Question

我对java List和arrayList并不熟悉。我只需要一些工作顺利地进行追加和排序。

我的算法很简单：

set a father string 
add father to speciesList
    mutate father to some new child
    make this new child the future father
    go to step 2

ga_和ga_struct的定义在此处给出

public class ga_struct {

    public String gene;
    public int fitness;

}


public class ga_{

    public List<ga_struct> vector= new ArrayList<ga_struct>();

        public void sortspecies()
        {
        Collections.sort(vector,new Comparator<ga_struct>() {
        @Override
        public int compare(ga_struct o1, ga_struct o2) {
            int res;
            if(o1.fitness<o2.fitness)
                res=-1;
            else if(o1.fitness>o2.fitness)
                res=1;
            else 
                res=0;
            return res;
                 }
              }
                  );

     }


    public ga_struct mutate(ga_struct parent)
    {
        Random r= new Random();
        ......     do some modification to the parent
        return parent;
    }
}

我一直这样做

        ga_ newSpecies = new ga_();
        Random r= new Random(10);
        ga_struct father= new ga_struct();
        father.gene="123";
        newSpecies.vector.add(father);

        for (int i = 1; i < 10; i++) {
            ga_struct ng = new ga_struct();        
            ng=newSpecies.mutate(father);
            ng.fitness=i;
            newSpecies.vector.add(ng);
            father=ng;          
            System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);

        }

        newSpecies.sortspecies();
        System.out.println("\ncurrent population\n");

        for (int i = 0; i < 10; i++) {
            System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);
        }

mutator函数一次只改变一个String(gene)个字符。我只是在第一个循环中从“父亲”变异了9个新物种。但是..我不知道为什么代码的输出给了我这个 -

133 with fitness factor 1
433 with fitness factor 2
433 with fitness factor 3
443 with fitness factor 4
453 with fitness factor 5
553 with fitness factor 6
563 with fitness factor 7
563 with fitness factor 8
573 with fitness factor 9

current population

573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9
573 with fitness factor 9

第一个循环证明突变正在缓慢进行..而且我也在突变后立即添加，那么为什么稍后所有这些都被最新版本覆盖？

Answer 1

首先，您的对象用法有点奇怪。

在变异中，你似乎正在改变并归还父亲。

这意味着您的列表将包含对同一实例的多个引用。

澄清：

public ga_struct mutate(ga_struct parent) //takes in reference to parent
{
    Random r= new Random(); //modifies parent
    ......     do some modification to the parent
    return parent; //return reference to parent
}

在你的主要：

    ga_ newSpecies = new ga_();
    Random r= new Random(10);
    ga_struct father= new ga_struct();//instantiate father
    father.gene="123";
    newSpecies.vector.add(father);

    for (int i = 1; i < 10; i++) {
        ga_struct ng = new ga_struct();//create new instance for child
        ng=newSpecies.mutate(father);//set ng as reference to same instance as father, instance instantiated on previous line is discarded
        ng.fitness=i;
        newSpecies.vector.add(ng);
        father=ng;          
        System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);

    }

尝试更像这样的东西：

    public ga_struct mutate(ga_struct parent)
{
    ga_struct ng = new ga_struct();
    ng.gene = father.gene;
    Random r= new Random();
    //do some modification to ng
    return ng;
}

并在你的主要：

a_ newSpecies = new ga_();
    Random r= new Random(10);
    ga_struct father= new ga_struct();
    father.gene="123";
    newSpecies.vector.add(father);

    for (int i = 1; i < 10; i++) {    
        ga_struct ng=newSpecies.mutate(father);
        ng.fitness=i;
        newSpecies.vector.add(ng);
        father=ng;          
        System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);

    }

    newSpecies.sortspecies();
    System.out.println("\ncurrent population\n");

    for (int i = 0; i < 10; i++) {
        System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);
    }

Answer 2

您没有创建新对象，您已将父对象添加了9次到向量。

基本上你得到的是

父亲 - ＆gt;物镜@ 123

您的List对象是什么样的 [obj @ 123，obj @ 123，obj @ 123，...]

您将需要创建新实例来记录此事件。我建议实现“clone（）”方法来执行此操作。

Answer 3

您正在处理任何地方的单个对象，您永远不会在列表中添加新的ga_struct实例。您的mutate()方法似乎只是修改parent参数并返回它 - 它仍然是同一个对象，只是修改过，这意味着它在任何地方都被修改过。

public ga_struct mutate(ga_struct parent)
{
    Random r= new Random();
    ......     do some modification to the parent
    return parent;
}

您确实创建了ga_struct的新实例，但是您可以通过设置对变异father的引用来立即覆盖它（它仍然是同一个实例，只是修改过）：

for (int i = 1; i < 10; i++) {
        ga_struct ng = new ga_struct();        
        ng=newSpecies.mutate(father); //the new ga_struct is overwritten
        ng.fitness=i;
        newSpecies.vector.add(ng);
        father=ng;          
        System.out.println(newSpecies.vector.get(i).gene+" with fitness factor "+newSpecies.vector.get(i).fitness);

    }

此循环中的输出似乎有效，因为您按顺序看到father的修改。 但是，您实际所做的只是将相同（已修改）对象的引用一遍又一遍地添加到List。

因此，当您最终全部打印出来时，您会在List中看到10个重复的条目。

我的建议是更改mutate()以返回ga_struct的新实例 - 您可以创建新对象并将其gene字段设置为变异gene来自parent的字段。或者您可以clone parent和然后更改克隆的基因字符串。在任何一种情况下，您最终都会返回ga_struct的新实例来解决问题。

public ga_struct mutate(ga_struct parent)
{
    Random r= new Random();
    ga_struct mutant = parent.clone(); 
   //or 
   //ga_struct mutant = new ga_struct();
   //mutant.gene = parent.gene;

    ......     do some modification to the mutant
    return mutant; //now you'll be returning a new object not just a modified one
}

java List＆lt;＆gt;的问题

3 个答案: