在Java 8中,我如何使用流API通过检查每个对象的属性的清晰度来过滤一个集合?

例如,我有一个Person对象列表,我想删除同名的人,

persons.stream().distinct();

将对Person对象使用默认的相等性检查,所以我需要这样的东西,

persons.stream().distinct(p -> p.getName());

不幸的是,distinct()方法没有这样的重载。如果不修改Person类内部的相等检查,是否可以简洁地做到这一点?


当前回答

Here is the example
public class PayRoll {

    private int payRollId;
    private int id;
    private String name;
    private String dept;
    private int salary;


    public PayRoll(int payRollId, int id, String name, String dept, int salary) {
        super();
        this.payRollId = payRollId;
        this.id = id;
        this.name = name;
        this.dept = dept;
        this.salary = salary;
    }
} 

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.Optional;
import java.util.stream.Collector;
import java.util.stream.Collectors;

public class Prac {
    public static void main(String[] args) {

        int salary=70000;
        PayRoll payRoll=new PayRoll(1311, 1, "A", "HR", salary);
        PayRoll payRoll2=new PayRoll(1411, 2    , "B", "Technical", salary);
        PayRoll payRoll3=new PayRoll(1511, 1, "C", "HR", salary);
        PayRoll payRoll4=new PayRoll(1611, 1, "D", "Technical", salary);
        PayRoll payRoll5=new PayRoll(711, 3,"E", "Technical", salary);
        PayRoll payRoll6=new PayRoll(1811, 3, "F", "Technical", salary);
        List<PayRoll>list=new ArrayList<PayRoll>();
        list.add(payRoll);
        list.add(payRoll2);
        list.add(payRoll3);
        list.add(payRoll4);
        list.add(payRoll5);
        list.add(payRoll6);


        Map<Object, Optional<PayRoll>> k = list.stream().collect(Collectors.groupingBy(p->p.getId()+"|"+p.getDept(),Collectors.maxBy(Comparator.comparingInt(PayRoll::getPayRollId))));


        k.entrySet().forEach(p->
        {
            if(p.getValue().isPresent())
            {
                System.out.println(p.getValue().get());
            }
        });



    }
}

Output:

PayRoll [payRollId=1611, id=1, name=D, dept=Technical, salary=70000]
PayRoll [payRollId=1811, id=3, name=F, dept=Technical, salary=70000]
PayRoll [payRollId=1411, id=2, name=B, dept=Technical, salary=70000]
PayRoll [payRollId=1511, id=1, name=C, dept=HR, salary=70000]

其他回答

Here is the example
public class PayRoll {

    private int payRollId;
    private int id;
    private String name;
    private String dept;
    private int salary;


    public PayRoll(int payRollId, int id, String name, String dept, int salary) {
        super();
        this.payRollId = payRollId;
        this.id = id;
        this.name = name;
        this.dept = dept;
        this.salary = salary;
    }
} 

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.Optional;
import java.util.stream.Collector;
import java.util.stream.Collectors;

public class Prac {
    public static void main(String[] args) {

        int salary=70000;
        PayRoll payRoll=new PayRoll(1311, 1, "A", "HR", salary);
        PayRoll payRoll2=new PayRoll(1411, 2    , "B", "Technical", salary);
        PayRoll payRoll3=new PayRoll(1511, 1, "C", "HR", salary);
        PayRoll payRoll4=new PayRoll(1611, 1, "D", "Technical", salary);
        PayRoll payRoll5=new PayRoll(711, 3,"E", "Technical", salary);
        PayRoll payRoll6=new PayRoll(1811, 3, "F", "Technical", salary);
        List<PayRoll>list=new ArrayList<PayRoll>();
        list.add(payRoll);
        list.add(payRoll2);
        list.add(payRoll3);
        list.add(payRoll4);
        list.add(payRoll5);
        list.add(payRoll6);


        Map<Object, Optional<PayRoll>> k = list.stream().collect(Collectors.groupingBy(p->p.getId()+"|"+p.getDept(),Collectors.maxBy(Comparator.comparingInt(PayRoll::getPayRollId))));


        k.entrySet().forEach(p->
        {
            if(p.getValue().isPresent())
            {
                System.out.println(p.getValue().get());
            }
        });



    }
}

Output:

PayRoll [payRollId=1611, id=1, name=D, dept=Technical, salary=70000]
PayRoll [payRollId=1811, id=3, name=F, dept=Technical, salary=70000]
PayRoll [payRollId=1411, id=2, name=B, dept=Technical, salary=70000]
PayRoll [payRollId=1511, id=1, name=C, dept=HR, salary=70000]

您可以在Eclipse Collections中使用distinct(HashingStrategy)方法。

List<Person> persons = ...;
MutableList<Person> distinct =
    ListIterate.distinct(persons, HashingStrategies.fromFunction(Person::getName));

如果可以重构人员以实现Eclipse Collections接口,则可以直接调用列表上的方法。

MutableList<Person> persons = ...;
MutableList<Person> distinct =
    persons.distinct(HashingStrategies.fromFunction(Person::getName));

HashingStrategy只是一个策略接口,允许您定义equals和hashcode的自定义实现。

public interface HashingStrategy<E>
{
    int computeHashCode(E object);
    boolean equals(E object1, E object2);
}

注意:我是Eclipse Collections的提交者。

我在这个清单中的解决方案:

List<HolderEntry> result ....

List<HolderEntry> dto3s = new ArrayList<>(result.stream().collect(toMap(
            HolderEntry::getId,
            holder -> holder,  //or Function.identity() if you want
            (holder1, holder2) -> holder1 
    )).values());

在我的情况下,我想找到不同的值,并把它们放在列表。

也许会对某人有用。我还有一个要求。从第三方对象A列表中删除所有具有相同A.b字段的相同A.id的对象(列表中具有相同A.id的多个A对象)。流分区的答案由Tagir Valeev启发我使用自定义收集器返回Map<A。id列表< > >。简单的flatMap将完成其余的工作。

 public static <T, K, K2> Collector<T, ?, Map<K, List<T>>> groupingDistinctBy(Function<T, K> keyFunction, Function<T, K2> distinctFunction) {
    return groupingBy(keyFunction, Collector.of((Supplier<Map<K2, T>>) HashMap::new,
            (map, error) -> map.putIfAbsent(distinctFunction.apply(error), error),
            (left, right) -> {
                left.putAll(right);
                return left;
            }, map -> new ArrayList<>(map.values()),
            Collector.Characteristics.UNORDERED)); }

如果你想要名单,下面是最简单的方法

Set<String> set = new HashSet<>(persons.size());
persons.stream().filter(p -> set.add(p.getName())).collect(Collectors.toList());

此外,如果您想要查找不同的或唯一的名称列表,而不是Person,您也可以使用以下两个方法。

方法一:使用区别

persons.stream().map(x->x.getName()).distinct.collect(Collectors.toList());

方法二:使用HashSet

Set<E> set = new HashSet<>();
set.addAll(person.stream().map(x->x.getName()).collect(Collectors.toList()));