Java List——lambda表达式的各种处理方式

根据某个属性分组

• 返回值：Map<key,List<>>
• 举例：根据userSex分组

Map<Integer, List<User>> collect = list.stream().collect(Collectors.groupingBy(User::getUserSex));

根据某个属性过滤

• 返回值：List<>
• 举例：查找男生（userSex == 1表示男生）

List<User> list = list.stream().filter(user -> user.getUserSex().toString().equals("1")).collect(Collectors.toList());

List转Map

• 返回值：Map
• 举例

Map<Long, User> userMap = 
userList.stream().collect(Collectors.toMap(User::getId,User->User);

Map<Long, String> userMap = 
userList.stream().collect(Collectors.toMap(User::getId,User::getUserName);

• 注意：如果集合对象有重复的key，会报错Duplicate key ....，可以用 (old,new)->old 来设置，如果有重复的key,则保留old,舍弃new

Map<Long, User> userMap = userList.stream().collect(Collectors.toMap(User::getId, user -> user,(old,new)->old));

• 转换时，把list多个元素拼接作为map的key值

Map<Long, String> userMap = 
userList.stream().collect(Collectors.toMap(user->user.getId()+"-"+user.getUserName(),user->user));

List排序

• 升序

list.sort(Comparator.comparing(User::getAge));

• 倒序

list.sort(Comparator.comparing(User::getAge).reversed());

 提取List某一属性

• 返回值：List<>
• 举例

List<String> userNames = userList.stream().map(User::getUserName).collect(Collectors.toList());