假设你有一些对象，它们有几个字段可以比较:
public class Person {
private String firstName;
private String lastName;
private String age;
/* Constructors */
/* Methods */
}
所以在这个例子中，当你问if:
a.compareTo(b) > 0
你可能会问a的姓是不是在b的姓之前，或者a的年龄是不是比b大，等等……
在不增加不必要的混乱或开销的情况下，在这些类型的对象之间进行多重比较的最干净的方法是什么?
comparable接口只允许通过一个字段进行比较 在我看来，添加大量的比较方法(如compareByFirstName()， compareByAge()等)是混乱的。
那么最好的解决办法是什么呢?
当前回答
使用java8:
Comparator.comparing((Person p)->p.firstName)
.thenComparing(p->p.lastName)
.thenComparingInt(p->p.age);
如果你有访问方法:
Comparator.comparing(Person::getFirstName)
.thenComparing(Person::getLastName)
.thenComparingInt(Person::getAge);
如果一个类实现Comparable，那么这个比较器可以用在compareTo方法中:
@Override
public int compareTo(Person o){
return Comparator.comparing(Person::getFirstName)
.thenComparing(Person::getLastName)
.thenComparingInt(Person::getAge)
.compare(this, o);
}
其他回答
在java中使用hashcode方法比较两个对象很容易。
public class Sample{
String a=null;
String b=null;
public Sample(){
a="s";
b="a";
}
public Sample(String a,String b){
this.a=a;
this.b=b;
}
public static void main(String args[]){
Sample f=new Sample("b","12");
Sample s=new Sample("b","12");
//will return true
System.out.println((s.a.hashCode()+s.b.hashCode())==(f.a.hashCode()+f.b.hashCode()));
//will return false
Sample f=new Sample("b","12");
Sample s=new Sample("b","13");
System.out.println((s.a.hashCode()+s.b.hashCode())==(f.a.hashCode()+f.b.hashCode()));
}
Java 8通过lambda方式我们可以通过方法引用进行比较。 学生POJO
public class Student {
int id;
String firstName;
String lastName;
String subject;
public Student(int id, String firstName, String lastName, String subject) {
this.id = id;
this.firstName = firstName;
this.lastName = lastName;
this.subject = subject;
}
enter code here
现在我们可以根据
1. id - > FirstName - > LastName - > 2。主题- > id - > FirstName - > LastName
我们将在数组Stream中使用Comparator
public class TestComprator {
public static void main(String[] args) {
Student s1= new Student(108, "James", "Testo", "Physics");
Student s2= new Student(101, "Fundu", "Barito", "Chem");
Student s3= new Student(105, "Sindhu", "Sharan", "Math");
Student s4= new Student(98, "Rechel", "Stephen", "Physics");
System.out.printf("----------id->FirstName->LastName->Subject-------------");
Arrays.asList(s1,s2,s3,s4).stream()
.sorted(Comparator.comparing(Student::getId)
.thenComparing(Student::getFirstName)
.thenComparing(Student::getLastName)
.thenComparing(Student::getSubject))
.forEach(System.out::println);
System.out.printf("----Subject->id->FirstName->LastName ------\n");
Arrays.asList(s1,s2,s3,s4).stream()
.sorted(Comparator. comparing(Student::getSubject)
.thenComparing(Student::getId)
.thenComparing(Student::getFirstName)
.thenComparing(Student::getLastName)
)
.forEach(System.out::println);
}
}
输出:
`----------id->FirstName->LastName->Subject-------------
Student{id=98, firstName='Rechel', lastName='Stephen', subject='Physics'}
Student{id=101, firstName='Fundu', lastName='Barito', subject='Chem'}
Student{id=105, firstName='Sindhu', lastName='Sharan', subject='Math'}
Student{id=108, firstName='James', lastName='Testo', subject='Physics'}
----Subject->id->FirstName->LastName ------
Student{id=101, firstName='Fundu', lastName='Barito', subject='Chem'}
Student{id=105, firstName='Sindhu', lastName='Sharan', subject='Math'}
Student{id=98, firstName='Rechel', lastName='Stephen', subject='Physics'}
Student{id=108, firstName='James', lastName='Testo', subject='Physics'}
我认为如果你的比较算法是“聪明的”，那就更令人困惑了。我会选择你建议的众多比较方法。
对我来说唯一的例外就是平等。对于单元测试，重写. equals(在.net中)对我来说很有用，以便确定两个对象之间的几个字段是否相等(而不是引用是否相等)。
对于那些能够使用Java 8流API的人来说，这里有一个更整洁的方法: lambda和排序
我正在寻找相当于c# LINQ:
.ThenBy(...)
我在Comparator上找到了Java 8的机制:
.thenComparing(...)
下面是演示算法的代码片段。
Comparator<Person> comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
请查看上面的链接，以获得更简洁的方法，并解释Java的类型推断如何使其与LINQ相比定义起来更笨拙。
下面是完整的单元测试供参考:
@Test
public void testChainedSorting()
{
// Create the collection of people:
ArrayList<Person> people = new ArrayList<>();
people.add(new Person("Dan", 4));
people.add(new Person("Andi", 2));
people.add(new Person("Bob", 42));
people.add(new Person("Debby", 3));
people.add(new Person("Bob", 72));
people.add(new Person("Barry", 20));
people.add(new Person("Cathy", 40));
people.add(new Person("Bob", 40));
people.add(new Person("Barry", 50));
// Define chained comparators:
// Great article explaining this and how to make it even neater:
// http://blog.jooq.org/2014/01/31/java-8-friday-goodies-lambdas-and-sorting/
Comparator<Person> comparator = Comparator.comparing(person -> person.name);
comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));
// Sort the stream:
Stream<Person> personStream = people.stream().sorted(comparator);
// Make sure that the output is as expected:
List<Person> sortedPeople = personStream.collect(Collectors.toList());
Assert.assertEquals("Andi", sortedPeople.get(0).name); Assert.assertEquals(2, sortedPeople.get(0).age);
Assert.assertEquals("Barry", sortedPeople.get(1).name); Assert.assertEquals(20, sortedPeople.get(1).age);
Assert.assertEquals("Barry", sortedPeople.get(2).name); Assert.assertEquals(50, sortedPeople.get(2).age);
Assert.assertEquals("Bob", sortedPeople.get(3).name); Assert.assertEquals(40, sortedPeople.get(3).age);
Assert.assertEquals("Bob", sortedPeople.get(4).name); Assert.assertEquals(42, sortedPeople.get(4).age);
Assert.assertEquals("Bob", sortedPeople.get(5).name); Assert.assertEquals(72, sortedPeople.get(5).age);
Assert.assertEquals("Cathy", sortedPeople.get(6).name); Assert.assertEquals(40, sortedPeople.get(6).age);
Assert.assertEquals("Dan", sortedPeople.get(7).name); Assert.assertEquals(4, sortedPeople.get(7).age);
Assert.assertEquals("Debby", sortedPeople.get(8).name); Assert.assertEquals(3, sortedPeople.get(8).age);
// Andi : 2
// Barry : 20
// Barry : 50
// Bob : 40
// Bob : 42
// Bob : 72
// Cathy : 40
// Dan : 4
// Debby : 3
}
/**
* A person in our system.
*/
public static class Person
{
/**
* Creates a new person.
* @param name The name of the person.
* @param age The age of the person.
*/
public Person(String name, int age)
{
this.age = age;
this.name = name;
}
/**
* The name of the person.
*/
public String name;
/**
* The age of the person.
*/
public int age;
@Override
public String toString()
{
if (name == null) return super.toString();
else return String.format("%s : %d", this.name, this.age);
}
}
晚做总比不到好——如果你正在寻找不必要的混乱或开销，那么很难同时在最少的代码/快速执行方面击败下面的方法。
数据类:
public class MyData {
int id;
boolean relevant;
String name;
float value;
}
比较器:
public class MultiFieldComparator implements Comparator<MyData> {
@Override
public int compare(MyData dataA, MyData dataB) {
int result;
if((result = Integer.compare(dataA.id, dataB.id)) == 0 &&
(result = Boolean.compare(dataA.relevant, dataB.relevant)) == 0 &&
(result = dataA.name.compareTo(dataB.name)) == 0)
result = Float.compare(dataA.value, dataB.value);
return result;
}
}
如果你只是想按自定义顺序对集合进行排序，那么下面的代码就更清晰了:
myDataList.sort((dataA, dataB) -> {
int result;
if((result = Integer.compare(dataA.id, dataB.id)) == 0 &&
(result = Boolean.compare(dataA.relevant, dataB.relevant)) == 0 &&
(result = dataA.name.compareTo(dataB.name)) == 0)
result = Float.compare(dataA.value, dataB.value);
return result;
});
