我有这样的问题，我想从哪里提取电影标题和评级

@Query(value ="SELECT m.title, avg(r.rating) " +
        "FROM movies m " +
        "INNER JOIN " +
        "ratings r " +
        "ON m.movieid = r.movieid " +
        "WHERE m.genres LIKE %:genre% " +
        "GROUP BY m.title " +
        "ORDER BY avg(r.rating) DESC " +
        "LIMIT 10",
        nativeQuery = true)
 List<Movies> topTenMoviesByGenreLikeIgnoreCase(@Param("genre") String genre);

但是这个列表显然不能存储有2个值的结果集
在了解了它之后，我发现这个链接spring data:jpa repository findall（）返回*map而不是list？但在我看来，它们的实施似乎并不清楚。
我不明白他们在第一个环节从哪里得到的长期价值

default Map<Long, TransactionModel> findByClientIdMap(Long id) {
    return findByClientId(id).stream().collect(Collectors.toMap(TransactionModel::getId, v -> v));
}

如何创建Map？

default Map<Movies, Float> topTenMoviesByGenreLikeIgnoreCaseMap(@Param("genre") String genre) {
    return topTenMoviesByGenreLikeIgnoreCase(genre).stream().collect(Collectors.toMap());
}

电影课：

@Entity
public class Movies {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer movieid;

private String title;
private String genres;

@OneToMany(mappedBy = "movieid")
private Set<Ratings> rates = new HashSet<>();

public Movies() {
}

public Movies(String title, String genres) {
    this.title = title;
    this.genres = genres;
}

/*Getters and Setters for variables*/

public Set<Ratings> getRates() {
    return rates;
}

public void setRates(Set<Ratings> rates) {
    this.rates = rates;
}
}

评级等级：

@Entity
public class Ratings {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer rateid;

@ManyToOne
@JoinColumn(name = "movieid")
private Movies movieid;

@ManyToOne
@JoinColumn(name = "userid")
private Usrs userid;

private float rating;
private Integer timestamp;

public Ratings() {
}

public Ratings(Movies movieid, Usrs userid, float rating, Integer timestamp)
{
    this.movieid = movieid;
    this.userid = userid;
    this.rating = rating;
    this.timestamp = timestamp;
}
/*Getters and Setters for variables*/
}