首页IT科技list过滤指定条件放到另一个list([进阶] list 过滤 stream 的lambda进阶操作, 这一篇就够了)

list过滤指定条件放到另一个list([进阶] list 过滤 stream 的lambda进阶操作, 这一篇就够了)

时间2025-05-26 20:37:09分类IT科技浏览3181
导读:目录...

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

1. list转化map基本操作

id最常用方式: 

-> 01  key-value值形式:

Map<Integer,String>map = list.stream().collect(Collectors.toMap(User::getId,User::getRealName))

-> 02 id-> 对象本身

Map<Integer,User> userMap3 = list.stream().collect(Collectors.toMap(User::getId, Function.identity()));

-> 03 id-> 对象本身的 lambda写法

Map<Integer,User> userMap2 = userList.stream().collect(Collectors.toMap(User::getId,User->User));

->04  解决冲突的key

Map<Integer,User> userMap4 = userList.stream().collect(Collectors.toMap(User::getId, Function.identity(),(key1,key2)->key2));

2. list计算操作

List<Long> testList = new ArrayList<>(Collections.nCopies(5, 0L)); testList.set(0,1L); testList.set(1,2L); testList.set(2,3L); testList.set(3,4L); testList.set(4,5L); System.out.println("sum1 is " + testList.stream().reduce(0L, (a, b) -> a + b)); // reduce根据初始值(参数1)和累积函数(参数2)依次对数据流进行操作,第一个值与初始值送入累积函数,后面计算结果和下一个数据流依次送入累积函数。 System.out.println("sum2 is " + testList.stream().reduce(0L, Long::sum)); System.out.println("sum3 is " + testList.stream().collect(Collectors.summingLong(Long::longValue))); // Collectors.summingLong()将流中所有元素视为Long类型,并计算所有元素的总和 System.out.println("sum4 is " + testList.stream().mapToLong(Long::longValue).sum()); System.out.println("***********************"); List<Person> testList1 = new ArrayList<>(Collections.nCopies(5, new Person(1))); System.out.println("class sum1 is " + testList1.stream().map(e -> e.getAge()).reduce(0, (a,b) -> a + b)); System.out.println("class sum2 is " + testList1.stream().map(e -> e.getAge()).reduce(0, Integer::sum)); System.out.println("class sum3 is " + testList1.stream().collect(Collectors.summingInt(Person::getAge))); System.out.println("class sum4 is " + testList1.stream().map(e -> e.getAge()).mapToInt(Integer::intValue).sum());

3. list转化泛型操作

List<User> list = userMapper.selectUserMessage(null); List<UserRespDTO> collect = list.stream().map(dto-> { UserRespDTO userRespDTO = new UserRespDTO(); BeanUtils.copyProperties(dto, respDTO); return userRespDTO; }).collect(Collectors.toList());

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)

import com.fasterxml.jackson.annotation.JsonFormat; import lombok.AllArgsConstructor; import lombok.Data; import lombok.NoArgsConstructor; import lombok.experimental.Accessors; import java.util.Date; /** * 模拟用户实体类对象 * * @author pzy * @version 0.1.0 */ @Data @NoArgsConstructor @AllArgsConstructor @Accessors(chain = true) public class User { private String username; private Integer age; private String telephone; @JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss", timezone = "GMT+8") private Date createTime; }

---> 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"]} }

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