Remove all text between two brackets

Question

Suppose I have some text like this,

text<-c(\"[McCain]: We need tax policies that respect the wage earners and job creators. [Obama]: It\'s harder to save         


        

Answer 1

I think this technically answers what you've asked, but you probably want to add a \\: to the end of the regex for prettier text (removing the colon and space).

library(stringr)
str_replace_all(text, "\\[.+?\\]", "")

#> [1] ": We need tax policies that respect the wage earners..."

vs...

str_replace_all(text, "\\[.+?\\]\\: ", "")
#> [1] "We need tax policies that respect the wage earners..." 

Created on 2018-08-16 by the reprex package (v0.2.0).

Answer 2

No need to use a PCRE regex with a negated character class / bracket expression, a "classic" TRE regex will work, too:

subject <- "Some [string] here and [there]"
gsub("\\[[^][]*]", "", subject)
## => [1] "Some  here and "

See the online R demo

Details:

• \\[ - a literal [ (must be escaped or used inside a bracket expression like [[] to be parsed as a literal [)
• [^][]* - a negated bracket expression that matches 0+ chars other than [ and ] (note that the ] at the start of the bracket expression is treated as a literal ])
• ] - a literal ] (this character is not special in both PCRE and TRE regexps and does not have to be escaped).

If you want to only replace the square brackets with some other delimiters, use a capturing group with a backreference in the replacement pattern:

gsub("\\[([^][]*)\\]", "{\\1}", subject)
## => [1] "Some {string} here and {there}"

See another demo

The (...) parenthetical construct forms a capturing group, and its contents can be accessed with a backreference \1 (as the group is the first one in the pattern, its ID is set to 1).

Answer 3

With this:

gsub("\\[[^\\]]*\\]", "", subject, perl=TRUE);

What the regex means:

  \[                       # '['
  [^\]]*                   # any character except: '\]' (0 or more
                           # times (matching the most amount possible))
  \]                       # ']'

Answer 4

The following should do the trick. The ? forces a lazy match, which matches as few . as possible before the subsequent ].

gsub('\\[.*?\\]', '', text)

Answer 5

Here'a another approach:

library(qdap)
bracketX(text, "square")