如何在Java中按两个字段排序？

Question

我有对象数组person (int age; String name;)。

如何按字母顺序按名称排序，然后按年龄排序？

您会使用哪种算法？

Answer 1

您可以按如下方式使用Collections.sort：

private static void order(List<Person> persons) {

    Collections.sort(persons, new Comparator() {

        public int compare(Object o1, Object o2) {

            String x1 = ((Person) o1).getName();
            String x2 = ((Person) o2).getName();
            int sComp = x1.compareTo(x2);

            if (sComp != 0) {
               return sComp;
            } 

            Integer x1 = ((Person) o1).getAge();
            Integer x2 = ((Person) o2).getAge();
            return x1.compareTo(x2);
    }});
}

List<Persons>现按名称排序，然后按年龄排序。

String.compareTo“按字典顺序比较两个字符串” - 来自docs。

Collections.sort是本机Collections库中的静态方法。它进行实际的排序，你只需要提供一个Comparator来定义列表中两个元素的比较方式：这是通过提供你自己的compare方法实现来实现的。

Answer 2

对于那些能够使用Java 8流API的人来说，有一种更简洁的方法，这里有详细记录： Lambdas and sorting

我正在寻找相当于C＃LINQ：

.ThenBy(...)

我在比较器上找到了Java 8中的机制：

.thenComparing(...)

所以这是演示算法的片段。

    Comparator<Person> comparator = Comparator.comparing(person -> person.name);
    comparator = comparator.thenComparing(Comparator.comparing(person -> person.age));

查看上面的链接，了解更简洁的方法以及有关Java类型推断如何使其与LINQ相比更加笨拙的解释。

以下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);
    }
}

Answer 3

使用Java 8 Streams方法......

//Creates and sorts a stream (does not sort the original list)       
persons.stream().sorted(Comparator.comparing(Person::getName).thenComparing(Person::getAge));

Java 8 Lambda方法......

//Sorts the original list Lambda style
persons.sort((p1, p2) -> {
        if (p1.getName().compareTo(p2.getName()) == 0) {
            return p1.getAge().compareTo(p2.getAge());
        } else {
            return p1.getName().compareTo(p2.getName());
        } 
    });

...最后

//This is similar SYNTAX to the Streams above, but it sorts the original list!!
persons.sort(Comparator.comparing(Person::getName).thenComparing(Person::getAge));

Answer 4

您需要实现自己的Comparator，然后使用它：例如

Arrays.sort(persons, new PersonComparator());

你的比较者看起来有点像这样：

public class PersonComparator implements Comparator<? extends Person> {

  public int compare(Person p1, Person p2) {
     int nameCompare = p1.name.compareToIgnoreCase(p2.name);
     if (nameCompare != 0) {
        return nameCompare;
     } else {
       return Integer.valueOf(p1.age).compareTo(Integer.valueOf(p2.age));
     }
  }
}

比较器首先比较名称，如果它们不相等则返回比较结果，否则在比较两个人的年龄时返回比较结果。

此代码只是一个草稿：因为该类是不可变的，您可以考虑构建它的单例，而不是为每个排序创建一个新实例。

Answer 5

让你的人员类实现Comparable<Person>，然后实现compareTo方法，例如：

public int compareTo(Person o) {
    int result = name.compareToIgnoreCase(o.name);
    if(result==0) {
        return Integer.valueOf(age).compareTo(o.age);
    }
    else {
        return result;
    }
}

首先按名称（不区分大小写）然后按年龄排序。然后，您可以在Person对象的集合或数组上运行Arrays.sort()或Collections.sort()。

Answer 6

Guava的ComparisonChain提供了一种干净的方式。请参阅此link。

用于执行链式比较语句的实用程序。例如：

   public int compareTo(Foo that) {
     return ComparisonChain.start()
         .compare(this.aString, that.aString)
         .compare(this.anInt, that.anInt)
         .compare(this.anEnum, that.anEnum, Ordering.natural().nullsLast())
         .result();
   }

Answer 7

使用Comparator，然后将对象放入Collection，然后Collections.sort();

class Person {

    String fname;
    String lname;
    int age;

    public Person() {
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getFname() {
        return fname;
    }

    public void setFname(String fname) {
        this.fname = fname;
    }

    public String getLname() {
        return lname;
    }

    public void setLname(String lname) {
        this.lname = lname;
    }

    public Person(String fname, String lname, int age) {
        this.fname = fname;
        this.lname = lname;
        this.age = age;
    }

    @Override
    public String toString() {
        return fname + "," + lname + "," + age;
    }
}

public class Main{

    public static void main(String[] args) {
        List<Person> persons = new java.util.ArrayList<Person>();
        persons.add(new Person("abc3", "def3", 10));
        persons.add(new Person("abc2", "def2", 32));
        persons.add(new Person("abc1", "def1", 65));
        persons.add(new Person("abc4", "def4", 10));
        System.out.println(persons);
        Collections.sort(persons, new Comparator<Person>() {

            @Override
            public int compare(Person t, Person t1) {
                return t.getAge() - t1.getAge();
            }
        });
        System.out.println(persons);

    }
}

Answer 8

根据需要创建尽可能多的比较器。之后，为每个订单类别调用方法“thenComparing”。这是Streams的一种做法。参见：

//Sort by first and last name
System.out.println("\n2.Sort list of person objects by firstName then "
                                        + "by lastName then by age");
Comparator<Person> sortByFirstName 
                            = (p, o) -> p.firstName.compareToIgnoreCase(o.firstName);
Comparator<Person> sortByLastName 
                            = (p, o) -> p.lastName.compareToIgnoreCase(o.lastName);
Comparator<Person> sortByAge 
                            = (p, o) -> Integer.compare(p.age,o.age);

//Sort by first Name then Sort by last name then sort by age
personList.stream().sorted(
    sortByFirstName
        .thenComparing(sortByLastName)
        .thenComparing(sortByAge)
     ).forEach(person->
        System.out.println(person));        

看：Sort user defined object on multiple fields – Comparator (lambda stream)

Answer 9

你可以这样做：

List<User> users = Lists.newArrayList(
  new User("Pedro", 12), 
  new User("Maria", 10), 
  new User("Rafael",12)
);

users.sort(
  Comparator.comparing(User::getName).thenComparing(User::getAge)
);

Answer 10

您可以使用Java 8 Lambda方法来实现此目的。 像这样：

persons.sort(Comparator.comparing(Person::getName).thenComparing(Person::getAge));

Answer 11

或者您可以利用Collections.sort()（或Arrays.sort()）稳定的事实（它不会重新排序相同的元素）并使用Comparator按年龄排序然后另一个按名称排序。

在这种特殊情况下，这不是一个好主意，但如果你必须能够在运行时更改排序顺序，那么它可能会有用。

Answer 12

您可以使用通用串行比较器按多个字段对集合进行排序。

import org.apache.commons.lang3.reflect.FieldUtils;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

/**
* @author MaheshRPM
*/
public class SerialComparator<T> implements Comparator<T> {
List<String> sortingFields;

public SerialComparator(List<String> sortingFields) {
    this.sortingFields = sortingFields;
}

public SerialComparator(String... sortingFields) {
    this.sortingFields = Arrays.asList(sortingFields);
}

@Override
public int compare(T o1, T o2) {
    int result = 0;
    try {
        for (String sortingField : sortingFields) {
            if (result == 0) {
                Object value1 = FieldUtils.readField(o1, sortingField, true);
                Object value2 = FieldUtils.readField(o2, sortingField, true);
                if (value1 instanceof Comparable && value2 instanceof Comparable) {
                    Comparable comparable1 = (Comparable) value1;
                    Comparable comparable2 = (Comparable) value2;
                    result = comparable1.compareTo(comparable2);
                } else {
                    throw new RuntimeException("Cannot compare non Comparable fields. " + value1.getClass()
                            .getName() + " must implement Comparable<" + value1.getClass().getName() + ">");
                }
            } else {
                break;
            }
        }
    } catch (IllegalAccessException e) {
        throw new RuntimeException(e);
    }
    return result;
}
}

Answer 13

使用番石榴的ComparisonChain时要小心，因为它会为每个被比较的元素创建一个实例，因此您将查看N x Log N比较链的创建，以比较是否要排序，或N个实例（如果您要迭代并检查是否相等）。

如果可能的话，我会使用最新的Java 8 API或使用允许您执行此操作的Guava的Comparator API创建静态Ordering，这是Java 8的示例：

import java.util.Comparator;
import static java.util.Comparator.naturalOrder;
import static java.util.Comparator.nullsLast;

private static final Comparator<Person> COMPARATOR = Comparator
  .comparing(Person::getName, nullsLast(naturalOrder()))
  .thenComparingInt(Person::getAge);

@Override
public int compareTo(@NotNull Person other) {
  return COMPARATOR.compare(this, other);
}

以下是如何使用番石榴的Ordering API：https://github.com/google/guava/wiki/OrderingExplained

Answer 14

Arrays.sort(persons, new PersonComparator());



import java.util.Comparator;

public class PersonComparator implements Comparator<? extends Person> {

    @Override
    public int compare(Person o1, Person o2) {
        if(null == o1 || null == o2  || null == o1.getName() || null== o2.getName() ){
            throw new NullPointerException();
        }else{
            int nameComparisonResult = o1.getName().compareTo(o2.getName());
            if(0 == nameComparisonResult){
                return o1.getAge()-o2.getAge();
            }else{
                return nameComparisonResult;
            }
        }
    }
}


class Person{
    int age; String name;

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

更新版本：

public class PersonComparator implements Comparator<? extends Person> {

   @Override
   public int compare(Person o1, Person o2) {

      int nameComparisonResult = o1.getName().compareToIgnoreCase(o2.getName());
      return 0 == nameComparisonResult?o1.getAge()-o2.getAge():nameComparisonResult;

   }
 }

Answer 15

对于这样的课程Book：

package books;

public class Book {

    private Integer id;
    private Integer number;
    private String name;

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public Integer getNumber() {
        return number;
    }

    public void setNumber(Integer number) {
        this.number = number;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public String toString() {
        return "book{" +
                "id=" + id +
                ", number=" + number +
                ", name='" + name + '\'' + '\n' +
                '}';
    }
}

使用模拟对象对主类进行排序

package books;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;


public class Main {

    public static void main(String[] args) {
        System.out.println("Hello World!");

        Book b = new Book();

        Book c = new Book();

        Book d = new Book();

        Book e = new Book();

        Book f = new Book();

        Book g = new Book();
        Book g1 = new Book();
        Book g2 = new Book();
        Book g3 = new Book();
        Book g4 = new Book();




        b.setId(1);
        b.setNumber(12);
        b.setName("gk");

        c.setId(2);
        c.setNumber(12);
        c.setName("gk");

        d.setId(2);
        d.setNumber(13);
        d.setName("maths");

        e.setId(3);
        e.setNumber(3);
        e.setName("geometry");

        f.setId(3);
        f.setNumber(34);
        b.setName("gk");

        g.setId(3);
        g.setNumber(11);
        g.setName("gk");

        g1.setId(3);
        g1.setNumber(88);
        g1.setName("gk");
        g2.setId(3);
        g2.setNumber(91);
        g2.setName("gk");
        g3.setId(3);
        g3.setNumber(101);
        g3.setName("gk");
        g4.setId(3);
        g4.setNumber(4);
        g4.setName("gk");





        List<Book> allBooks = new ArrayList<Book>();

        allBooks.add(b);
        allBooks.add(c);
        allBooks.add(d);
        allBooks.add(e);
        allBooks.add(f);
        allBooks.add(g);
        allBooks.add(g1);
        allBooks.add(g2);
        allBooks.add(g3);
        allBooks.add(g4);



        System.out.println(allBooks.size());


        Collections.sort(allBooks, new Comparator<Book>() {

            @Override
            public int compare(Book t, Book t1) {
                int a =  t.getId()- t1.getId();

                if(a == 0){
                    int a1 = t.getNumber() - t1.getNumber();
                    return a1;
                }
                else
                    return a;
            }
        });
        System.out.println(allBooks);

    }


   }

Answer 16

在这种情况下，我不确定在Person类中编写隔离区是否很难看。它是这样的：

public class Person implements Comparable <Person> {

    private String lastName;
    private String firstName;
    private int age;

    public Person(String firstName, String lastName, int BirthDay) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.age = BirthDay;
    }

    public int getAge() {
        return age;
    }

    public String getFirstName() {
        return firstName;
    }

    public String getLastName() {
        return lastName;
    }

    @Override
    public int compareTo(Person o) {
        // default compareTo
    }

    @Override
    public String toString() {
        return firstName + " " + lastName + " " + age + "";
    }

    public static class firstNameComperator implements Comparator<Person> {
        @Override
        public int compare(Person o1, Person o2) {
            return o1.firstName.compareTo(o2.firstName);
        }
    }

    public static class lastNameComperator implements Comparator<Person> {
        @Override
        public int compare(Person o1, Person o2) {
            return o1.lastName.compareTo(o2.lastName);
        }
    }

    public static class ageComperator implements Comparator<Person> {
        @Override
        public int compare(Person o1, Person o2) {
            return o1.age - o2.age;
        }
    }
}
public class Test {
    private static void print() {
       ArrayList<Person> list = new ArrayList();
        list.add(new Person("Diana", "Agron", 31));
        list.add(new Person("Kay", "Panabaker", 27));
        list.add(new Person("Lucy", "Hale", 28));
        list.add(new Person("Ashley", "Benson", 28));
        list.add(new Person("Megan", "Park", 31));
        list.add(new Person("Lucas", "Till", 27));
        list.add(new Person("Nicholas", "Hoult", 28));
        list.add(new Person("Aly", "Michalka", 28));
        list.add(new Person("Adam", "Brody", 38));
        list.add(new Person("Chris", "Pine", 37));
        Collections.sort(list, new Person.lastNameComperator());
        Iterator<Person> it = list.iterator();
        while(it.hasNext()) 
            System.out.println(it.next().toString()); 
     }  
}