三种将list转换为map的方法

1）传统方法

假设有某个类如下

Java代码

class Movie {  

    private Integer rank;

    private String description;  

    public Movie(Integer rank, String description) {

        super();

        this.rank = rank;

        this.description = description;

    }  

    public Integer getRank() {

        return rank;

    }  

    public String getDescription() {

        return description;

    }  

    @Override

    public String toString() {

        return Objects.toStringHelper(this)

                .add("rank", rank)

                .add("description", description)

                .toString();

    }

}

1、使用传统的方法：

public void convert_list_to_map_with_java () {  

    List<Movie> movies = new ArrayList<Movie>();

    movies.add(new Movie(1, "The Shawshank Redemption"));

    movies.add(new Movie(2, "The Godfather"));  

    Map<Integer, Movie> mappedMovies = new HashMap<Integer, Movie>();

    for (Movie movie : movies) {

        mappedMovies.put(movie.getRank(), movie);

    }  

    logger.info(mappedMovies);  

    assertTrue(mappedMovies.size() == 2);

    assertEquals("The Shawshank Redemption", mappedMovies.get(1).getDescription());

}

2、JAVA 8直接用流的方法

public void convert_list_to_map_with_java8_lambda () {  

    List<Movie> movies = new ArrayList<Movie>();

    movies.add(new Movie(1, "The Shawshank Redemption"));

    movies.add(new Movie(2, "The Godfather"));  

    Map<Integer, Movie> mappedMovies = movies.stream().collect(

            Collectors.toMap(Movie::getRank, (p) -> p));  

    logger.info(mappedMovies);  

    assertTrue(mappedMovies.size() == 2);

    assertEquals("The Shawshank Redemption", mappedMovies.get(1).getDescription());

}

（2）转换过程中的两个问题

a、key重复

重复时用后面的value 覆盖前面的value

Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(key1 , key2)-> key2 ));

重复时将前面的value 和后面的value拼接起来

Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(key1 , key2)-> key1+","+key2 ));

重复时将重复key的数据组成集合

Map<String, List<String>> map = list.stream().collect(Collectors.toMap(Person::getId,

                p ->  {

                     List<String> getNameList = new ArrayList<>();

                         getNameList.add(p.getName());

                         return getNameList;

                     },

                     (List<String> value1, List<String> value2) -> {

                         value1.addAll(value2);

                         return value1;

                     }

));

b、valve为null

在转换流中加上判空，即便value为空,依旧输出

Map<String, List<String>> map = list.stream().collect(Collectors.toMap(Person::getId,

                p ->  {

                     List<String> getNameList = new ArrayList<>();

                         getNameList.add(p.getName());

                         return getNameList;

                     },

                     (List<String> value1, List<String> value2) -> {

                         value1.addAll(value2);

                         return value1;

                     }

))

3、使用guava 工具类库

public void convert_list_to_map_with_guava () {  

    List<Movie> movies = Lists.newArrayList();

    movies.add(new Movie(1, "The Shawshank Redemption"));

    movies.add(new Movie(2, "The Godfather"));  

    Map<Integer,Movie> mappedMovies = Maps.uniqueIndex(movies, new Function <Movie,Integer> () {

          public Integer apply(Movie from) {

            return from.getRank();

    }});  

    logger.info(mappedMovies);  

    assertTrue(mappedMovies.size() == 2);

    assertEquals("The Shawshank Redemption", mappedMovies.get(1).getDescription());

}

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一、
list转map
Map<Long, User> maps = userList.stream().collect(Collectors.toMap(User::getId,Function.identity()));

二、
另外，转换成map的时候，可能出现key一样的情况，如果不指定一个覆盖规则，上面的代码是会报错的。转成map的时候，最好使用下面的方式：
Map<Long, User> maps = userList.stream().collect(Collectors.toMap(User::getId, Function.identity(), (key1, key2) -> key2));

三、
有时候，希望得到的map的值不是对象，而是对象的某个属性，那么可以用下面的方式：
Map<Long, String> maps = userList.stream().collect(Collectors.toMap(User::getId, User::getAge, (key1, key2) -> key2));

四、
//List 以ID分组 Map<Integer,List>
Map<Integer, List> groupBy = appleList.stream().collect(Collectors.groupingBy(Apple::getId));

System.err.println(“groupBy:”+groupBy);
{1=[Apple{id=1, name=‘苹果1’, money=3.25, num=10}, Apple{id=1, name=‘苹果2’, money=1.35, num=20}], 2=[Apple{id=2, name=‘香蕉’, money=2.89, num=30}], 3=[Apple{id=3, name=‘荔枝’, money=9.99, num=40}]}

巴特西