looks for characters I specify, which can include leading or trailing spaces, and replaces them with something I specify, which can include nothing at all, and if it does not find the specific characters, the case is unchanged.
Yes, I will end up repeating the same syntax many times, it's probably easier to create a vector but if possible I'd like to get the syntax I described, as there are other similar operations I need to do as well.
Thank you for looking.
> x <- c("Acadia Parish", "Fifth District")
> x2 <- gsub("^(\\w*).*$", "\\1", x)
> x2
[1] "Acadia" "Fifth"
Legend:
^ Start of pattern.
() Group (or token).
\w* One or more occurrences of word character more than 1 times.
.* one or more occurrences of any character except new line \n.
$ end of pattern.
\1 Returns group from regexp
Maybe I'm missing something but I don't see why you can't simply use conditionals in your regex expression, then trim out the annoying white space.
string <- c("Arcadia Parish", "Fifth District")
bad_words <- c("Parish", "District") # Write all the words you want removed here!
bad_regex <- paste(bad_words, collapse = "|")
trimws( sub(bad_regex, "", string) )
# [1] "Arcadia" "Fifth"
dataframename$varname <- gsub(" Parish","", dataframename$varname)
I need to change "/" into "\" in my R code. I have something like this:
tmp <- paste(getwd(),"tmp.xls",sep="/")
so my tmp is c:/Study/tmp.xls
and I want it to be: c:\Study\tmp.xls
Is it possible to change it in R?
Update as per comments.
If this is simply to save the file, then as #sgibb suggested, you are better off using file.path():
file.path(getwd(), "tmp.xls")
Update 2: You want double back-slashes.
tmp is a string and if you want to have an actual backslash you need to escape it -- with a backslash.
However, when R interprets the double slashes (for example, when looking for a file with the path indicated by the string), it will treat the seemingly double slashes as one.
Take a look at what happens when you output the string with cat()
cat("c:\\Study\\tmp.xls")
c:\Study\tmp.xls
The second slash has "disappeared"
Original Answer:
in R, \ is an escape character, thus if you want to print it literally, you need to escape the escape character: \\. This is what you want to put in your paste statement.
You can also use .Platform$file.sep as your sep argument, which will make your code much more portable.
tmp <- paste(getwd(),"tmp.xls",sep=.Platform$file.sep)
If you already have a string you would like to replace, you can use
gsub("/", "\\", tmp, fixed=TRUE)
I want to split a line in an R script over multiple lines (because it is too long). How do I do that?
Specifically, I have a line such as
setwd('~/a/very/long/path/here/that/goes/beyond/80/characters/and/then/some/more')
Is it possible to split the long path over multiple lines? I tried
setwd('~/a/very/long/path/here/that/goes/beyond/80/characters/and/
then/some/more')
with return key at the end of the first line; but that does not work.
Thanks.
Bah, comments are too small. Anyway, #Dirk is very right.
R doesn't need to be told the code starts at the next line. It is smarter than Python ;-) and will just continue to read the next line whenever it considers the statement as "not finished". Actually, in your case it also went to the next line, but R takes the return as a character when it is placed between "".
Mind you, you'll have to make sure your code isn't finished. Compare
a <- 1 + 2
+ 3
with
a <- 1 + 2 +
3
So, when spreading code over multiple lines, you have to make sure that R knows something is coming, either by :
leaving a bracket open, or
ending the line with an operator
When we're talking strings, this still works but you need to be a bit careful. You can open the quotation marks and R will read on until you close it. But every character, including the newline, will be seen as part of the string :
x <- "This is a very
long string over two lines."
x
## [1] "This is a very\nlong string over two lines."
cat(x)
## This is a very
## long string over two lines.
That's the reason why in this case, your code didn't work: a path can't contain a newline character (\n). So that's also why you better use the solution with paste() or paste0() Dirk proposed.
You are not breaking code over multiple lines, but rather a single identifier. There is a difference.
For your issue, try
R> setwd(paste("~/a/very/long/path/here",
"/and/then/some/more",
"/and/then/some/more",
"/and/then/some/more", sep=""))
which also illustrates that it is perfectly fine to break code across multiple lines.
Dirk's method above will absolutely work, but if you're looking for a way to bring in a long string where whitespace/structure is important to preserve (example: a SQL query using RODBC) there is a two step solution.
1) Bring the text string in across multiple lines
long_string <- "this
is
a
long
string
with
whitespace"
2) R will introduce a bunch of \n characters. Strip those out with strwrap(), which destroys whitespace, per the documentation:
strwrap(long_string, width=10000, simplify=TRUE)
By telling strwrap to wrap your text to a very, very long row, you get a single character vector with no whitespace/newline characters.
For that particular case there is file.path :
File <- file.path("~",
"a",
"very",
"long",
"path",
"here",
"that",
"goes",
"beyond",
"80",
"characters",
"and",
"then",
"some",
"more")
setwd(File)
The glue::glue function can help. You can write a string on multiple lines in a script but remove the line breaks from the string object by ending each line with \\:
glue("some\\
thing")
something
I know this post is old, but I had a Situation like this and just want to share my solution. All the answers above work fine. But if you have a Code such as those in data.table chaining Syntax it becomes abit challenging. e.g. I had a Problem like this.
mass <- files[, Veg:=tstrsplit(files$file, "/")[1:4][[1]]][, Rain:=tstrsplit(files$file, "/")[1:4][[2]]][, Roughness:=tstrsplit(files$file, "/")[1:4][[3]]][, Geom:=tstrsplit(files$file, "/")[1:4][[4]]][time_[s]<=12000]
I tried most of the suggestions above and they didn´t work. but I figured out that they can be split after the comma within []. Splitting at ][ doesn´t work.
mass <- files[, Veg:=tstrsplit(files$file, "/")[1:4][[1]]][,
Rain:=tstrsplit(files$file, "/")[1:4][[2]]][,
Roughness:=tstrsplit(files$file, "/")[1:4][[3]]][,
Geom:=tstrsplit(files$file, "/")[1:4][[4]]][`time_[s]`<=12000]
There is no coinvent way to do this because there is no operator in R to do string concatenation.
However, you can define a R Infix operator to do string concatenation:
`%+%` = function(x,y) return(paste0(x,y))
Then you can use it to concat strings, even break the code to multiple lines:
s = "a" %+%
"b" %+%
"c"
This will give you "abc".
This will keep the \n character, but you can also just wrap the quote in parentheses. Especially useful in RMarkdown.
t <- ("
this is a long
string
")
I do this all of the time. Use the paste0() function.
Rootdir = "/myhome/thisproject/part1/"
Subdir = "subdirectory1/subsubdir2/"
fullpath = paste0( Rootdir, Subdir )
fullpath
> fullpath
[1] "/myhome/thisproject/part1/subdirectory1/subsubdir2/"