使用java8[复制]合并对象

Question

我有一张这样的清单

    List<Objeto> myList = new ArrayList<Objeto>();
    Objeto o1 = new Objeto("1", "bbb");
    myList.add(o1);
    Objeto o2 = new Objeto("1", "rrrr");
    myList.add(o2);
    Objeto o3 = new Objeto("2", "eee");
    myList.add(o3);
    Objeto o4 = new Objeto("2", "wwww");
    myList.add(o4);
    Objeto o5 = new Objeto("3", "iiii");
    myList.add(o5);

其中Objecto是此类型的对象

class Objeto{
private String contentId;
private String address;

Objeto(String id, String address){
    this.contentId = id;
    this.address = address;
}

//接球手和接球手

}

我想将列表合并成如下的哈希图

(1, {"bbb","rrrr"})
(2, {"eee","wwww"})
(3, {"iiii"})

我可以使用java 8 lambda来实现它吗？或者通过任何其他方式？

非常感谢！

Answer 1

package com.company;

import java.util.ArrayList;
import java.util.stream.Collectors;

public class Main {
public static class Objeto {
    private String contentId;
    private String address;

    Objeto(String id, String address) {
        this.contentId = id;
        this.address = address;
    }

    public String getContentId() {
        return contentId;
    }

    public String getAddress() {
        return address;
    }
}

public static void main(String[] args) {
    java.util.List<Objeto> myList = new ArrayList<Objeto>();
    Objeto o1 = new Objeto("1", "bbb");
    myList.add(o1);
    Objeto o2 = new Objeto("1", "rrrr");
    myList.add(o2);
    Objeto o3 = new Objeto("2", "eee");
    myList.add(o3);
    Objeto o4 = new Objeto("2", "wwww");
    myList.add(o4);
    Objeto o5 = new Objeto("3", "iiii");
    myList.add(o5);

    myList.stream().collect(
            Collectors.groupingBy(
                    Objeto::getContentId,
                    Collectors.mapping(
                            Objeto::getAddress,
                            Collectors.toList()))
    ).forEach((id, addresses) -> System.out.printf("(%s, {%s})\n", id, addresses.stream().collect(Collectors.joining(","))));
}
}

输出：

(1, {bbb,rrrr})
(2, {eee,wwww})
(3, {iiii})

Answer 2

您可以使用收集器。groupingBy可按对象的ID和收集器对对象进行分组。映射将每个对象映射到其相应的地址：

Map<String,List<String>>
    map = myList.stream()
                .collect(Collectors.groupingBy(Objeto::getContentID,
                                               Collectors.mapping(Object::getAddress
                                                                  Collectors.toList())));