Преобразовать строку в рваный массив с помощью регулярных выражений

(1,1,'potato '),(2,2,'meat '),(3,3,'bean '),(4,4,'bean '),(5,5,'onions '),(6,1,'cheese '),(7,2,'canned food '),(8,3,'bean '),(9,4,'margarine '),(10,5,'milk '),(11,1,'sugar '),(12,2,'butter '),(13,3,'sugar '),(14,4,'cheese '),(15,5,'beet '),(16,4,'fish ');

string str = @"(1,1,'potato '),(2,2,'meat '),(3,3,'bean '),(4,4,'bean '),(5,5,'onions '),(6,1,'cheese '),(7,2,'canned food '),(8,3,'bean '),(9,4,'margarine '),(10,5,'milk '),(11,1,'sugar '),(12,2,'butter '),(13,3,'sugar '),(14,4,'cheese '),(15,5,'beet '),(16,4,'fish ');";
MatchCollection Matches = Regex.Matches(str, @"\(\d+,(?<key>\d+),'\s*(?<value>[\w\s]+)\s*'\)");
Dictionary<int, List<string>> RaggedArray = new Dictionary<int, List<string>>(Matches.Count);
foreach (Match match in Matches)
{
    if (RaggedArray.ContainsKey(Int32.Parse(match.Groups["key"].Value)))
        RaggedArray[Int32.Parse(match.Groups["key"].Value)].Add(match.Groups["value"].Value.Trim());
    else
        RaggedArray.Add(Int32.Parse(match.Groups["key"].Value), new List<string> { match.Groups["value"].Value.Trim() });
}