Question

基于https://stackoverflow.com/a/29671501/2517622的问题

给出具有ID，姓名和智商的员工列表：

List<Employee> employee = Arrays.asList(new Employee(1, "John", 80), new Employee(1, "Bob", 120), Employee(1, "Roy", 60), new Employee(2, "Alice", 100));

我要输出：

[Employee{id=1, name='Bob', iq=120}, Employee{id=2, name='Alice', iq=100}]

因此，出于显而易见的原因，请根据雇员的id属性从列表中删除重复项，并选择智商最高的雇员。：）

特别是，我有兴趣调整此解决方案，该解决方案仅根据id删除重复项：

    import static java.util.Comparator.comparingInt;
    import static java.util.stream.Collectors.collectingAndThen;
    import static java.util.stream.Collectors.toCollection;

    ...
    List<Employee> unique = employee.stream()
                                    .collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparingInt(Employee::getId))),
                                                               ArrayList::new));

有办法吗？

Answer 1

怎么样，

Collection<Employee> distinctEmps = employee.stream()
    .collect(Collectors.toMap(Employee::getId, Function.identity(), 
        (e1, e2) -> e1.getIq() >= e2.getIq() ? e1 : e2))
    .values();

仅遵循@Holgers方法的另一个变体是

Collection<Employee> distinctEmps = employee.stream()
    .collect(Collectors.toMap(Employee::getId, Function.identity(), 
        BinaryOperator.maxBy(Comparator.comparing(Employee::getIq))))
    .values();

根据属性和Java 8中的谓词删除重复项

1 个答案: