关于Jdk8对Collections进行分组

一.功能

1、 跟据某个属性分组OfficeId：

 Map<String, List<IncomeSumPojo>> collect = list.stream().collect(Collectors.groupingBy(IncomeSumPojo::getOfficeId));

2、根据某个属性分组OfficeId，汇总某个属性Money：

Map<String, Double> collect = list.stream().collect(Collectors.groupingBy(IncomeSumPojo::getOfficeId,Collectors.summingDouble(IncomeSumPojo::getMoney)));

3、根据某个属性添加条件过滤数据：

list = list.stream().filter(u -> !u.getAmount().equals("0.00")).collect(Collectors.toList());

4、判断一组对象里面有没有属性值是某个值：

List<Menu> menuList = UserUtils.getMenuList();

boolean add = menuList.stream().anyMatch(m -> "plan:ctPlan:add".equals(m.getPermission()));

5、取出一组对象的某个属性组成一个新集合：

List<String> tableNames=list.stream().map(User::getMessage).collect(Collectors.toList());

6、list去重复：

list = list.stream().distinct().collect(Collectors.toList());

二.例子

先准备一点数据：

public class User {
    private Integer id;
    private String type;
    private String name;
    private String date;

// omit setter and getter

}

User user1 = new User(1,"张三","小学", "2017-03");
User user2 = new User(2,"李四","小学", "2017-03");
User user3 = new User(3,"王五","初中", "2017-02");
User user4 = new User(4,"马六","高中", "2017-04");
User user5= new User(5,"jack","大学", "2017-04");

List<User > list = new ArrayList<>();

list.add(user1 );
list.add(user2 );
list.add(user3 );
list.add(user4 );
list.add(user5);

现在想对用户列表根据日期进行分组，得到一个map: date -> List<User>。

Java7以前也许是这样的：

List<User> userList = Arrays.asList(user1, user2, user3, user5, user4);
        Map<String, List<User>> groupUserMap = Maps.newHashMap();
        for (User user : userList) {
            String month = user.getDate();
            if (groupUserMap.containsKey(month)) {
                groupUserMap.get(month).add(user);
            } else {
                groupUserMap.put(month, Lists.newArrayList(user));
            }
        }

有了Java8之后，一切很简单了：

Map<String, List<User>> groupUserMapOne =userList.stream().collect(Collectors.groupingBy(User::getDate));

如果想分组的List里是User的name呢：

 Map<String, List<String>> userByDate = Stream.of(user1, user2, user3, user5, user4).collect(Collectors.groupingBy(User::getDate,
               Collectors.mapping(User::getName, Collectors.toList())));
        System.out.println(userByDate);