高效地将项目添加到列表 <class> 并行 C#

Efficiently add items to a list<class> in parallel C#

我有功能代码,可以在 class 列表中拆分一个 属性 的字符串:由 string, string, string 组成的数据框。

现在我正在声明一个空的 Dataframe2 (string,string[], string) 并使用 Add

将项目附加到列表中
class Program

{


    public static string[] SPString(string text)
    {
        string[] elements;
        elements = text.Split(' ');
        return elements;
    }

    //Structures
    public class Dataframe
    {

        public string Name { get; set; }
        public string Text { get; set; }
        public string Cat { get; set; }
    }

    public class Dataframe2
    {

        public string Name { get; set; }
        public string[] Text { get; set; }
        public string Cat { get; set; }
    }



    static void Main(string[] args)
    {

        List<Dataframe> doc = new List<Dataframe>{new Dataframe { Name = "Doc1", Text = "The quick brown cat", Cat = ""},
            new Dataframe { Name = "Doc2", Text = "The big fat cat", Cat = "Two"},
            new Dataframe { Name = "Doc4", Text = "The quick brown rat", Cat = "One"},
            new Dataframe { Name = "Doc3", Text = "Its the cat in the hat", Cat = "Two"},
            new Dataframe { Name = "Doc5", Text = "Mice and rats eat seeds", Cat = "One"},
        };

        // Can this be made more efficient?
        ConcurrentBag<Dataframe2> doc2 = new ConcurrentBag<Dataframe2>();
        Parallel.ForEach(doc, entry =>
        {
            string s = entry.Text;
            string[] splitter = SPString(s);
            doc2.Add(new Dataframe2 {Name = entry.Name, Text = splitter, Cat =entry.Cat});
        } );

    }
}

是否有更有效的方法使用并行 LINQ 将内容添加到列表中,其中 Dataframe2 继承了我没有修改的属性?

您可以尝试使用 PLinq 添加 并行性 并同时保留 List<T>:

// Do NOT create and then fill the List<T> (which is not thread-safe) in parallel manually,
// Let PLinq do it for you
List<Dataframe2> doc2 = doc
  .AsParallel()
  .Select(entry => {
     //TODO: make Dataframe2 from given Dataframe (entry)
     ...
     return new Dataframe2 {Name = entry.Name, Text = splitter, Cat = entry.Cat};
  }) 
  .ToList();