java List<>的问题

Question

我对java List和arrayList不是很熟悉。我只是需要一些东西来顺利地工作，以追加和排序。

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

这里给出了ga_和ga_struct的定义

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;
    }
}

我一直在这么做

        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);
        }

赋值函数一次只修改String(gene)中的一个字符。我刚刚从第一个循环中的“父亲”变异了9个新物种。但是..。我不知道为什么代码的输出会给我这个-

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

第一个循环证明变异进行得很慢。我还在突变后立即添加了，那么为什么后来所有这些都被最新版本覆盖了呢？

Answer 1

首先，你的对象用法有点奇怪。

在变异中，你似乎在改变和回归父亲。

这意味着您的列表将包含对同一实例的多个引用。

澄清一下：

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
}

在你的主要内容中：

    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);

    }

尝试更多像这样的东西：

    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;
}

在你的主要内容中：

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

你不是在创建一个新的对象，你已经在向量中添加了9次父对象。

从本质上讲，你得到的是

->父亲电子邮件: obj@123

你的列表对象看起来像obj@123，...

您将需要创建新的实例来记录此操作。我建议实现"clone()“方法来实现这一点。

Answer 3

您在任何地方都在使用单个对象，您永远不会向列表中添加新的ga_struct实例。您的mutate()方法似乎只是简单地修改了parent参数并返回它-它仍然是同一个对象，只是被修改了，这意味着它在任何地方都被修改了。

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

您确实创建了一个新的ga_struct实例，但通过设置对变异的father (它仍然是相同的实例，只是修改过)的引用来立即覆盖它：

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);

    }

您在此循环中的输出似乎是有效的，因为您可以按顺序查看对father的修改。然而，您实际要做的只是在 List**.**中一次又一次地添加对相同(修改过的)对象的引用

因此，当您最终将它们全部打印出来时，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-您可以创建一个新对象，并将其gene字段设置为parent中的变异gene字段。或者你可以clone parent然后改变克隆人的基因字符串。在任何一种情况下，您最终都将返回一个新的ga_struct实例，它应该可以解决这个问题。

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
}