使用 Comparable 在 Lambda 中处理 java 8 Collection Sort 方法中的空检查

Question

我有一个 Member 的列表，如果我有空值，需要在 java 8 collection.sort 方法中对 Person firstName 进行空检查处理。目前我有 Member 实现 Comparable 接口以便 compareTo 另一个成员的全名。我从 TimSort binarySort 得到一个 NullPointerException。好像是 bug in OpenJDK. Found a solution to the NullPointerException in this question。上述问题的任何推荐解决方案以及我在下面提到的内容。

主程序

import java.util.Collections;
import java.util.List;

public class SortMembersByFirstName {
    public static void main(String[] args) {
        List<Member> members = getMembers();
        List<Member> sorted = setMemberCacheOrderByFirstName(members);
        for (Member mem: sorted) {
            System.out.println(mem.toString());
        }
    }

    private static List<Member> getMembers() {
        Person tim = new Person("Tim", "Southee");
        Member member1 = new Member(tim);
        Person eoin = new Person("Eoin", "Morgan");
        Member member2 = new Member(eoin);
        Person moeenAli = new Person("Moeen", "Ali");
        Member member3 = new Member(moeenAli);
        Person timFirstNameNull = new Person(null, "Tim");
        Member member4 = new Member(timFirstNameNull);
        Person timFirstNameLastNameNull = new Person(null, null);
        Member member5 = new Member(timFirstNameLastNameNull);
        Person nullPerson = null;
        Member member6 = new Member(nullPerson);

        List<Member> members = new ArrayList<>();
        members.add(member1);
        members.add(member2);
        members.add(member3);
        members.add(member4);
        members.add(member5);
        members.add(member6);
        return members;
    }

    /**
     * Order list members by first name
     */
    private static List<Member> setMemberCacheOrderByFirstName(List<Member> members) {
        // TimSort.binarySort NPE
        Collections.sort(members, (m1, m2) ->
            m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName())
        );
        return members;
    }
}

会员class

class Member implements Comparable<Member>{
    private Person person;
    public Member(Person person){
        this.person = person;
    }
    public Person getPerson(){
        return this.person;
    }
    public void setPerson(Person person){
        this.person = person;
    }
    public String getFullName(){
        return this.getPerson() != null ? this.getPerson().getFirstName() + " " + this.getPerson().getLastName() : "";
    }
    public int compareTo(Member o) {
        return this.getFullName().compareTo(o.getFullName());
    }

    @Override
    public String toString() {
        return "Member{" +
                "person=" + person +
                '}';
    }
}

人class

class Person{
    private String firstName;
    private String lastName;
    public String getFirstName(){
        return this.firstName;
    }
    public String getLastName(){
        return this.firstName;
    }
    public void setFirstName(String firstName){
        this.firstName = firstName;
    }

    @Override
    public String toString() {
        return "Person{" +
                "firstName='" + firstName + '\'' +
                ", lastName='" + lastName + '\'' +
                '}';
    }

    public Person(String firstname, String lastname){
        this.firstName = firstname;
        this.lastName= lastname;
    }

}

一个解决方案是捕获抛出的空指针异常。像这样

private static List<Member> setMemberCacheOrderByFirstName(List<Member> members) {
        Collections.sort(members, (m1, m2) -> {
            try {
                return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
            }catch (NullPointerException e){
                return 0;
            }
        });
        return members;
    }

或者手动检查是否存在不理想的空值。

private static List<Member> setMemberCacheOrderByFirstName(List<Member> members) {
        Collections.sort(members, (m1, m2) -> {
            // sort the list for firstnames
            if (m1.getPerson() != null && m2.getPerson() != null && m1.getPerson().getFirstName() != null
                    && m2.getPerson().getFirstName() != null) {
                return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
            } else {
                // Assuming any firstname is null
                if (m1.getPerson().getFirstName() == null || m2.getPerson().getFirstName() == null
                        && (m1.getPerson() != null && m2.getPerson() != null)) {
                    if (m1.getPerson().getFirstName() == null) {
                        m1.getPerson().setFirstName("");
                        return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
                    }
                    m2.getPerson().setFirstName("");
                    return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
                }
                // Assuming any member or person is null from schedulers
                return "".compareTo("");
            }
        });
        return members;
    }

Answer 1

使用 Comparator.nullsFirst 或 Comparator.nullsLast 将是正确的选择。

如果您需要 first 处的空值。你可以使用这个

Collections.sort(members, (m1, m2) -> {
            if(m1.getPerson() == null || m2.getPerson() == null ||
                    m1.getPerson().getFirstName() == null || m2.getPerson().getFirstName() == null){
                return -1;
            }

            if(m1.getPerson() != null && m2.getPerson() != null &&
                    m1.getPerson().getFirstName() != null && m2.getPerson().getFirstName() != null){
                return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
            }
            return 0;
        });

如果您需要 last 处的空值。你可以使用这个

Collections.sort(members, (m1, m2) -> {
            if(m1.getPerson() == null && m2.getPerson() == null &&
                    m1.getPerson().getFirstName() == null && m2.getPerson().getFirstName() == null){
                return -1;
            }

            if(m1.getPerson() != null && m2.getPerson() != null &&
                    m1.getPerson().getFirstName() != null && m2.getPerson().getFirstName() != null){
                return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
            }
            return 0;
        });

优化的方法是。

Collections.sort(members, (m1, m2) -> {
            if (m1.getPerson() == null || m2.getPerson() == null ||
                    m1.getPerson().getFirstName() == null ||
                    m2.getPerson().getFirstName() == null) {
                return -1;
            } else {
                return m1.getPerson().getFirstName().compareTo(m2.getPerson().getFirstName());
            }
        });

使用 Comparable 在 Lambda 中处理 java 8 Collection Sort 方法中的空检查

Handling null checks in java 8 Collection Sort method within a Lambda using Comparable

java

collections

lambda