2016-06-14 67 views
2

我试图以可读格式(XML)保存类。 问题是,生成的文件只能输出为:XML序列化结果空的XML

<?xml version="1.0" encoding="Windows-1252"?> 
<ExtremeLearningMachine xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" /><?xml version="1.0" encoding="Windows-1252"?> 
<ExtremeLearningMachine xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" /><?xml version="1.0" encoding="Windows-1252"?> 
<ExtremeLearningMachine xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" /><?xml version="1.0" encoding="Windows-1252"?> 
<ExtremeLearningMachine xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" /> 

这里是我的类:

public class ExtremeLearningMachine { 
    public ExtremeLearningMachine() 
    { 

    } 
    int input, hidden; //only have 1 output neuron 
    double[,] W1, W2; 
    public ExtremeLearningMachine(int inputNeuron, int hiddenNeuron) { input = inputNeuron; hidden = hiddenNeuron; } 
    public void train(int dataCount, double[,] trainingSet) { 
     //set matrix 
     double[,] trainInput = new double[input, dataCount], desireOutput = new double[1, dataCount]; 
     for (int i = 0; i < dataCount; i++) { 
      for (int j = 0; j < input; j++) trainInput[j, i] = trainingSet[i, j]; 
      desireOutput[0, i] = trainingSet[i, input]; 
     } 
     //W1 
     W1 = new double[hidden, input]; 
     for (int i = 0; i < hidden; i++) { for (int j = 0; j < input; j++)W1[i, j] = Random.value; } 
     //hidden 
     //double[,] H = new double[hidden, dataCount]; 
     double[,] H = Matrix.Multiply(W1, trainInput); 
     //activation function(binary sigmoid) 
     for (int i = 0; i < hidden; i++) { 
      for (int j = 0; j < dataCount; j++) H[i, j] = 1f/(1f + Mathf.Exp((float)-H[i, j])); 
     } 
     //W2 
     W2 = Matrix.Multiply(desireOutput, H.PseudoInverse()); 
    } 
    public double test(double[,] set) {//only [~,1] allowed 
     double[,] H = Matrix.Multiply(W1, set.Transpose()); 
     //activation function(binary sigmoid) 
     for (int i = 0; i < hidden; i++) H[i, 0] = 1f/(1f + Mathf.Exp((float)-H[i, 0])); 
     H = Matrix.Multiply(W2, H); 
     return H[0, 0]; 
    } 
} 

这里是我的节省代码:

void save() 
{ 
    System.Xml.Serialization.XmlSerializer writer = 
     new System.Xml.Serialization.XmlSerializer(typeof(ExtremeLearningMachine)); 

    string path = Directory.GetCurrentDirectory() + "\\ElmTrain.xml"; 
    System.IO.FileStream file = System.IO.File.Create(path); 
    for(int i=0;i<elm.Length;i++) 
    writer.Serialize(file, elm[i]); 
    file.Close(); 
} 

而且我的加载代码,在遇到任何问题(我还没有测试过,因为我无法保存):

void load() 
{ 
    System.Xml.Serialization.XmlSerializer reader = 
    new System.Xml.Serialization.XmlSerializer(typeof(ExtremeLearningMachine)); 
    System.IO.StreamReader file = new System.IO.StreamReader("//ElmTrain.xml"); 
    elm = (ExtremeLearningMachine[])reader.Deserialize(file); 
    file.Close(); 
} 

我也开到任何其他的想法来保存这个类在其他可读的格式,如果它的建议

非常感谢您

回答

0

首先,ExtremeLearningMachine没有任何公共成员序列化,所以是:期待它是空的; XmlSerializer仅序列化公共字段和属性。例如,尝试添加公共属性以补充您的私人字段。

其次:不要将多个片段序列化到同一个文档。相反,请创建一个容器,并对其进行序列化。坦率地说,你可以只使用ExtremeLearningMachine[]作为容器,因为你已经有:

var writer = new XmlSerializer(typeof(ExtremeLearningMachine[])); 
string path = Path.Combine(Directory.GetCurrentDirectory(), "ElmTrain.xml"); 
using(var file = File.Create(path)) { 
    writer.Serialize(file, elm); 
} 

和:

var reader = new XmlSerializer(typeof(ExtremeLearningMachine[])); 
using(var file = File.OpenRead(path)) { 
    elm = (ExtremeLearningMachine[])reader.Deserialize(file); 
} 
+0

我用你试过代码(我也已经公开了这些变量),但现在它导致空文件和异常:Array不是一维数组。 – christantoan

+0

@christantoan right;所以这个异常告诉你,如果XmlSerializer是一维的,那么它只是快乐的序列化数组。你有一个二维数组。所以......它告诉你问题是什么。选项:1:压平数据(只要存储尺寸,所有数组都可以进行矢量化); 2:使用不同的序列化程序 –

+0

注意:我强烈建议为序列化创建单独的DTO模型;你的域模型可能有一个二维数组,而DTO模型可以有一维数组,一个高度和一个宽度(尽管技术上至少有一个是多余的,因为1D数组包含一个长度) –

0

虽然转换尝试下面的代码:

  XmlSerializer serializer = new XmlSerializer(typeof(ExtremeLearningMachine)); 
      MemoryStream memStream = new MemoryStream(); 
      serializer.Serialize(memStream, elm); 
      FileStream file = new FileStream(folderName + "\\ElmTrain.xml", FileMode.Create, FileAccess.ReadWrite);//Provide correct path as foldername 
      memStream.WriteTo(file); 
      file.Close();