Str_Replace (Package Stringr) Cannot Replace Brackets in R

str_replace (package stringr) cannot replace brackets in r?

Escaping the parentheses does it...

str_replace(fruit,"\\(\\)","")
# [1] "goodapple"

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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"

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Of course, basic gsub handles this just fine too:

gsub("()", "", fruit, fixed=TRUE)
# [1] "goodapple"

Problem escaping the parentheses when using str_replace

If it is a fixed match, we can use replace from base R along with ==

library(dplyr)
DF%>%
   mutate(b= replace(b,
     b == "Discharged home or self-care (routine discharge) AHR","Home"))

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Or in str_replace, specify the fixed because the () are regex metacharacters for capturing groups and by default it is in regex mode .

library(stringr)
DF %>%
  mutate(b=str_replace(b,
     fixed("Discharged home or self-care (routine discharge) AHR"),"Home"))

TO literally evaluate the (, we need to escape (\\() or place it inside square brackets or with fixed wrapper

DF%>%
   mutate(b=str_replace(b,
    "Discharged home or self-care \\(routine discharge\\) AHR","Home"))

How to replace a caret with str_replace from stringr

An option is to wrap with fixed and should be fine

library(stringr)
str_replace_all(code, fixed("^"), "")
#[1] "GSPC"      "FTSE"      "000001.SS" "HSI"       "FCHI"      "KS11"      "TWII"      "GDAXI"     "STI"  

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Also, as we are replacing with blank (""), an option is str_remove

str_remove(code, fixed("^"))

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Regarding why the OP's code didn't, inside the square brackets, if we use ^, it is not reading the literal character, instead the metacharacter in it looks for characters other than and here it is blank ([^])

str_replace doesn't replace all occurrences, but gsub does?

It only matches the first occurency, whereas gsub does it all. Use str_replace_all instead:

str_replace(string = "aa", pattern = "a", replacement = "b") # only first

str_replace_all(string = "aa", pattern = "a", replacement = "b") # all

Trying to replace a () in a string in R using str_replace

Sub, gsub

\\ identify special characters

If you want to replace ONLY the parenthesis that are in the middle of the string (that is not at the start or at the end):

text <- "tBodyAcc-mean()-X"
sub("#\\(\\)#", "", text)
[1] "tBodyAcc-mean-X"

text <- "tBodyAcc-mean-X()"
sub("#\\(\\)#", "", text)
[1] "tBodyAcc-mean-X()"

If you want to replace ANY parenthesis (including those at the end and at the start of the string)

text <- "tBodyAcc-mean()-X"
sub("\\(\\)", "", text)

EDIT, as pointed out in several comments using gsub instead of sub will replace all the "()" in a string, while sub only replace the first "()"

text <- "t()BodyAcc-mean()-X"

sub("\\(\\)", "", text)
[1] "tBodyAcc-mean()-X"

> gsub("\\(\\)", "", text)
[1] "tBodyAcc-mean-X"

str_replace: replacement depending on wildcard value [A-Z]

It needs to be captured as a group (([A-Z])) and replace with the backreference (\\1) of the captured group i.e. regex interpretation is in the pattern and not in the replacement

stringr::str_replace("PrinceofWales", "of([A-Z])", " of \\1")
[1] "Prince of Wales"

According to ?str_replace

replacement - A character vector of replacements. Should be either length one, or the same length as string or pattern. References of the form \1, \2, etc will be replaced with the contents of the respective matched group (created by ()).

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Or another option is a regex lookaround

stringr::str_replace("PrinceofWales", "of(?=[A-Z])", " of ")
[1] "Prince of Wales"

Removing/replacing brackets from R string using gsub

Using the correct regex works:

gsub("[()]", "", "(abc)")

The additional square brackets mean "match any of the characters inside".

str_replace() escaping question marks in R with a vector from a SPDF, stringr package

library(stringr)
str_replace_all("Is Nuevo Leon in Mexico?", 
                c("M.xico" = "México", "Nuevo Le.n" = "Nuevo León"))

# [1] "Is Nuevo León in México?"

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