有很多方法，这一个也会有帮助-简单，干净和清晰

    List<Employee> employees = new ArrayList<>();

    employees.add(new Employee(11, "Ravi"));
    employees.add(new Employee(12, "Stalin"));
    employees.add(new Employee(23, "Anbu"));
    employees.add(new Employee(24, "Yuvaraj"));
    employees.add(new Employee(35, "Sena"));
    employees.add(new Employee(36, "Antony"));
    employees.add(new Employee(47, "Sena"));
    employees.add(new Employee(48, "Ravi"));

    List<Employee> empList = new ArrayList<>(employees.stream().collect(
                    Collectors.toMap(Employee::getName, obj -> obj,
                    (existingValue, newValue) -> existingValue))
                   .values());

    empList.forEach(System.out::println);


    //  Collectors.toMap(
    //  Employee::getName, - key (the value by which you want to eliminate duplicate)
    //  obj -> obj,  - value (entire employee object)
    //  (existingValue, newValue) -> existingValue) - to avoid illegalstateexception: duplicate key

Output - toString()重载

Employee{id=35, name='Sena'}
Employee{id=12, name='Stalin'}
Employee{id=11, name='Ravi'}
Employee{id=24, name='Yuvaraj'}
Employee{id=36, name='Antony'}
Employee{id=23, name='Anbu'}

我做了一个通用版本:

private <T, R> Collector<T, ?, Stream<T>> distinctByKey(Function<T, R> keyExtractor) {
    return Collectors.collectingAndThen(
            toMap(
                    keyExtractor,
                    t -> t,
                    (t1, t2) -> t1
            ),
            (Map<R, T> map) -> map.values().stream()
    );
}

每年的例子:

Stream.of(new Person("Jean"), 
          new Person("Jean"),
          new Person("Paul")
)
    .filter(...)
    .collect(distinctByKey(Person::getName)) // return a stream of Person with 2 elements, jean and Paul
    .map(...)
    .collect(toList())

您可以将person对象包装到另一个类中，该类只比较person的名称。之后，您将打开被包装的对象以再次获得人员流。流操作可能如下所示:

persons.stream()
    .map(Wrapper::new)
    .distinct()
    .map(Wrapper::unwrap)
    ...;

类Wrapper可能看起来如下所示:

class Wrapper {
    private final Person person;
    public Wrapper(Person person) {
        this.person = person;
    }
    public Person unwrap() {
        return person;
    }
    public boolean equals(Object other) {
        if (other instanceof Wrapper) {
            return ((Wrapper) other).person.getName().equals(person.getName());
        } else {
            return false;
        }
    }
    public int hashCode() {
        return person.getName().hashCode();
    }
}

扩展Stuart Marks的回答，这可以用更短的方式完成，不需要并发映射(如果你不需要并行流):

public static <T> Predicate<T> distinctByKey(Function<? super T, ?> keyExtractor) {
    final Set<Object> seen = new HashSet<>();
    return t -> seen.add(keyExtractor.apply(t));
}

然后调用:

persons.stream().filter(distinctByKey(p -> p.getName());

我们也可以使用RxJava(非常强大的响应式扩展库)

Observable.from(persons).distinct(Person::getName)

or

Observable.from(persons).distinct(p -> p.getName())

你可以使用StreamEx库:

StreamEx.of(persons)
        .distinct(Person::getName)
        .toList()