'Remove parentheses and text within from strings in R

In R, I have a list of companies such as:

companies  <-  data.frame(Name=c("Company A Inc (COMPA)","Company B (BEELINE)", "Company C Inc. (Coco)", "Company D Inc.", "Company E"))

I want to remove the text with parenthesis, ending up with the following list: 

                  Name
1        Company A Inc 
2            Company B
3       Company C Inc.
4       Company D Inc.
5            Company E

One approach I tried was to split the string and then use ldply:

companies$Name <- as.character(companies$Name)
c<-strsplit(companies$Name, "\\(")
ldply(c)

But because not all company names have parentheses portions, it fails: 

Error in list_to_dataframe(res, attr(.data, "split_labels"), .id, id_as_factor) : 
  Results do not have equal lengths

I'm not married to the strsplit solution. Whatever removes that text and the parentheses would be fine.

rregex


Solution 1:[1]

A gsub should work here

gsub("\\s*\\([^\\)]+\\)","",as.character(companies$Name))
# or using "raw" strings as of R 4.0
gsub(r"{\s*\([^\)]+\)}","",as.character(companies$Name))

# [1] "Company A Inc"  "Company B"      "Company C Inc."
# [4] "Company D Inc." "Company E"

Here we just replace occurrences of "(...)" with nothing (also removing any leading space). R makes it look worse than it is with all the escaping we have to do for the parenthesis since they are special characters in regular expressions.

Solution 2:[2]

You could use stringr::str_replace. It's nice because it accepts factor variables.

companies <- data.frame(Name=c("Company A Inc (COMPA)","Company B (BEELINE)", 
                               "Company C Inc. (Coco)", "Company D Inc.", 
                               "Company E"))

library(stringr)
str_replace(companies$Name, " \\s*\\([^\\)]+\\)", "")
# [1] "Company A Inc"  "Company B"      "Company C Inc." 
# [4] "Company D Inc." "Company E"

And if you still want to use strsplit, you could do

companies$Name <- as.character(companies$Name)
unlist(strsplit(companies$Name, " \\(.*\\)"))
# [1] "Company A Inc"  "Company B"      "Company C Inc."
# [4] "Company D Inc." "Company E"

Solution 3:[3]

You could also use:

library(qdap)
companies$Name <-  genX(companies$Name, " (", ")")

companies
        Name
1  Company A Inc
2       CompanyB
3 Company C Inc.
4 Company D Inc.
5       CompanyE

Solution 4:[4]

If the parentheses are paired and balanced, you can use

gsub("\\s*(\\([^()]*(?:(?1)[^()]*)*\\))", "", x, perl=TRUE)

See the regex and R demo online:

companies  <-  data.frame(Name=c("Company A Inc (COMPA)","Company B (BEELINE)", "Company C Inc. (Coco)", "Company D Inc.", "Company E"))
gsub("\\s*(\\([^()]*(?:(?1)[^()]*)*\\))", "", companies$Name, perl=TRUE)

Output:

[1] "Company A Inc"  "Company B"      "Company C Inc." "Company D Inc."
[5] "Company E"

Regex details

  • \s* - zero or more whitespaces
  • (\([^()]*(?:(?1)[^()]*)*\)) - Capturing group 1 (required to recurse the pattern part between parentheses):
    • \( - a ( char
    • [^()]* - zero or more chars other than ( and )
    • (?:(?1)[^()]*)* - zero or more occurrences of the whole Group 1 pattern ((?1) is a regex subroutine recursing Group 1 pattern) and then zero or more chars other than ( and )
    • \) - a ) char.

Solution 5:[5]

In your case it will come to the desired result, wenn you remove everything starting with (.

sub(" \\(.*", "", companies$Name)
#[1] "Company A Inc"  "Company B"      "Company C Inc." "Company D Inc." "Company E"

To remove parentheses and text within from a strings you can use.

sub("\\(.*)", "", c("ab (cd) ef", "(ij) kl"))
#[1] "ab  ef" " kl"

If there are more than one parentheses:

gsub("\\(.*?)", "", c("ab (cd) ef (gh)", "(ij) kl"))
#[1] "ab  ef " " kl"

( needs to be escaped \\(, . means everything, * means repeated 0 to n, ? means non greedy to remove not everything from the first to the last match.

As an alternative you can use [^)] what means everything but not a ).

sub("\\([^)]*)", "", c("ab (cd) ef", "(ij) kl"))
#[1] "ab  ef" " kl"   

gsub("\\([^)]*)", "", c("ab (cd) ef (gh)", "(ij) kl"))
#[1] "ab  ef " " kl"

If there are nested parentheses:

gsub("\\(([^()]|(?R))*\\)", "", c("ab ((cd) ef) gh (ij)", "(ij) kl"), perl=TRUE)
#[1] "ab  gh " " kl"

Where a(?R)z is a recursion which match one or more letters a followed by exactly the same number of letters z.

Solution 6:[6]

library(qdap)
bracketX(companies$Name) -> companies$Name

Solution 7:[7]

Another gsub solution: replace the term in the parens preceded by an optional space by "", i.e. empty string

gsub("(\\s*\\(\\w+\\))", "", companies$Name)

[1] "Company A Inc"  "Company B"      "Company C Inc." "Company D Inc."
[5] "Company E"

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Gregor Thomas
Solution 3
Solution 4 Wiktor Stribiżew
Solution 5
Solution 6 Thushara Dulam
Solution 7 Eyayaw

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