Question

将遗传编程与ECJ 21结合使用，我正在寻找有关如何序列化ec.Individual（成功（？）进化后）的建议。

诀窍是，我需要在稍后阶段对此GP进行反序列化，并执行它...最好在ECJ框架之外（在执行工作时似乎有很多脚手架参与“执行”GP ECJ框架，因为那实际上是为了使事物发展，而不是“运行”它。）

我几乎就有了这样的东西：

public static void main (String [] args) {
    File f = new File("./my.params");
        if (!f.exists() ) {
            throw new InvalidParameterException(f.getName() + " does NOT exist");
        }
        ParameterDatabase pd = new ParameterDatabase(f, new String []{f.getCanonicalPath()});

        Output output = ec.Evolve.buildOutput();

        EvolutionState evs = ec.Evolve.initialize(pd, 0,output);

        evs.run(EvolutionState.C_STARTED_FRESH);
        Individual [] individuals =  ((SimpleStatistics)evs.statistics).getBestSoFar();

        String bestIndividStr = "";
        for (Individual individual : individuals) {
            bestIndividStr = printToLog(evs, individual);
        }

        Species s = individuals[0].species;
        s = new GPSpecies();

        Individual gpInd = s.newIndividual(evs,  new LineNumberReader(new StringReader(bestIndividStr)));
}

private static String printToLog(EvolutionState evs, Individual individual) {
    String bestIndividStr;
    ByteArrayOutputStream baos  = new ByteArrayOutputStream();
    PrintWriter pw = new PrintWriter(baos);
    individual.printIndividual(evs, pw);
    pw.append(System.lineSeparator());
    pw.close();
    bestIndividStr = baos.toString();
    LoggerFactory.getLogger(Evolve.class).info("Best Dude: \n{}",bestIndividStr);
    return bestIndividStr;
}

好的，有些问题：

我现在如何输入一些变量并评估gpInd？
在此示例情况下，我有原始个人可用，所以我可以从那里得到物种参考（然后用于从原始输出中创建一个新的个体）。在我的真实情况，我不会那样。

为printIndividual(..)生成的文本输出编写自己的解析器和评估堆栈应该不会那么难，但如果有更简单的（内置）方式，我宁愿不这样做。

PS：到目前为止我所基于的是this tutorial

Answer 1

我有类似的问题。我的解决方案是通过继承GPNode来创建个人的XML表示。以下来源显示了相应的实现。

public abstract class XMLGPNode extends GPNode {

/**
 * 
 */
private static final long serialVersionUID = 2732707537997825895L;


public StringBuilder makeXMLTree() {
    DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder docBuilder = null;
    StringBuilder sb = new StringBuilder();

    try {
        docBuilder = docFactory.newDocumentBuilder();
        Document doc = docBuilder.newDocument();

        makeXMLTree(doc, null);

        TransformerFactory transformerFactory = TransformerFactory.newInstance();
        Transformer transformer;

        transformer = transformerFactory.newTransformer();

        DOMSource source = new DOMSource(doc);
        StringWriter sw = new StringWriter();

        StreamResult result = new StreamResult(sw);

        // Output to console for testing
        // StreamResult result = new StreamResult(System.out);

        transformer.transform(source, result);
        sb.append(new String(sw.getBuffer()));
    } catch (TransformerConfigurationException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (ParserConfigurationException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (TransformerException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }


    return sb;
}


private void makeXMLTree(Document doc, Element parent)  {

    Element child = doc.createElement(toString());              
    if (parent==null) {
        doc.appendChild(child);
    } else {
        parent.appendChild(child);
    }

    if (children.length > 0) {                  
        for(int x=0; x<children.length; x++)   {
            ((XMLGPNode)children[x]).makeXMLTree(doc, child);

        }
    }

}

然后可以在问题中对最终个人进行序列化，以便将其保存或转移到另一个应用程序或加载它以进行新的评估。

以下代码段显示了makeXMLTree - [{1}} - evaluate实施方法中根节点的GPProblem - 方法调用。

    public void evaluate(final EvolutionState state, 
        final Individual ind, 
        final int subpopulation,
        final int threadnum)  {

    if (!ind.evaluated) {  // don't bother reevaluating

        int hits = 0;
        ...
        ...

        transformationType = new TransformationType(TransformationType.TYPE.values()[state.random[threadnum].nextInt(TransformationType.MAX_INPUT_TYPES)]);

        ((GPIndividual)ind).trees[0].child.eval(state,threadnum,input,stack,((GPIndividual)ind),this);

        XMLGPNode child = (XMLGPNode)((GPIndividual)ind).trees[0].child;


        **String individual = child.makeXMLTree().toString();**

... ...

序列化ECJ GP Individual以便以后执行

1 个答案: