I have a lot of error messages that I am trying to clean up.
some of the errors end with the text "(sec): 0.xxx"
i'm trying to use gsub to remove everything after (sec)
data$Message <- gsub("(sec).*", "", data$Message, perl = TRUE)
this returns everything after (
I know it would be easy to just use ":" or ")" but then it effects other errors that I do not want to change.
Is there a way to use gsub to look at several characters -like "(sec)"- instead of just one?
on a related note is their a symbol that represents any number (excludes text) similiar to "."?
You can use regex look behind ?<= to avoid sec being removed and at the same time assert the removed pattern follows sec, so (?<=sec\\)).* will remove everything after sec) but not sec) itself:
gsub("(?<=sec\\)).*", "", "(sec): 0.xxx", perl = TRUE)
# [1] "(sec)"
You can select the first part of the expression (between brackets) and omit the rest:
gsub('(^.*\\(sec\\)).*', '\\1', '(sec): 0.xxx')
## [1] "(sec)"
Related
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.
I have tried gsub as following to remove everything before the first space but it didn't work.
lagl2$SUSPENSE <- gsub(pattern = "(.*)\\s",replace=" ", lagl2$SUSPENSE)
example of the row data: 64400/GL WORKERS COMPENSATION
and I want the result to be like that: WORKERS COMPENSATION
This is just an example but I have many observations and one column and need to delete everything before the first space.
I am new to R and to programming but I started loving it.
You can remove everything before first space using sub as -
sub(".*?\\s", "", "64400/GL WORKERS COMPENSATION")
#[1] "WORKERS COMPENSATION"
To apply to the whole column you can do -
lagl2$SUSPENSE <- sub(".*?\\s", "", lagl2$SUSPENSE)
You could also assert the start of the string ^ , and match optional non whitespace chars followed by one or more whitespace chars using \S*\s+ that you want to remove.
sub("^\\S*\\s+", "", "64400/GL WORKERS COMPENSATION")
Output
[1] "WORKERS COMPENSATION"
You can match everything before first space using lookarounds
/^[^\s]+(?=\s)\s+/gm
Demo
I have:`
String="(anthony,2019, p.485)"
Desidered Output:
String="(anthony,2019)"
i want removed only p.485.
I use the regex: `
gsub("\\( \\,p\\.[0-9]\\)","",String)
but it does not work.
Thanks!
We can use sub to match , followed by a space, 'p' and one or more digits (\\d+), replace by blank ("") in the replacement
sub(", p\\.\\d+", "", String)
#[1] "(anthony,2019)"
We could also try this slightly inefficient regex:
String="(anthony,2019, p.485)"
gsub(",\\s\\w.\\d{1,}","",String,perl=TRUE)
#[1] "(anthony,2019)"
This seems like it might be a bit fragile, since it depends on there not being any spaces after commas except for the last one:
gsub("[,] [^,)]+","", String)
[1] "(anthony,2019)"
That fragility afflicts akrun's answer as well. A more robust solution would be to match on the last comma but leave in the closing paren. This solution will locate that last comma and excise everything up to but not included the closing paren:
gsub("(.+)([,] [^,]+)([)])","\\1\\3", String) # 3 capture classes
[1] "(anthony,2019)" # return only first and third
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 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/"