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

?

1	Map<Long, User> maps = userList.stream().collect(Collectors.toMap(User::getId,Function.identity()));

二、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：

?

1	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：

?

1	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>

?

1	Map<Integer, List> groupBy = appleList.stream().collect(Collectors.groupingBy(Apple::getId));

五、List 以ID分组 Map<Integer,List>，并对list中某一列做特殊处理

?

1 2 3 4 5 6 7

Map<Integer, List<Apple>> appleMap = appleList.stream()                         .peek(t -> {                             t.setSku(t.getSku().toUpperCase());  //sku转大写                             t.setQty(Objects.isNull(t.getQty) ? BigDecimal.ZERO : t.getQty);  //数量为null则设为0                             t.setPrice(Optional.ofNullable(t.getPrice).orElse(BigDecimal.ZERO));  //价格非null取当前值 如果是null则取0                         })                         .collect(Collectors.groupingBy(Apple::getId));