Split string in Ruby, ignoring contents of parentheses?

Question

I need to split a string into a list of parts in Ruby, but I need to ignore stuff inside paramentheses. For example:

A +4, B +6, C (hello, goodbye) +5, D +3

I'd like the resulting list to be:

[0]A +4
[1]B +6
[2]C (hello, goodbye) +5
[3]D +3

But I can't simply split on commas, because that would split the contents of the parentheses. Is there a way to split stuff out without pre-parsing the commas in the braces into something else?

Thanks.

Answer 1

Try this:

s = 'A +4, B +6, C (hello, goodbye) +5, D +3'
tokens = s.scan(/(?:\(.*?\)|[^,])+/)
tokens.each {|t| puts t.strip}

Output:

A +4
B +6
C (hello, goodbye) +5
D +3

A short explanation:

(?:        # open non-capturing group 1
  \(       #   match '('
  .*?      #   reluctatly match zero or more character other than line breaks
  \)       #   match ')'
  |        #   OR
  [^,]     #   match something other than a comma
)+         # close non-capturing group 1 and repeat it one or more times

Another option is to split on a comma followed by some spaces only when the first parenthesis that can be seen when looking ahead is an opening parenthesis (or no parenthesis at all: ie. the end of the string):

s = 'A +4, B +6, C (hello, goodbye) +5, D +3'
tokens = s.split(/,\s*(?=[^()]*(?:\(|$))/)
tokens.each {|t| puts t}

will produce the same output, but I find the scan method cleaner.

Answer 2

string = "A +4, B +6, C (hello, goodbye) +5, D +3"
string.split(/ *, *(?=[^\)]*?(?:\(|$))/)
# => ["A +4", "B +6", "C (hello, goodbye) +5", "D +3"]

How this regex works:

/
   *, *        # find comma, ignoring leading and trailing spaces.
  (?=          # (Pattern in here is matched against but is not returned as part of the match.)
    [^\)]*?    #   optionally, find a sequence of zero or more characters that are not ')'
    (?:        #   <non-capturing parentheses group>
      \(       #     left paren ')'
      |        #     - OR -
      $        #     (end of string)
    )
  )
/