Java的8个流地图<整数，列表<String>>

Question

这是Java对象在Java中处理8个流

private static List<Person> getPersons() { 
     List<Person> results = new ArrayList<Person>(); 

     results.add(new Person("Elsdon", "Jaycob", "Java programmer", "male", 43, 2000)); 
     results.add(new Person("Tamsen", "Brittany", "Java programmer", "female", 23, 1500)); 
     results.add(new Person("Floyd", "Donny", "Java programmer", "male", 33, 1800)); 
     results.add(new Person("Sindy", "Jonie", "Java programmer", "female", 32, 1600)); 
     results.add(new Person("Vere", "Hervey", "Java programmer", "male", 22, 1200)); 
     results.add(new Person("Maude", "Jaimie", "Java programmer", "female", 27, 1900)); 
     results.add(new Person("Shawn", "Randall", "Java programmer", "male", 30, 2300)); 
     results.add(new Person("Jayden", "Corrina", "Java programmer", "female", 35, 1700)); 
     results.add(new Person("Palmer", "Dene", "Java programmer", "male", 33, 2000)); 
     results.add(new Person("Addison", "Pam", "Java programmer", "female", 34, 1300)); 

     results.add(new Person("Jarrod", "Pace", "PHP programmer", "male", 34, 1550)); 
     results.add(new Person("Clarette", "Cicely", "PHP programmer", "female", 23, 1200)); 
     results.add(new Person("Victor", "Channing", "PHP programmer", "male", 32, 1600)); 
     results.add(new Person("Tori", "Sheryl", "PHP programmer", "female", 21, 1000)); 
     results.add(new Person("Osborne", "Shad", "PHP programmer", "male", 32, 1100)); 
     results.add(new Person("Rosalind", "Layla", "PHP programmer", "female", 25, 1300)); 
     results.add(new Person("Fraser", "Hewie", "PHP programmer", "male", 36, 1100)); 
     results.add(new Person("Quinn", "Tamara", "PHP programmer", "female", 21, 1000)); 
     results.add(new Person("Alvin", "Lance", "PHP programmer", "male", 38, 1600)); 
     results.add(new Person("Evonne", "Shari", "PHP programmer", "female", 40, 1800)); 

     return results; 
    } 

我生成键值“年龄” Map对象。这是码

Function<Person, Map<String, String>> nameNsalary = (Person p) -> { 
       Map<Int, String> map = new HashMap<Int,String>(); 

       if (p.getAge()>=40) 
        map.put(40 , p.getFirstName() + " " + p.getLastName() +":" + p.getSalary()); 
       else if(p.getAge()>=30 && p.getAge()<40) 
        map.put(30 , p.getFirstName() + " " + p.getLastName() +":" + p.getSalary()); 
       else if(p.getAge()<20) 
        map.put(20 , p.getFirstName() + " " + p.getLastName() +":" + p.getSalary()); 
       else 
        map.put(10 , p.getFirstName() + " " + p.getLastName() +":" + p.getSalary()); 

       return map; 
      }; 

persons.stream().map(nameNsalary).forEach(System.out::println); 

输出地图对象是成功的像下面，

{40=Elsdon Jaycob:2000} 
{20=Tamsen Brittany:1500} 
{30=Floyd Donny:1800} 
{30=Sindy Jonie:1600} 
{20=Vere Hervey:1200} 
{20=Maude Jaimie:1900} 
{30=Shawn Randall:2300} 
{30=Jayden Corrina:1700} 
{30=Palmer Dene:2000} 
{30=Addison Pam:1300} 
{30=Jarrod Pace:1625} 
{20=Clarette Cicely:1260} 
{30=Victor Channing:1680} 
{20=Tori Sheryl:1050} 
{30=Osborne Shad:1155} 
{20=Rosalind Layla:1365} 
{30=Fraser Hewie:1155} 
{20=Quinn Tamara:1050} 
{30=Alvin Lance:1680} 
{40=Evonne Shari:1890} 

但我不改变输出类型从Map<Integer, String>到Map<Integer, List<String>>，例如，

30 = [Jayden Corrina:1700, Palmer Dene:2000, Addison Pam:1300] 

我不知道如何处理这些Map对象数据到Map<Int, List<String>>。我做了这个代码，

Map<Integer, List<String>> result = persons.stream().map(nameNsalary) 
       .collect(Collectors.toMap(mapper -> (Integer)mapper.getKey(), mapper -> mapper.getValue())); 

但这代码引发以下异常

Exception in thread "main" java.lang.Error: Unresolved compilation problems: 
    Type mismatch: cannot convert from Map<Object,Object> to Map<Integer,List<String>> 
    The method getKey() is undefined for the type Map<Integer,String> 
    The method getValue() is undefined for the type Map<Integer,String> 

我坚持这个程序。

Answer 1

您的映射器效率很低，它会为每个条目创建一个新映射。

使用groupingBy收集器会更简单。

Map<Integer, List<String>> result = persons.stream() 
     .collect(Collectors.groupingBy(
      p -> (p.getAge()/10) * 10, // some integer division trick 
      Collectors.mapping(
       p -> p.getFirstName() + " " + p.getLastName() +":" + p.getSalary(), 
       Collectors.toList() 
      ) 
     )); 

将返回：

20 = [Tamsen Brittany:1500, Vere Hervey:1200, Maude Jaimie:1900, Clarette Cicely:1200, Tori Sheryl:1000, Rosalind Layla:1300, Quinn Tamara:1000] 
40 = [Elsdon Jaycob:2000, Evonne Shari:1800] 
30 = [Floyd Donny:1800, Sindy Jonie:1600, Shawn Randall:2300, Jayden Corrina:1700, Palmer Dene:2000, Addison Pam:1300, Jarrod Pace:1550, Victor Channing:1600, Osborne Shad:1100, Fraser Hewie:1100, Alvin Lance:1600] 

---

你也可以放弃额外的映射步骤，并创建一个Map<Integer, List<Person>>代替：

Map<Integer, List<Person>> result = persons.stream() 
    .collect(Collectors.groupingBy(p -> (p.getAge()/10) * 10));