位置: IT常识 - 正文
目录
1. list转化map基本操作
id最常用方式:
-> 01 key-value值形式:
-> 02 id-> 对象本身
-> 03 id-> 对象本身的 lambda写法
->04 解决冲突的key
2. list计算操作
3. list转化泛型操作
4. List转成map的进阶操作
->4.1 有实体类的进阶操作(常用)
--->准备工作01 : 基础数据(User实体类)
--->4.1.1: (1对1关系) 分组: 用户名-> 用户实体类对象User(String->T)
--->4.1.2 (1对1关系) 用户名 -> 电话号(String-> String)
--->4.1.301: (1对多) 分组01: 根据年龄分组 Integer-> List
--->4.1.302: (1对多) 分组01: 根据年龄分组 Integer-> List 方法二
--->4.1.4: (1对多) 分组 根据年龄分出 年龄-> 用户 Integer -> List
-> 4.2 没有实体类的进阶操作
[待续未完....时间有限]
-> 5. 总结: 4整个的main方法
-> 5..1 不啰嗦 直接上代码
推荐整理分享[进阶] list 过滤 stream 的lambda进阶操作, 这一篇就够了(list集合过滤),希望有所帮助,仅作参考,欢迎阅读内容。
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---> 准备工作02:
//1. 创建基础数据 List<User> list = new ArrayList<>(); User user1 = new User("张三", 33, "13345678913", new Date()); User user2 = new User("李四", 44, "13345678914", new Date()); User user3 = new User("王五", 55, "13345678915", new Date()); User user4 = new User("李六", 55, "13345678916", new Date()); list.add(user1); list.add(user2); list.add(user3); list.add(user4);--->4.1.1: (1对1关系) 分组: 用户名-> 用户实体类对象User(String->T)Map<String, User> map1 = list.stream().collect(Collectors.toMap(User::getUsername, each -> each, (value1, value2) -> value1)); System.out.println(JSON.toJSONString(map1));//{"李四":{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"},"张三":{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"},"李六":{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"},"王五":{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"}}
--->4.1.2 (1对1关系) 用户名 -> 电话号(String-> String) Map<String,String> map2 = list.stream().collect(Collectors.toMap(User::getUsername,User::getTelephone,(value1, value2) -> value1)); System.out.println(JSON.toJSONString(map2));//{"李四":"13345678914","张三":"13345678913","李六":"13345678916","王五":"13345678915"}
--->4.1.301: (1对多) 分组01: 根据年龄分组 Integer-> List<T> Map<Integer, List<User>> map301 = list.stream().collect(Collectors.groupingBy(User::getAge)); System.out.println(JSON.toJSONString(map301));//{33:[{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"}],55:[{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"},{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"}],44:[{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"}]}--->4.1.302: (1对多) 分组01: 根据年龄分组 Integer-> List<T> 方法二Map<Integer,List<User>> map302 = list.stream().collect(Collectors.toMap(User::getAge, Collections::singletonList,(value1, value2) -> { List<User> union = new ArrayList<>(value1); union.addAll(value2); return union; })); System.out.println(JSON.toJSONString(map302));//{33:[{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"}],55:[{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"},{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"}],44:[{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"}]}--->4.1.4: (1对多) 分组 根据年龄分出 年龄-> 用户 Integer -> List<String>业务: 想查看每个年龄下都有谁(姓名即可)
Map<Integer,List<String>> map4 = list.stream().collect(Collectors.toMap(User::getAge,each->Collections.singletonList(each.getUsername()),(value1, value2) -> { List<String> usernameList = new ArrayList<>(value1); usernameList.addAll(value2); return usernameList; })); System.out.println(JSON.toJSONString(map4));//{33:["张三"],55:["王五","李六"],44:["李四"]}-> 4.2 没有实体类的进阶操作[待续未完....时间有限]-> 5. 总结: 4整个的main方法-> 5..1 不啰嗦 直接上代码 public static void main(String[] args) { //1. 创建基础数据 List<User> list = new ArrayList<>(); User user1 = new User("张三", 33, "13345678913", new Date()); User user2 = new User("李四", 44, "13345678914", new Date()); User user3 = new User("王五", 55, "13345678915", new Date()); User user4 = new User("李六", 55, "13345678916", new Date()); list.add(user1); list.add(user2); list.add(user3); list.add(user4); /*1. (1对1关系) 分组: 用户名-> 用户实体类对象User*/ Map<String, User> map1 = list.stream().collect(Collectors.toMap(User::getUsername, each -> each, (value1, value2) -> value1)); System.out.println(JSON.toJSONString(map1)); //{"李四":{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"},"张三":{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"},"李六":{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"},"王五":{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"}} /*2. (1对1关系) 用户名 -> 电话号 */ Map<String,String> map2 = list.stream().collect(Collectors.toMap(User::getUsername,User::getTelephone,(value1, value2) -> value1)); System.out.println(JSON.toJSONString(map2)); //{"李四":"13345678914","张三":"13345678913","李六":"13345678916","王五":"13345678915"} /*301. (1对多关系) 分组01: 根据年龄分组 目的: 将年龄相等的进行分组 年龄->用户实体类 */ Map<Integer, List<User>> map301 = list.stream().collect(Collectors.groupingBy(User::getAge)); System.out.println(JSON.toJSONString(map301)); //{33:[{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"}],55:[{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"},{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"}],44:[{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"}]} /*302. (1对多关系) 分组02: 根据年龄分组 目的: 将年龄相等的进行分组 年龄->用户实体类 */ Map<Integer,List<User>> map302 = list.stream().collect(Collectors.toMap(User::getAge, Collections::singletonList,(value1, value2) -> { List<User> union = new ArrayList<>(value1); union.addAll(value2); return union; })); System.out.println(JSON.toJSONString(map302)); //{33:[{"age":33,"createTime":1679536527360,"telephone":"13345678913","username":"张三"}],55:[{"age":55,"createTime":1679536527360,"telephone":"13345678915","username":"王五"},{"age":55,"createTime":1679536527360,"telephone":"13345678916","username":"李六"}],44:[{"age":44,"createTime":1679536527360,"telephone":"13345678914","username":"李四"}]} /*4. (1对多关系) 分组 根据年龄分出 年龄-> 用户 业务: 想查看每个年龄下都有谁(姓名即可)*/ Map<Integer,List<String>> map4 = list.stream().collect(Collectors.toMap(User::getAge,each->Collections.singletonList(each.getUsername()),(value1, value2) -> { List<String> usernameList = new ArrayList<>(value1); usernameList.addAll(value2); return usernameList; })); System.out.println(JSON.toJSONString(map4)); //{33:["张三"],55:["王五","李六"],44:["李四"]} }创作不易, 点个赞再走呗~~~
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