Rumah >Java >javaTutorial >Cara menggunakan ungkapan lambda untuk mengisih dalam Java
Mari kita lihat beberapa contoh pengisihan biasa, pengisihan jenis data asas
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]
Kita dapat melihat bahawa hasil pelaksanaan adalah seperti yang dijangkakan, tetapi Dalam kebanyakan senario, kita mungkin perlu mengisih atribut tertentu objek, jadi bagaimana kita harus melakukannya? Mari lihat contoh di bawah:
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}]
Bagaimana jika kita perlu mengumpulkan dan menyusun mengikut jantina? Mari lihat contoh di bawah
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 }] }
Kami melihat bahawa terdapat masalah dengan hasil output di atas Jika umur adalah sama, mereka tidak disusun mengikut nama. Mari lihat contoh di bawah
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 }] }
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