I would like to delete the lines which contain the opening bracket "(" from my dataframe.
I tried the following:
df[!grepl("(", df$Name),]
But this does not track down the ( sign
You have to double-escape the ( with \\.
x <- c("asdf", "asdf", "df", "(as")
x[!grepl("\\(", x)]
# [1] "asdf" "asdf" "df"
Just apply this to your df like df[!grepl("\\(", df$Name), ]
You could also think about removing all puctuation characters by using regex:
x[!grepl("[[:punct:]]", x)]
As pointed out by #CSquare in the comments, here is a great summary about special characters in R regex
Additional input from the comments:
#Sotos: Gaining performance with pattern='(' and fixed = TRUE since the regex could be bypassed.
x[!grepl('(', x, fixed = TRUE)]
Related
Is there an R function to get only the part of a string before the 2nd capital character appears?
For example:
Example <- "MonkeysDogsCats"
Expected output should be:
"Monkeys"
Maybe something like
stringr::str_extract("MonkeysDogsCats", "[A-Z][a-z]*")
#[1] "Monkeys"
Here is an alternative approach:
Here we first put a space before all uppercase and then extract the first word:
library(stringr)
word(gsub("([a-z])([A-Z])","\\1 \\2", Example), 1)
[1] "Monkeys"
A base solution with sub():
x <- "MonkeysDogsCats"
sub("(?<=[a-z])[A-Z].*", "", x, perl = TRUE)
# [1] "Monkeys"
Another way using stringr::word():
stringr::word(x, 1, sep = "(?=[A-Z])\\B")
# [1] "Monkeys"
If the goal is strictly to capture any string before the 2nd capital character, one might want pick a solution it'll also work with all types of strings including numbers and special characters.
strings <- c("MonkeysDogsCats",
"M4DogsCats",
"M?DogsCats")
stringr::str_remove(strings, "(?<=.)[A-Z].*")
Output:
[1] "Monkeys" "M4" "M?"
It depends on what you want to allow to match. You can for example match an uppercase char [A-Z] optionally followed by any character that is not an uppercase character [^A-Z]*
If you don't want to allow whitespace chars, you can exclude them [^A-Z\\s]*
library(stringr)
str_extract("MonkeysDogsCats", "[A-Z][^A-Z]*")
Output
[1] "Monkeys"
R demo
If there should be an uppercase character following, and there are only lowercase characters allowed:
str <- "MonkeysDogsCats"
regmatches(str, regexpr("[A-Z][a-z]*(?=[A-Z])", str, perl = TRUE))
Output
[1] "Monkeys"
R demo
I have a table with a string column formatted like this
abcdWorkstart.csv
abcdWorkcomplete.csv
And I would like to extract the last word in that filename. So I think the beginning pattern would be the word "Work" and ending pattern would be ".csv". I wrote something using grepl but not working.
grepl("Work{*}.csv", data$filename)
Basically I want to extract whatever between Work and .csv
desired outcome:
start
complete
I think you need sub or gsub (substitute/extract) instead of grepl (find if match exists). Note that when not found, it will return the entire string unmodified:
fn <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')
out <- sub(".*Work(.*)\\.csv$", "\\1", fn)
out
# [1] "start" "complete" "abcdNothing.csv"
You can work around this by filtering out the unchanged ones:
out[ out != fn ]
# [1] "start" "complete"
Or marking them invalid with NA (or something else):
out[ out == fn ] <- NA
out
# [1] "start" "complete" NA
With str_extract from stringr. This uses positive lookarounds to match any character one or more times (.+) between "Work" and ".csv":
x <- c("abcdWorkstart.csv", "abcdWorkcomplete.csv")
library(stringr)
str_extract(x, "(?<=Work).+(?=\\.csv)")
# [1] "start" "complete"
Just as an alternative way, remove everything you don't want.
x <- c("abcdWorkstart.csv", "abcdWorkcomplete.csv")
gsub("^.*Work|\\.csv$", "", x)
#[1] "start" "complete"
please note:
I have to use gsub. Because I first remove ^.*Work then \\.csv$.
For [\\s\\S] or \\d\\D ... (does not work with [g]?sub)
https://regex101.com/r/wFgkgG/1
Works with akruns approach:
regmatches(v1, regexpr("(?<=Work)[\\s\\S]+(?=[.]csv)", v1, perl = T))
str1<-
'12
.2
12'
gsub("[^.]","m",str1,perl=T)
gsub(".","m",str1,perl=T)
gsub(".","m",str1,perl=F)
. matches also \n when using the R engine.
Here is an option using regmatches/regexpr from base R. Using a regex lookaround to match all characters that are not a . after the string 'Work', extract with regmatches
regmatches(v1, regexpr("(?<=Work)[^.]+(?=[.]csv)", v1, perl = TRUE))
#[1] "start" "complete"
data
v1 <- c('abcdWorkstart.csv', 'abcdWorkcomplete.csv', 'abcdNothing.csv')
I have the next vector of strings
[1] "/players/playerpage.htm?ilkidn=BRYANPHI01"
[2] "/players/playerpage.htm?ilkidhh=WILLIROB027"
[3] "/players/playerpage.htm?ilkid=THOMPWIL01"
I am looking for a way to retrieve the part of the string that is placed after the equal sign meaning I would like to get a vector like this
[1] "BRYANPHI01"
[2] "WILLIROB027"
[3] "THOMPWIL01"
I tried using substr but for it to work I have to know exactly where the equal sign is placed in the string and where the part i want to retrieve ends
We can use sub to match the zero or more characters that are not a = ([^=]*) followed by a = and replace it with ''.
sub("[^=]*=", "", str1)
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
data
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
Using stringr,
library(stringr)
word(str1, 2, sep = '=')
#[1] "BRYANPHI01" "WILLIROB027" "THOMPWIL01"
Using strsplit,
strsplit(str1, "=")[[1]][2]
# [1] "BRYANPHI01"
With Sotos comment to get results as vector:
sapply(str1, function(x){
strsplit(x, "=")[[1]][2]
})
Another solution based on regex, but extracting instead of substituting, which may be more efficient.
I use the stringi package which provides a more powerful regex engine than base R (in particular, supporting look-behind).
str1 <- c("/players/playerpage.htm?ilkidn=BRYANPHI01",
"/players/playerpage.htm?ilkidhh=WILLIROB027",
"/players/playerpage.htm?ilkid=THOMPWIL01")
stri_extract_all_regex(str1, pattern="(?<==).+$", simplify=T)
(?<==) is a look-behind: regex will match only if preceded by an equal sign, but the equal sign will not be part of the match.
.+$ matches everything until the end. You could replace the dot with a more precise symbol if you are confident about the format of what you match. For example, '\w' matches any alphanumeric character, so you could use "(?<==)\\w+$" (the \ must be escaped so you end up with \\w).
I have a list of strings
/temp/123/afedcgid/abc.csv
/temp/123/4388dkfa/abc1.csv
/temp/123/4388dkfa/ab1.csv
I want to remove name of the file from the strings
The results desired are
/temp/123/afedcgid
/temp/123/4388dkfa
/temp/123/4388dkfa
How can i do it. Thanks.
You could try the below,
sub("/[^/]*$", "", x)
It removes all the chars from the last / symbol.
OR
> x <- "/temp/123/afedcgid/abc.csv"
> sub("(.*)/.*", "\\1", x)
[1] "/temp/123/afedcgid"
captures all the chars from the start upto the last / symbol (excluding /). Then the following chars are matched by .*. Replacing the matched chars with chars inside group 1 will give you the desired output.
Example:
> x <- "/temp/123/afedcgid/abc.csv"
> sub("/[^/]*$", "", x)
[1] "/temp/123/afedcgid"
OR
regmatches(x, gregexpr(".+(?=/)", x, perl=TRUE))
Use this regex to catch character you want to replace
\/\w+\.\w+$
try this demo
Demo
files <- c("/temp/123/afedcgid/abc.csv" ,
"/temp/123/4388dkfa/abc1.csv" , "/temp/123/4388dkfa/ab1.csv")
sub("\\/\\w+\\.\\w+$" , "" , files)
as you may know you need to \\ for escaping sequences in R
I have a string, say
fruit <- "()goodapple"
I want to remove the brackets in the string. I decide to use stringr package because it usually can handle this kind of issues. I use :
str_replace(fruit,"()","")
But nothing is replaced, and the following is replaced:
[1] "()good"
If I only want to replace the right half bracket, it works:
str_replace(fruit,")","")
[1] "(good"
However, the left half bracket does not work:
str_replace(fruit,"(","")
and the following error is shown:
Error in sub("(", "", "()good", fixed = FALSE, ignore.case = FALSE, perl = FALSE) :
invalid regular expression '(', reason 'Missing ')''
Anyone has ideas why this happens? How can I remove the "()" in the string, then?
Escaping the parentheses does it...
str_replace(fruit,"\\(\\)","")
# [1] "goodapple"
You may also want to consider exploring the "stringi" package, which has a similar approach to "stringr" but has more flexible functions. For instance, there is stri_replace_all_fixed, which would be useful here since your search string is a fixed pattern, not a regex pattern:
library(stringi)
stri_replace_all_fixed(fruit, "()", "")
# [1] "goodapple"
Of course, basic gsub handles this just fine too:
gsub("()", "", fruit, fixed=TRUE)
# [1] "goodapple"
The accepted answer works for your exact problem, but not for the more general problem:
my_fruits <- c("()goodapple", "(bad)apple", "(funnyapple")
str_replace(my_fruits,"\\(\\)","")
## "goodapple" "(bad)apple", "(funnyapple"
This is because the regex exactly matches a "(" followed by a ")".
Assuming you care only about bracket pairs, this is a stronger solution:
str_replace(my_fruits, "\\([^()]{0,}\\)", "")
## "goodapple" "apple" "(funnyapple"
Building off of MJH's answer, this removes all ( or ):
my_fruits <- c("()goodapple", "(bad)apple", "(funnyapple")
str_replace_all(my_fruits, "[//(//)]", "")
[1] "goodapple" "badapple" "funnyapple"