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java - 序列化 ECJ GP Individual 以供以后执行

转载 作者:行者123 更新时间:2023-11-30 09:06:36 26 4
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使用遗传编程 ECJ 21 ,我正在寻找有关如何序列化 ec.Individual(在成功(?)进化之后)的建议。

诀窍是,然后我需要在稍后阶段反序列化这个 GP,然后执行它......最好在 ECJ 框架之外(在工作时“执行”GP 似乎涉及很多脚手架ECJ 框架,因为它实际上旨在发展事物,而不是“运行”它)。

我有一些关于这个的东西:

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,一些问题:

  1. 我现在如何输入一些变量并计算 gpInd
  2. 在这个示例情况下,我有原始个人可用,所以我可以从中获得物种引用(然后用于从原始输出创建一个新的个体)。在我真实的情况,我不会那样做。

printIndividual(..) 生成的文本输出编写我自己的解析器和评估堆栈不应该那么难,但如果有更简单的(内置的)我宁愿不这样做-in) 方式。

PS:到目前为止,我所拥有的是基于 this tutorial 构建的

最佳答案

我遇到了类似的问题。我的解决方案是通过子类化 GPNode 创建个体的 XML 表示。以下来源显示了适当的实现。

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体进行序列化,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。

以下代码片段显示了 GPProblem 实现的 evaluate 方法中根节点的 makeXMLTree 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();**

......

关于java - 序列化 ECJ GP Individual 以供以后执行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24373981/

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