有人可以幫助我提高這個LINQ

Question

我這裏有一個代碼，使用LINQ工程師列表生成XML。我的問題是，是否有改進和加快這一方法

public static string CreateXMLforEngineersByLinq(List<Engineers> lst) 
    { 
     string x = "<Engineers>\n"; 
     x += string.Concat(lst.Select(s => 
       string.Format("<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n", 
       s.LicenseID, s.LastName, s.FirstName, s.MiddleName))); 
     return x + "</Engineers>"; 
    } 

結果的任何方式：

嗨，下面的代碼被添加爲我展示我公司生產多種答案的方法實際速度和修訂再次歡迎感謝那些誰的人誰幫助這裏:)

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using test.Classes; 

namespace test.LINQ 
{ 
public static class BatchOperations 
{ 
    delegate string Operations(List<Engineers> i);   
    public static List<int> BatchAddition(List<int> lstNumbers) 
    { 
     lstNumbers = (from nl in lstNumbers 
         select nl + 2).ToList(); 

     return lstNumbers; 
    } 

    public static string CreateXMLforEngineersTheSimpleWay(IEnumerable<Engineers> lst) 
    { 
     StringBuilder x = new StringBuilder(); 
     x.AppendLine("<Engineers>"); 
     foreach (var s in lst) 
     { 
      x.AppendFormat("<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n", 
          s.LicenseID, 
          s.LastName, 
          s.FirstName, 
          s.MiddleName); 
     } 
     x.AppendLine("</Engineers1>"); 
     return x.ToString(); 
    } 

    public static string CreateXMLforEngineersByLinq(List<Engineers> lst) 
    { 
     string x = "<Engineers>\n"; 
     x += string.Concat(lst.Select(s => 
       string.Format("<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n", 
       s.LicenseID, s.LastName, s.FirstName, s.MiddleName))); 
     return x + "</Engineers2>"; 
    } 

    public static string CreateXMLforEngineersByLoop(List<Engineers> lst) 
    { 
     string XmlForEngineers = "<Engineers>"; 
     foreach (Engineers item in lst) 
     { 
      XmlForEngineers += string.Format("<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n" 
       , item.LicenseID, item.LastName, item.FirstName, item.MiddleName); 
     } 
     XmlForEngineers += "</Engineers3>"; 
     return XmlForEngineers; 
    } 

    public static void ShowEngineersByLicense() 
    { 
     List<Engineers> lstEngr = new List<Engineers>(); 

     Engineers tom = new Engineers(); 
     tom.FirstName = "Tom"; 
     tom.MiddleName = "Brook"; 
     tom.LastName = "Crook"; 
     tom.LicenseID = "1343-343434"; 

     Engineers ken = new Engineers(); 
     ken.FirstName = "ken"; 
     ken.MiddleName = "Brook"; 
     ken.LastName = "Crook"; 
     ken.LicenseID = "1343-343434"; 

     Engineers ben = new Engineers(); 
     ben.FirstName = "ben"; 
     ben.MiddleName = "Brook"; 
     ben.LastName = "Crook"; 
     ben.LicenseID = "1343-343434"; 

     for (int y = 0; y <= 1000; y++) 
     { 
      lstEngr.Add(tom); 
      lstEngr.Add(ken); 
      lstEngr.Add(ben); 
     } 

     List<Operations> j = new List<Operations>(); 
     j.Add(a => CreateXMLforEngineersTheSimpleWay(lstEngr)); 
     j.Add(i => CreateXMLforEngineersByLinq(lstEngr)); 
     j.Add(i => CreateXMLforEngineersByLoop(lstEngr)); 

     DateTime start, end; 
     TimeSpan diff1 = new TimeSpan(); 

     foreach (Operations currentMethod in j) 
     { 
      start = DateTime.Now; 
      Console.Write(currentMethod(lstEngr)); 
      end = DateTime.Now; 
      diff1 = end - start; 
      Console.WriteLine(diff1); 
      Console.Write("\n\n"); 
     } 
    } 
} 

}

Answer 1

所有的其他例子是使用字符串連接來生成XML。這樣做的缺點是，如果您的任何值無法正確轉義XML中的不受支持的字符（如<和&），它將會失敗。更好地使用XML進行更原生的工作。考慮以下幾點：

public static string CreateXMLforEngineersByLinq(List<Engineers> lst) 
{ 
    string x = New XElement("Engineers", 
        lst.Select(new XElement, "Engineer", 
         new XElement("LicenseID", s.LicenseID), 
         new XElement("LastName", s.LastName), 
         new XElement("FirstName", s.FirstName), 
         new XElement("MiddleName", s.MiddleName) 
        ) 
       ); 
    return x.ToString(); 
} 

如果你不想把它推出的字符串之前在內存中創建整個XML的內存開銷，你可能要考慮使用XStreamingElement（http://msdn.microsoft.com/en-us/library/system.xml.linq.xstreamingelement.aspx）。有關如何使用它的快速視頻，請參見http://www.microsoft.com/uk/msdn/nuggets/nugget/295/LINQ-to-XML-Streaming-Large-Data-Files-Out-of-Memory.aspx。

Answer 2

最明顯加快：在使用StringBuilder代替一個字符串。對StringBuilder的性能與簡單的拼接比較全面的討論，參見this article。

而且，這是你用LINQ共分配得到一個簡單，更快的解決方案一種情況;

public static string CreateXMLforEngineersTheSimpleWay(IEnumerable<Engineers> lst) 
    { 
     StringBuilder x = new StringBuilder(); 
     x.AppendLine("<Engineers>"); 
     foreach(var s in lst) 
     { 
       x.AppendFormat("<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n", 
          s.LicenseID, 
          s.LastName, 
          s.FirstName, 
          s.MiddleName); 
     } 
     x.AppendLine("</Engineers>"); 
     return x.ToString(); 
    } 

如果您有心在Linq上設置，您可以使用String.Join方法。不過，我懷疑這會不會執行，以及第一個解決方案，如低於此解決方案必須創建一個臨時數組。

public static string CreateXMLforEngineersByLinq(List<Engineers> lst) 
{ 
    var x = new StringBuilder(); 
    x.AppendLine("<Engineers>)"; 
    x.Append(
       string.Join(
          "\n", 
          lst.Select(s => 
           string.Format(
              "<Engineer>\n<LicenseID>{0}</LicenseID>\n<LastName>{1}</LastName>\n<FirstName>{2}</FirstName>\n<MiddleName>{3}</MiddleName>\n</Engineer>\n", 
              s.LicenseID, 
              s.LastName, 
              s.FirstName, 
              s.MiddleName 
              ) 
          ).ToArray() 
         ); 
     x.AppendLine("</Engineers>"); 
     return x.ToString(); 

} 

Answer 3

我認爲這將是worthed以測試是否使用XmlSeriaizer整個列表的序列化是不是更快。

像這樣：

using (var fileStream = new FileStream("engineers.xml", FileMode.OpenOrCreate)) 
{ 
    var xmlSerializer = new XmlSerializer(typeof(List<Engineer>)) 
    xmlSerializer.Serialize(fileStream, list); 
} 

Answer 4

使用XmlSerializer，並建立了方便的擴展方法。

public static class XmlExtensions 
{ 
    public static string ToXml<T>(this T instance) 
    { 
     var xmlSerializer = new XmlSerializer(typeof(T)); 
     var stringWriter = new StringWriter(); 
     xmlSerializer.Serialize(stringWriter, instance); 
     return stringWriter.ToString(); 
    } 
} 

public static string CreateXMLforEngineersByLinqMyWay(List<Engineers> lst) 
{ 
    return string.Format("<Engineers>{0}</Engineers>" 
     , string.Join("", 
      lst.Select(s => s.ToXml()) // Might have to put `.ToArray()` here 
      ) 
     ); 
} 

或者你可以簡單地這樣做：

return lst.ToXml(); 

如果這是一個更大的對象樹序列化的一部分，那麼就放棄這整個方法，以及頂層對象上做ToXml。

你可能需要編輯ToXml方法得到它砸XML命名空間等