java8 lambda list map 便捷操作记录

1.list的直接forEach

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

//{}内可以执行逻辑代码
list.forEach((UserAccount user) -> {
            System.out.println(user.getId());
});

//只有一行代码时 {}可以省略
list.forEach((UserAccount user) -> System.out.println(user.getId()));

2.list int 排序

List<Integer> list = Arrays.asList(1, 2, 3, 9, 11, 6);

//正序
list.sort(Comparator.naturalOrder());
//倒序
list.sort(Comparator.reverseOrder());

3.list 对象 属性排序

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

//正序
list.sort(Comparator.comparing(UserAccount::getId));

//倒序
list.sort(Comparator.comparing(UserAccount::getId).reversed())

//正序流操作
List<UserAccount> collect = list.stream().sorted(Comparator.comparing(UserAccount::getId)).collect(Collectors.toList());

//倒序流操作
List<UserAccount> collect1 = listuserpaixu.stream().sorted((p1, p2) -> p2.getAddTime().compareTo(p1.getAddTime())).collect(Collectors.toList());

4.list 对象 根据对象属性过滤 

//过滤 list中 用户邮箱为111的用户
List<UserAccount> lis = list2.stream().filter(user -> user.getEmail().equals("111")).collect(Collectors.toList());

//返回符合表达式的集合的第一个对象
Optional<UserAccount> first = list2.stream().filter(user -> user.getEmail().equals("111")).findFirst();

//防止空指针
first.ifPresent(a-> System.out.println(a));

5.list对象  根据对象的某个属性生成一个新的list

//获取邮箱字段集合
List<String> list = list.stream().map(UserAccount::getEmail).collect(Collectors.toList());

6.分组

//根据id分组
final Map<Integer, List<UserAccount>> collect = list.stream().collect((Collectors.groupingBy(UserAccount::getId)));

7.list  转  map

//map 的key 和value 都是属性值
Map<String, String> map = list.stream().collect(Collectors.toMap(User::getId, User::getName));

//key为属性 value为对象本身
Map<String, User> map = userList.stream().collect(Collectors.toMap(User::getId, t->t));
//或
Map<String, User> map = userList.stream().collect(Collectors.toMap(User::getId, Function.identity()));

//如果在转换的过程中, list对象的属性作为map的key时有重复 会报错,java.lang.IllegalStateException: Duplicate key
//可以用下面的方法解决
//1.拼接
Map<String, String> map = list.stream().collect(Collectors.toMap(User::getId, User::getName, (old,newK)->old+","+newK));
Map<String, User> map = list.stream().collect(Collectors.toMap(User::getId, t->t, (old, newK)->old+","+newK));

//或取新值或老值
Map<String, String> map = list.stream().collect(Collectors.toMap(User::getId, User::getName, (old,newK)->newK));
Map<String, User> map = list.stream().collect(Collectors.toMap(User::getId, t->t, (old, newK)->newK));

//还可以排序 这里根据key排序 注意这里的返回值不同
TreeMap<String, User> collect = list.stream().collect(Collectors.toMap(User::getId, t->t, (old, newK)->newK, TreeMap::new));

待续....

标签：map,Collectors,stream,list,getId,collect,java8
来源： https://www.cnblogs.com/loveCrane/p/15918898.html