Java 8 lambda过滤基于条件和顺序

Question

我试图根据多个条件筛选列表,排序.

class Student{
        private int Age;
        private String className;
        private String Name;

        public Student(int age, String className, String name) {
            Age = age;
            this.className = className;
            Name = name;
        }

        public int getAge() {
            return Age;
        }

        public void setAge(int age) {
            Age = age;
        }

        public String getClassName() {
            return className;
        }

        public void setClassName(String className) {
            this.className = className;
        }

        public String getName() {
            return Name;
        }

        public void setName(String name) {
            Name = name;
        }
    }

现在,如果我有一个清单,请说

List<Student> students = new ArrayList<>();
        students.add(new Student(24, "A", "Smith"));
        students.add(new Student(24, "A", "John"));
        students.add(new Student(30, "A", "John"));
        students.add(new Student(20, "B", "John"));
        students.add(new Student(24, "B", "Prince"));

我怎样才能获得具有不同名称的最年长学生的名单？在C#中,通过使用System.Linq GroupBy然后比较然后使用select进行展平,这将非常简单,我不太确定如何在Java中实现相同的功能.

Answer 1

使用toMap收藏家:

Collection<Student> values = students.stream()
                .collect(toMap(Student::getName,
                        Function.identity(),
                        BinaryOperator.maxBy(Comparator.comparingInt(Student::getAge))))
                .values();

说明

我们正在使用这个重载toMap:

toMap?(Function<? super T,? extends K> keyMapper,
      Function<? super T,? extends U> valueMapper,
      BinaryOperator<U> mergeFunction)

• Student::getName上面是keyMapper用于提取地图键值的函数.
• Function.identity()上面是valueMapper用于提取地图值的值的函数,其中Function.identity()简单地返回源自己的元素即Student对象.
• BinaryOperator.maxBy(Comparator.comparingInt(Student::getAge))上面是合并函数,用于"决定在关键合并的情况下返回哪个学生对象,即当两个给定的学生具有相同的名称时",在这种情况下取​​最旧的Student.
• 最后,调用values()返回给我们一组学生.

等效的C#代码是:

var values = students.GroupBy(s => s.Name, v => v,
                          (a, b) => b.OrderByDescending(e => e.Age).Take(1))
                      .SelectMany(x => x);

解释(对于那些不熟悉.NET的人)

我们正在使用以下扩展方法GroupBy:

System.Collections.Generic.IEnumerable<TResult> GroupBy<TSource,TKey,TElement,TResult> 
       (this System.Collections.Generic.IEnumerable<TSource> source, 
         Func<TSource,TKey> keySelector, 
         Func<TSource,TElement> elementSelector, 
     Func<TKey,System.Collections.Generic.IEnumerable<TElement>,TResult> resultSelector);

• s => s.Name以上是keySelector用于将值提取到group by 的函数.
• v => v上面是elementSelector用于提取值的函数,即Student它们自己的对象.
• b.OrderByDescending(e => e.Age).Take(1)上面是resultSelector给出一个IEnumerable<Student>代表b最年长的学生.
• 最后,我们申请.SelectMany(x => x);将结果折叠IEnumerable<IEnumerable<Student>>成一个IEnumerable<Student>.

Answer 2

或者没有溪流:

Map<String, Student> map = new HashMap<>();
students.forEach(x -> map.merge(x.getName(), x, (oldV, newV) -> oldV.getAge() > newV.getAge() ? oldV : newV));
Collection<Student> max = map.values();