有人可以帮我改进这个 linq

Question

我这里有一个代码，它使用 linq 为工程师列表生成 XML。我的问题是有什么方法可以改进和加速这种方法

    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>";
    }

结果：

嗨，下面的代码是为我添加的，以显示我产生更多答案的方法的实际速度，欢迎修改再次感谢那些在这里提供帮助的人:)

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

使用XmlSerializer，为方便起见构建一个扩展方法。

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
            )
        );
}

或者你可以简单地这样做：

return lst.ToXml();

如果这是更大的对象树序列化的一部分，那么就放弃整个方法，并在顶级对象上执行ToXml。

您可能需要编辑 ToXml 方法以使其删除 XML 命名空间等。

Answer 2

最明显的加速：使用StringBuilder 而不是字符串。请参阅this article，全面讨论 StringBuilder 与简单连接相比的性能。

此外，在这种情况下，您可以通过完全放弃 linq 获得更简单、更快速的解决方案；

 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();
    }

如果您对 Linq 有信心，可以使用 String.Join 方法。但是，我怀疑这不会像第一个解决方案那样好，因为下面的这个解决方案必须创建一个临时数组。

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

所有其他示例都使用字符串 concat 来生成您的 XML。这样做的缺点是，如果您的任何值没有正确转义 XML 中不受支持的字符（如

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();
} 

如果您不想在将整个 XML 推送到字符串之前在内存中创建整个 XML 的内存开销，您可能需要考虑使用 XStreamingElement (http://msdn.microsoft.com/en-us/library/system.xml.linq.xstreamingelement.aspx)。有关如何使用它的快速视频，请参阅 http://www.microsoft.com/uk/msdn/nuggets/nugget/295/LINQ-to-XML-Streaming-Large-Data-Files-Out-of-Memory.aspx。

Answer 4

我认为值得测试一下使用XmlSeriaizer 对整个列表进行序列化是否更快。

像这样：

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