简单的accord.net机器学习示例

Question

简单的accord.net机器学习示例

c#machine-learningaccord.net

9

我对机器学习和accord.net（我编写C＃代码）都很陌生。

我想创建一个简单的项目，其中我查看振荡的简单时间序列数据，然后让accord.net学习它并预测下一个值将是什么。

这是数据（时间序列）应该看起来的样子：

X - Y

然后我希望它能预测以下内容：

X - Y

你们能否给我一些关于如何解决它的示例?

- RHC

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- reddal · Accepted Answer

一种简单的方法是使用Accord ID3决策树。关键在于确定要使用哪些输入 - 你不能仅仅训练X - 树将无法从中学习到X未来的值 - 然而，你可以构建一些从X（或Y的先前值）派生的特征，这将非常有用。

通常对于这样的问题 - 你会基于先前Y的值派生的特征进行每个预测，而不是X。然而，这假设你可以在每个预测之间按顺序观察Y（然后你就无法为任意X进行预测），所以我将坚持原问题。

下面是我尝试构建Accord ID3决策树来解决此问题的代码。我使用了几个不同的x％n值作为特征 - 希望树可以从中找出答案。实际上，如果我将（x-1）％4添加为一个特征，则可以在单个级别中仅使用该属性完成它 - 但我想重点是让树发现模式。

以下是该代码：

    // this is the sequence y follows
    int[] ysequence = new int[] { 1, 2, 3, 2 };

    // this generates the correct Y for a given X
    int CalcY(int x) => ysequence[(x - 1) % 4];

    // this generates some inputs - just a few differnt mod of x
    int[] CalcInputs(int x) => new int[] { x % 2, x % 3, x % 4, x % 5, x % 6 };


    // for https://dev59.com/4Jzha4cB1Zd3GeqPFHLP
    [TestMethod]
    public void AccordID3TestStackOverFlowQuestion2()
    {
        // build the training data set
        int numtrainingcases = 12;
        int[][] inputs = new int[numtrainingcases][];
        int[] outputs = new int[numtrainingcases];

        Console.WriteLine("\t\t\t\t x \t y");
        for (int x = 1; x <= numtrainingcases; x++)
        {
            int y = CalcY(x);
            inputs[x-1] = CalcInputs(x);
            outputs[x-1] = y;
            Console.WriteLine("TrainingData \t " +x+"\t "+y);
        }

        // define how many values each input can have
        DecisionVariable[] attributes =
        {
            new DecisionVariable("Mod2",2),
            new DecisionVariable("Mod3",3),
            new DecisionVariable("Mod4",4),
            new DecisionVariable("Mod5",5),
            new DecisionVariable("Mod6",6)
        };

        // define how many outputs (+1 only because y doesn't use zero)
        int classCount = outputs.Max()+1;

        // create the tree
        DecisionTree tree = new DecisionTree(attributes, classCount);

        // Create a new instance of the ID3 algorithm
        ID3Learning id3learning = new ID3Learning(tree);

        // Learn the training instances! Populates the tree
        id3learning.Learn(inputs, outputs);

        Console.WriteLine();
        // now try to predict some cases that werent in the training data
        for (int x = numtrainingcases+1; x <= 2* numtrainingcases; x++)
        {
            int[] query = CalcInputs(x);

            int answer = tree.Decide(query); // makes the prediction

            Assert.AreEqual(CalcY(x), answer); // check the answer is what we expected - ie the tree got it right
            Console.WriteLine("Prediction \t\t " + x+"\t "+answer);
        }
    }

这是它生成的输出结果：

                 x   y
TrainingData     1   1
TrainingData     2   2
TrainingData     3   3
TrainingData     4   2
TrainingData     5   1
TrainingData     6   2
TrainingData     7   3
TrainingData     8   2
TrainingData     9   1
TrainingData     10  2
TrainingData     11  3
TrainingData     12  2

Prediction       13  1
Prediction       14  2
Prediction       15  3
Prediction       16  2
Prediction       17  1
Prediction       18  2
Prediction       19  3
Prediction       20  2
Prediction       21  1
Prediction       22  2
Prediction       23  3
Prediction       24  2

希望对你有所帮助。

编辑：根据评论，下面的示例被修改为在目标变量（Y）的先前值上进行训练 - 而不是从时间索引（X）派生的特征。这意味着您不能从系列的开始开始训练 - 因为您需要先前的Y值的历史记录。在此示例中，我从x = 9开始，只是因为这样保持了相同的序列。

        // this is the sequence y follows
    int[] ysequence = new int[] { 1, 2, 3, 2 };

    // this generates the correct Y for a given X
    int CalcY(int x) => ysequence[(x - 1) % 4];

    // this generates some inputs - just a few differnt mod of x
    int[] CalcInputs(int x) => new int[] { CalcY(x-1), CalcY(x-2), CalcY(x-3), CalcY(x-4), CalcY(x - 5) };
    //int[] CalcInputs(int x) => new int[] { x % 2, x % 3, x % 4, x % 5, x % 6 };


    // for https://dev59.com/4Jzha4cB1Zd3GeqPFHLP
    [TestMethod]
    public void AccordID3TestTestStackOverFlowQuestion2()
    {
        // build the training data set
        int numtrainingcases = 12;
        int starttrainingat = 9;
        int[][] inputs = new int[numtrainingcases][];
        int[] outputs = new int[numtrainingcases];

        Console.WriteLine("\t\t\t\t x \t y");
        for (int x = starttrainingat; x < numtrainingcases + starttrainingat; x++)
        {
            int y = CalcY(x);
            inputs[x- starttrainingat] = CalcInputs(x);
            outputs[x- starttrainingat] = y;
            Console.WriteLine("TrainingData \t " +x+"\t "+y);
        }

        // define how many values each input can have
        DecisionVariable[] attributes =
        {
            new DecisionVariable("y-1",4),
            new DecisionVariable("y-2",4),
            new DecisionVariable("y-3",4),
            new DecisionVariable("y-4",4),
            new DecisionVariable("y-5",4)
        };

        // define how many outputs (+1 only because y doesn't use zero)
        int classCount = outputs.Max()+1;

        // create the tree
        DecisionTree tree = new DecisionTree(attributes, classCount);

        // Create a new instance of the ID3 algorithm
        ID3Learning id3learning = new ID3Learning(tree);

        // Learn the training instances! Populates the tree
        id3learning.Learn(inputs, outputs);

        Console.WriteLine();
        // now try to predict some cases that werent in the training data
        for (int x = starttrainingat+numtrainingcases; x <= starttrainingat + 2 * numtrainingcases; x++)
        {
            int[] query = CalcInputs(x);

            int answer = tree.Decide(query); // makes the prediction

            Assert.AreEqual(CalcY(x), answer); // check the answer is what we expected - ie the tree got it right
            Console.WriteLine("Prediction \t\t " + x+"\t "+answer);
        }
    }

您还可以考虑对Y的以前值之间的差异进行培训 - 这在绝对值不如相对变化重要的情况下效果更好。