Stream实现List和Map互转总结

前言

本篇介绍Stream流List和Map互转，同时在转换过程中遇到的问题分析。

一、Map转List

1.1 分析

按照默认顺序

1	java复制代码mapToList.entrySet().stream().map(a -> new User(a.getKey(), a.getValue())).collect(Collectors.toList());

根据key排序

1	java复制代码mapToList.entrySet().stream().sorted(Comparator.comparing(a -> a.getKey())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());

根据key排序

1	java复制代码mapToList.entrySet().stream().sorted(Map.Entry.comparingByKey()).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());

根据key倒序排序

1	java复制代码mapToList.entrySet().stream().sorted(Map.Entry.comparingByKey(Comparator.reverseOrder())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());

根据value排序

1	java复制代码mapToList.entrySet().stream().sorted(Comparator.comparing(Map.Entry::getValue)).map(a -> new User(a.getKey(), a.getValue())).collect(Collectors.toList());

根据value倒序排序

1	java复制代码mapToList.entrySet().stream().sorted(Collections.reverseOrder(Map.Entry.comparingByValue())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());

1.2 完整代码

实体类

java复制代码/**
 * @author lilinchao
 * @date 2021/7/28
 * @description 1.0
 **/
public class User {
    private Integer id;
    private String name;

    public User() {
    }

    public User(Integer id, String name) {
        this.id = id;
        this.name = name;
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public String toString() {
        return "User{" +
                "id=" + id +
                ", name='" + name + '\'' +
                '}';
    }
}

Map转List代码

java复制代码import java.util.*;
import java.util.stream.Collectors;

/**
 * @author lilinchao
 * @date 2021/7/28
 * @description 1.0
 **/
public class StreamMapToList {

    /**
     * 数据初始化
     */
    private static final Map<Integer, String> mapToList;
    static
    {
        mapToList = new HashMap<Integer, String>();
        mapToList.put(1003, "Thymee");
        mapToList.put(1001, "Leefs");
        mapToList.put(1002, "Jeyoo");
    }
    public static void main(String ages[]){
        List<User> userList = defaultOrder();
        System.out.println(userList);
        List<User> userList2and = orderByKeyMethodOne();
        System.out.println(userList2and);
        List<User> userList3and = orderByKeyMethodTwo();
        System.out.println(userList3and);
        List<User> userList4and = reverseOrderByKey();
        System.out.println(userList4and);
        List<User> userList5and = orderByValue();
        System.out.println(userList5and);
        List<User> userList6and = reverseOrderByValue();
        System.out.println(userList6and);
    }

    /**
     * 按照默认顺序
     */
    public static List<User> defaultOrder(){
        List<User> userList = mapToList.entrySet().stream().map(a -> new User(a.getKey(), a.getValue())).collect(Collectors.toList());
        return userList;
    }
    /**
     *根据key排序，方法1
     */
    public static List<User> orderByKeyMethodOne(){
        List<User> userList = mapToList.entrySet().stream().sorted(Comparator.comparing(a -> a.getKey())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());
        return userList;
    }
    /**
     *根据key排序，方法2
     */
    public static List<User> orderByKeyMethodTwo(){
        List<User> userList = mapToList.entrySet().stream().sorted(Map.Entry.comparingByKey()).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());
        return userList;
    }
    /**
     *根据key倒序排序
     */
    public static List<User> reverseOrderByKey(){
        List<User> userList = mapToList.entrySet().stream().sorted(Map.Entry.comparingByKey(Comparator.reverseOrder())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());
        return userList;
    }
    /**
     * 根据value排序
     */
    public static List<User> orderByValue(){
        List<User> userList = mapToList.entrySet().stream().sorted(Comparator.comparing(Map.Entry::getValue)).map(a -> new User(a.getKey(), a.getValue())).collect(Collectors.toList());
        return userList;
    }
    /**
     *根据value倒序排序
     */
    public static List<User> reverseOrderByValue(){
        List<User> userList = mapToList.entrySet().stream().sorted(Collections.reverseOrder(Map.Entry.comparingByValue())).map(a -> new User(a.getKey(),a.getValue())).collect(Collectors.toList());
        return userList;
    }
}

运行结果

json复制代码[User{id=1001, name='Leefs'}, User{id=1002, name='Jeyoo'}, User{id=1003, name='Thymee'}]
[User{id=1001, name='Leefs'}, User{id=1002, name='Jeyoo'}, User{id=1003, name='Thymee'}]
[User{id=1001, name='Leefs'}, User{id=1002, name='Jeyoo'}, User{id=1003, name='Thymee'}]
[User{id=1003, name='Thymee'}, User{id=1002, name='Jeyoo'}, User{id=1001, name='Leefs'}]
[User{id=1002, name='Jeyoo'}, User{id=1001, name='Leefs'}, User{id=1003, name='Thymee'}]
[User{id=1003, name='Thymee'}, User{id=1001, name='Leefs'}, User{id=1002, name='Jeyoo'}]

二、List转Map

2.1 分析

指定key-value，value是对象中的某个属性值

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, User::getName));

指定key-value，value是对象本身

User->User 是一个返回本身的lambda表达式

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, User->User));

指定key-value，value是对象本身

Function.identity()是简洁写法，也是返回对象本身

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, Function.identity()));

指定key-value,key 冲突的解决办法

(key1,key2)->key2:第二个key覆盖第一个key
(key1,key2)->key1:保留第一个key

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, Function.identity(),(key1,key2)->key2));

2.2 完整代码

java复制代码import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

/**
 * @author lilinchao
 * @date 2021/7/28
 * @description 1.0
 **/
public class StreamListToMap {

    private static final List<User> userList;
    static
    {
        userList = Arrays.asList(
                new User(1003,"keko"),
                new User(1001,"jeek"),
                new User(1002,"mack")
        );
    }

    public static void main(String ages[]){

        Map<Integer, String> map = method01();
        System.out.println(map);
        Map<Integer, User> map2and = method02();
        System.out.println(map2and);
        Map<Integer, User> map3and = method03();
        System.out.println(map3and);
        Map<Integer, User> map4and = method04();
        System.out.println(map4and);
    }

    /**
     * 指定key-value，value是对象中的某个属性值
     */
    public static Map<Integer,String> method01(){
        Map<Integer, String> userMap = userList.stream().collect(Collectors.toMap(User::getId, User::getName));
        return userMap;
    }
    /**
     *指定key-value，value是对象本身，User->User 是一个返回本身的lambda表达式
     */
    public static Map<Integer,User> method02(){
        Map<Integer, User> userMap = userList.stream().collect(Collectors.toMap(User::getId, User->User));
        return userMap;
    }
    /**
     * 指定key-value，value是对象本身，Function.identity()是简洁写法，也是返回对象本身
     */
    public static Map<Integer,User> method03(){
        Map<Integer, User> userMap = userList.stream().collect(Collectors.toMap(User::getId, Function.identity()));
        return userMap;
    }
    /**
     * 指定key-value,key 冲突的解决办法
     * (key1,key2)->key2:第二个key覆盖第一个key
     * (key1,key2)->key1:保留第一个key
     */
    public static Map<Integer,User> method04(){
        Map<Integer, User> userMap = userList.stream().collect(Collectors.toMap(User::getId, Function.identity(),(key1,key2)->key2));
        return userMap;
    }
}

运行结果

json复制代码{1001=jeek, 1002=mack, 1003=keko}
{1001=User{id=1001, name='jeek'}, 1002=User{id=1002, name='mack'}, 1003=User{id=1003, name='keko'}}
{1001=User{id=1001, name='jeek'}, 1002=User{id=1002, name='mack'}, 1003=User{id=1003, name='keko'}}
{1001=User{id=1001, name='jeek'}, 1002=User{id=1002, name='mack'}, 1003=User{id=1003, name='keko'}}

三、List转Map常见问题

3.1 常见问题

问题一

报错Duplicate key xxxx

该问题是因为在生成Map集合时key值重复造成的

解决方案

1. 后面的value覆盖前面的value

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, User::getName, (key1, key2) -> key2));

也可以保留前面的value,将key2换成key1即可

2. value进行拼接

1	java复制代码userList.stream().collect(Collectors.toMap(User::getId, User::getName, (key1, key2) -> key1+","+key2));

3. key重复时，返回集合

java复制代码userList2and.stream().collect(Collectors.toMap(User::getId, p -> {
                    List<String> getNameList = new ArrayList<>();
                    getNameList.add(p.getName());
                    return getNameList;
                },
                (List<String> value1, List<String> value2) -> {
                    value1.addAll(value2);
                    return value1;
                }
        ));

问题二

报错：Exception in thread "main" java.lang.NullPointerException

该问题是因为在存入Map集合时value值为null

解决方案

在转换流中加上判空，即便value为空,依旧输出。(与上面方法三相同)

3.2 完整代码

java复制代码import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

/**
 * @author lilinchao
 * @date 2021/7/28
 * @description 1.0
 **/
public class StreamListToMapAnalyze {

    private static final List<User> userList;
    private static final List<User> userList2and;
    static
    {
        //重复的key
        userList = Arrays.asList(
                new User(1003,"keko"),
                new User(1001,"jeek"),
                new User(1001,"teek"),
                new User(1002,"mack")
        );
        //重复的key,value为null
        userList2and = Arrays.asList(
                new User(1003,"keko"),
                new User(1001,"jeek"),
                new User(1001,"teek"),
                new User(1002,null)
        );
    }

    public static void main(String ages[]){
        Map<Integer, String> map = keyRepeatMethod01();
        System.out.println(map);
        Map<Integer, String> map2and = keyRepeatMethod02();
        System.out.println(map2and);
        Map<Integer,List<String>> map3and = keyRepeatMethod03();
        System.out.println(map3and);
    }

    /**
     * key重复时，后面的value覆盖前面的value
     */
    public static Map<Integer,String> keyRepeatMethod01(){
        Map<Integer, String> userMap = userList.stream().collect(Collectors.toMap(User::getId, User::getName, (key1, key2) -> key2));
        return userMap;
    }
    /**
     * key重复时，value进行拼接
     */
    public static Map<Integer,String> keyRepeatMethod02(){
        Map<Integer, String> userMap = userList.stream().collect(Collectors.toMap(User::getId, User::getName, (key1, key2) -> key1+","+key2));
        return userMap;
    }
    /**
     * key重复时，返回集合
     * value值为空，在转换流中加上判空，即便value为空,依旧输出
     */
    public static Map<Integer,List<String>> keyRepeatMethod03(){
        Map<Integer, List<String>> userListMap = userList2and.stream().collect(Collectors.toMap(User::getId, p -> {
                    List<String> getNameList = new ArrayList<>();
                    getNameList.add(p.getName());
                    return getNameList;
                },
                (List<String> value1, List<String> value2) -> {
                    value1.addAll(value2);
                    return value1;
                }
        ));
        return userListMap;
    }
}

运行结果

1
2
3

json复制代码{1001=teek, 1002=mack, 1003=keko}
{1001=jeek,teek, 1002=mack, 1003=keko}
{1001=[jeek, teek], 1002=[null], 1003=[keko]}

附：参考文章链接

www.cnblogs.com/fugitive/p/…

www.jb51.net/article/170…

本文转载自: 掘金

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