Java8 - 将 List 集合转为 Map

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@Data
@AllArgsConstructor
@NoArgsConstructor
public class User {

    private Integer userId;
    private String name;
    private String email;
}

1. 将List转为Map：key=userId, value=name

public class Main {
    public static void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User(1, "user1", "email1@demo.com"));
        users.add(new User(2, "user2", "email2@demo.com"));
        users.add(new User(3, "user3", "email3@demo.com"));
        users.add(new User(4, "user4", "email4@demo.com"));

        Map<Integer, String> userIdAndName = users.stream().collect(Collectors.toMap(User::getUserId, User::getName));

        System.out.println(userIdAndName);
    }
}

输出：

{1=user1, 2=user2, 3=user3, 4=user4}

2. 将List转为Map：key=userId, value=User对象

public class Main {
    public static void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User(1, "user1", "email1@demo.com"));
        users.add(new User(2, "user2", "email2@demo.com"));
        users.add(new User(3, "user3", "email3@demo.com"));
        users.add(new User(4, "user4", "email4@demo.com"));

        Map<Integer, User> userIdAndName = users.stream().collect(Collectors.toMap(User::getUserId, user->user));

        System.out.println(userIdAndName);
    }
}

输出：

{
    1=User(userId=1, name=user1, email=email1@demo.com), 
    2=User(userId=2, name=user2, email=email2@demo.com), 
    3=User(userId=3, name=user3, email=email3@demo.com), 
    4=User(userId=4, name=user4, email=email4@demo.com)
}

3. 重复的 key 处理

如果出现相同的key，那么会抛出重复key的异常，比如我们将userId为3的修改为2，则会和第二个用户冲突：

public class Main {
    public static void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User(1, "user1", "email1@demo.com"));
        users.add(new User(2, "user2", "email2@demo.com"));
        users.add(new User(2, "user3", "email3@demo.com"));
        users.add(new User(4, "user4", "email4@demo.com"));

        Map<Integer, User> userIdAndName = users.stream().collect(Collectors.toMap(User::getUserId, user->user));

        System.out.println(userIdAndName);
    }
}

抛出异常：

Exception in thread "main" java.lang.IllegalStateException: Duplicate key User(userId=2, name=user2, email=email2@demo.com)
	at java.util.stream.Collectors.lambda$throwingMerger$113(Collectors.java:133)
	at java.util.stream.Collectors$$Lambda$3/780237624.apply(Unknown Source)
	at java.util.HashMap.merge(HashMap.java:1245)
	at java.util.stream.Collectors.lambda$toMap$171(Collectors.java:1320)
	at java.util.stream.Collectors$$Lambda$5/1560911714.accept(Unknown Source)
	at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
	at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1374)
	at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:512)
	at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:502)
	at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
	at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
	at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)
	at com.imooc.uua.rest.Main.main(Main.java:16)

Collectors.toMap支持第三个参数，对重复值进行选取：

public class Main {
    public static void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User(1, "user1", "email1@demo.com"));
        users.add(new User(2, "user2", "email2@demo.com"));
        users.add(new User(2, "user3", "email3@demo.com"));
        users.add(new User(4, "user4", "email4@demo.com"));

        Map<Integer, User> userIdAndName = users.stream().collect(Collectors.toMap(
                User::getUserId,
                user->user,
                (oldValue,newValue)->newValue));

        System.out.println(userIdAndName);
    }
}

输出：

{
     1=User(userId=1, name=user1, email=email1@demo.com), 
     2=User(userId=2, name=user3, email=email3@demo.com), 
     4=User(userId=4, name=user4, email=email4@demo.com)
}

可以看到，只会出现 user3，而 user2 被覆盖掉了

4. 将 id 和用户列表映射：groupingBy

刚才上面出现重复的ID，是根据值进行覆盖，在某些情况下需要映射成列表。即：List -> Map<Integer, List>的情况。

public class Main {
    public static void main(String[] args) {
        List<User> users = new ArrayList<>();
        users.add(new User(1, "user1", "email1@demo.com"));
        users.add(new User(2, "user2", "email2@demo.com"));
        users.add(new User(2, "user3", "email3@demo.com"));
        users.add(new User(4, "user4", "email4@demo.com"));

        Map<Integer, List<User>> userIdAndName = users.stream().collect(Collectors.groupingBy(User::getUserId));

        System.out.println(userIdAndName);
    }
}

输出：文章来源地址https://www.toymoban.com/news/detail-478192.html

{
    1=[User(userId=1, name=user1, email=email1@demo.com)], 
    2=[User(userId=2, name=user2, email=email2@demo.com), 
       User(userId=2, name=user3, email=email3@demo.com)], 
    4=[User(userId=4, name=user4, email=email4@demo.com)]
}