javascript - 仅查找和替换匹配的右括号

Question

I'm wondering if there's an elegant way to find and replace the matching brackets for code depending on what is adjacent to the opening bracket. For example, I might want to convert:

// code for random_number() function
for(i in 1:5){
   print(i);
   mean(c(1,2,random_number()))
   sd(c(4,5,6))
}

into:

// code for random_number() function
for(i in 1:5){
   print(i);
   mean([1,2,random_number()])
   sd([4,5,6])
}

because all I want to do is convert "c(1,2,3)" into "[1,2,3]" (and "c(4,5,6)" into "[4,5,6]"). I want to identify the closing bracket opposite "c(" in a script like this.

I don't want to change any of the other opening and closing brackets. Is there an elegant way of doing this? The closest thing I found was Javascript replace opening and closing brackets , but it seemed more like a method to replace all brackets with the occasional exception, rather than specific brackets but ignoring most.

My only thoughts are to count the opening and closing of brackets, but that seems quite ugly, to say the least. Is there a tidier way?

Answer 1

You could use RegExp for that, something like: c\((([^)]*\([^)]*\))*[^)]*)\) Here is an example:

 const regExp = /c\((([^)]*\([^)]*\))*[^)]*)\)/g; const source = `// code for random_number() function for(i in 1:5){ print(i); mean(c(1,2,random_number())) sd(c(4,5,6)) }`; const result = source.replace(regExp, '[$1]') console.log(result);

This will probably fail with more complex source strings.

What it does, is to look for a c character followed by a ( character, then ignore all opening brackets that have a corresponding closing bracket and put all that along with any characters that are not a ) into a group and then look for a ) character.

Balancing parenthesis with regex is not a good idea, this works with only one level (fails when there are two opening parenthesis without a closing parenthesis between them). A more reliable approach will be looping the string and keeping track of all the opening and closing brackets.

Beside all that, the regex doesn't distinguish between brackets that are part of code or those that are part of string literals.