最初に、基本的なデータ型の並べ替えであるいくつかの一般的な並べ替えの例を見てみましょう
List list = Arrays.asList(1,3,2,5,4); list.sort(Comparator.naturalOrder()); System.out.println(list); list.sort(Comparator.reverseOrder()); System.out.println(list); 输出结果: [1, 2, 3, 4, 5] [5, 4, 3, 2, 1]
実行結果が期待どおりであることがわかりますが、ほとんどのシナリオでは、オブジェクトの特定の属性を並べ替える必要がある場合がありますが、どのように実行すればよいでしょうか?以下の例を見てみましょう:
public class Student { private String name; private String sexual; private Integer age; public Student(String name, String sexual,Integer age) { this.name = name; this.sexual = sexual; this.age = age; } public String getName() { return name; } public void setName(String name) { this.name = name; } public String getSexual() { return sexual; } public void setSexual(String sexual) { this.sexual = sexual; } public Integer getAge() { return age; } public void setAge(Integer age) { this.age = age; } @Override public String toString() { return "Student{" + "name='" + name + '\'' + ", sexual='" + sexual + '\'' + ", age=" + age + '}'; } public class Starter { public static void main(String[] args) { List<Student> list = Arrays.asList( new Student("jack", 12), new Student("john", 13), new Student("lily", 11), new Student("lucy", 10) ); list.sort(Comparator.comparing(Student::getAge)); System.out.println(list); list.sort(Comparator.comparing(Student::getAge).reversed()); System.out.println(list); } } 输出结果: [Student{name='lucy', age=10}, Student{name='lily', age=11}, Student{name='jack', age=12}, Student{name='john', age=13}] [Student{name='john', age=13}, Student{name='jack', age=12}, Student{name='lily', age=11}, Student{name='lucy', age=10}]
性別ごとにグループ化して並べ替える必要がある場合、どうすればよいでしょうか?以下の例を見てみましょう
public class Starter { public static void main(String[] args) { List<Student> list = Arrays.asList( new Student("jack", "male", 12), new Student("john", "male", 13), new Student("lily", "female", 11), new Student("david", "male", 14), new Student("luck", "female", 13), new Student("jones", "female", 15), new Student("han", "male", 13), new Student("alice", "female", 11), new Student("li", "male", 12) ); Map<String, List<Student>> groupMap = list.stream().sorted(Comparator.comparing(Student::getAge)) .collect(Collectors.groupingBy(Student::getSexual, Collectors.toList())); System.out.println(groupMap.toString()); } } 输出结果: { female = [ Student { name = 'lily', sexual = 'female', age = 11 }, Student { name = 'alice', sexual = 'female', age = 11 }, Student { name = 'luck', sexual = 'female', age = 13 }, Student { name = 'jones', sexual = 'female', age = 15 }], male = [ Student { name = 'jack', sexual = 'male', age = 12 }, Student { name = 'li', sexual = 'male', age = 12 }, Student { name = 'john', sexual = 'male', age = 13 }, Student { name = 'han', sexual = 'male', age = 13 }, Student { name = 'david', sexual = 'male', age = 14 }] }
上記の出力結果には問題があります。年齢が同じ場合、名前順にソートされていません。この関数はどのように実装すればよいでしょうか?以下の例を見てみましょう
Map<String, List<Student>> groupMap = list.stream().sorted(Comparator.comparing(Student::getAge) .thenComparing(Student::getName)).collect(Collectors.groupingBy(Student::getSexual, Collectors.toList())); 输出结果: { female = [ Student { name = 'alice', sexual = 'female', age = 11 }, Student { name = 'lily', sexual = 'female', age = 11 }, Student { name = 'luck', sexual = 'female', age = 13 }, Student { name = 'jones', sexual = 'female', age = 15 }], male = [ Student { name = 'jack', sexual = 'male', age = 12 }, Student { name = 'li', sexual = 'male', age = 12 }, Student { name = 'han', sexual = 'male', age = 13 }, Student { name = 'john', sexual = 'male', age = 13 }, Student { name = 'david', sexual = 'male', age = 14 }] }
以上がJavaでラムダ式を使用してソートする方法の詳細内容です。詳細については、PHP 中国語 Web サイトの他の関連記事を参照してください。