Java8函数式编程中比较实用的操作语法

[Leeyom Wang]

分组（一对多）

假如有如下的一个数据结构：

[
  {
    "userId": 1,
    "name": "王二狗",
    "className": "classA"
  },
  {
    "userId": 2,
    "name": "李老四",
    "className": "classA"
  },
  {
    "userId": 3,
    "name": "张翠花",
    "className": "classB"
  },
  {
    "userId": 4,
    "name": "李雷",
    "className": "classB"
  }
]

需要将它按班级className进行分组，即如下的数据结构：

{
  "classA": [
    {
      "userId": 1,
      "name": "王二狗",
      "className": "classA"
    },
    {
      "userId": 2,
      "name": "李老四",
      "className": "classA"
    }
  ],
  "classB": [
    {
      "userId": 3,
      "name": "张翠花",
      "className": "classB"
    },
    {
      "userId": 4,
      "name": "李雷",
      "className": "classB"
    }
  ]
}

学生实体类 Student.java：

@Data
@AllArgsConstructor
public class Student {

    private long userId;

    private String name;

    private String className;
}

采用Java8函数式编程 groupingBy 语法进行快速分组：

Student s1 = new Student(1, "王二狗", "classA");
Student s2 = new Student(2, "李老四", "classA");
Student s3 = new Student(3, "张翠花", "classB");
Student s4 = new Student(4, "李雷", "classB");
List<Student> list = new ArrayList<>();
list.add(s1);
list.add(s2);
list.add(s3);
list.add(s4);
Map<String, List<Student>> map = list.stream().collect(Collectors.groupingBy(Student::getClassName));
System.out.println(JSONUtil.toJsonStr(map));

分组（一对一）

还是这个数据结构：

[
  {
    "userId": 1,
    "name": "王二狗",
    "className": "classA"
  },
  {
    "userId": 2,
    "name": "李老四",
    "className": "classA"
  },
  {
    "userId": 3,
    "name": "张翠花",
    "className": "classB"
  },
  {
    "userId": 4,
    "name": "李雷",
    "className": "classB"
  }
]

只不过要按用户的 userId 进行分组，分组后的数据格式如下：

{
  "1": {
    "className": "classA",
    "userId": 1,
    "name": "王二狗"
  },
  "2": {
    "className": "classA",
    "userId": 2,
    "name": "李老四"
  },
  "3": {
    "className": "classB",
    "userId": 3,
    "name": "张翠花"
  },
  "4": {
    "className": "classB",
    "userId": 4,
    "name": "李雷"
  }
}

采用Java8函数式编程的 toMap 语法进行快速分组：

Map<Long, Student> studentMap = list.stream().collect(Collectors.toMap(Student::getUserId, student -> student, (k1, k2) -> k1));

对集合中重复的元素进行去重

对于简单的集合，比如字符串，整型类的集合，采用 distinct 进行快速去重：

List<String> strList = Arrays.asList("a", "b", "c", "c", "d");
List<String> distinctList = strList.stream().distinct().collect(Collectors.toList());
System.out.println(JSONUtil.toJsonStr(distinctList));

对于集合元素是对象的，根据对象指定的属性进行去重：

// 根据className去重
List<Student> unique = list.stream().collect(Collectors.collectingAndThen(
    Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(Student::getClassName))), ArrayList::new));
System.out.println(JSONUtil.toJsonStr(unique));

// 根据userId和className去重
List<Student> unique2 = list.stream().collect(Collectors.collectingAndThen(
    Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(o -> o.getUserId() + ";" + o.getClassName()))), ArrayList::new));
System.out.println(JSONUtil.toJsonStr(unique2));

对集合元素进行快速排序

Student s1 = new Student(1, "王二狗", "classA",10);
Student s2 = new Student(2, "李老四", "classA",9);
Student s3 = new Student(3, "张翠花", "classB",8);
Student s4 = new Student(4, "李雷", "classB", 12);
List<Student> list = new ArrayList<>();
list.add(s1);
list.add(s2);
list.add(s3);
list.add(s4);
// 按照年龄进行升序
list.sort(Comparator.comparing(Student::getAge));
// 按照年龄进行降序
list.sort(Comparator.comparing(Student::getAge).reversed());

快速判断对象集合中是否存在指定的元素

// 查找学生当中是否存在一个叫王二狗的同学
boolean matchResult = list.stream().anyMatch(student -> "王二狗".equals(student.getName()));

将List转变为逗号分隔的字符串

List<String> cities = Arrays.asList("Milan", "London", "New York", "San Francisco");
String citiesCommaSeparated = String.join(",", cities);
System.out.println(citiesCommaSeparated);
// 输出: Milan,London,New York,San Francisco

参考文章

• 一文带你玩转 Java8 Stream 流，从此操作集合 So Easy