Which regex removes punctuation from quotation marks in text

Question

I have a database and throughout the text there are some quotes that are in quotation marks. I would like to remove all the dots \".\" that are enclosed in quotation marks i

Answer 1

mystring <-'"é preciso olhar para o futuro. vou atuar" no front em que posso 
fazer alguma coisa "para .frente", disse jose.'

You can use the following pattern with gsub:

gsub('(?!(([^"]*"){2})*[^"]*$)\\.', "", mystring, perl = T)

Same with stringr:

str_replace_all(mystring, '(?!(([^"]*"){2})*[^"]*$)\\.', '')

Output:

#> "é preciso olhar para o futuro vou atuar" no front em que posso 
#> fazer alguma coisa "para frente", disse jose.

Answer 2

You may simply use str_replace_all with a mere "[^"]*" pattern and use a callback function as the replacement argument to remove all dots with a gsub call:

str_replace_all(string, '"[^"]*"', function(x) gsub(".", "", x, fixed=TRUE))

So,

• "[^"]*" matches all substrings in string starting with ", then having 0+ chars other than " and then a "
• Once the match is found, it is passed to the callback as x where gsub(".", "", x, fixed=TRUE) replaces all . (fixed=TRUE makes it a literal dot, not a regex pattern) with an empty string.