Java/Kotlin如何流式处理两个共有ID属性的集合生成匹配对象Map

• 需求：通过简洁的流式处理方案，对两个共有同ID属性的集合做关联匹配，生成键值对为<第一个集合元素, 第二个集合匹配元素>的Map，替代嵌套双重forEach的低效率写法，后续可直接对生成的Map执行批量业务操作。
• 示例基础代码：

data class FirstNameModel(val idNumber: String, val firstName: String)
data class LastNameModel(val idNumber: String, val lastName: String)

val randomFirstNameList = listOf(
   FirstNameModel("5631ab", "Bob"),
   FirstNameModel("ca790a", "George"),
   FirstNameModel("j8f1sa", "Alice")
)

val randomLastNameList = listOf(
   LastNameModel("j8f1sa", "Smith"),
   LastNameModel("5631ab", "Johnson"),
   LastNameModel("ca790a", "Takai")
)

// stream function to correctly create map (not just a null one like below).
val map: Map<FirstNameModel, LastNameModel>? = null

fun printIt() {
   map?.forEach {
       println("Name for id ${it.key.idNumber} is ${it.key.firstName} ${it.value.lastName}")
   }
//    should print something like:
//            Name for id 5631ab is Bob Johnson
//            Name for id ca790a is George Takai
//            Name for id j8f1sa is Alice Smith
}

核心思路：先对被关联的集合（即LastNameModel集合）按共有ID建立哈希索引，将单次ID查询的时间复杂度降到O(1)，整体关联逻辑时间复杂度从双重循环的O(n*m)降到O(n+m)，全流程采用流式链式调用，逻辑清晰易维护。

Kotlin 实现 #

直接使用Kotlin标准库提供的集合操作函数即可完成，不需要额外依赖：

// 1. 先将LastName集合按idNumber建立哈希索引
val lastNameIndex = randomLastNameList.associateBy { it.idNumber }
// 2. 流式关联生成目标Map，自动过滤无匹配ID的项
val map: Map<FirstNameModel, LastNameModel> = randomFirstNameList
    .mapNotNull { firstItem ->
        lastNameIndex[firstItem.idNumber]?.let { matchedLast -> firstItem to matchedLast }
    }
    .toMap()

fun printIt() {
   map.forEach {
       println("Name for id ${it.key.idNumber} is ${it.key.firstName} ${it.value.lastName}")
   }
}

执行printIt()将输出预期结果：

Name for id 5631ab is Bob Johnson
Name for id ca790a is George Takai
Name for id j8f1sa is Alice Smith

• 如果需要保留ID不匹配的项（值为null），将mapNotNull替换为map，取值逻辑改为lastNameIndex[firstItem.idNumber]即可。

Java 实现 #

基于Java 8及以上版本的Stream API实现，逻辑和Kotlin版本一致：

import java.util.List;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

// 模型类
class FirstNameModel {
    private String idNumber;
    private String firstName;
    public FirstNameModel(String idNumber, String firstName) {
        this.idNumber = idNumber;
        this.firstName = firstName;
    }
    public String getIdNumber() { return idNumber; }
    public String getFirstName() { return firstName; }
}
class LastNameModel {
    private String idNumber;
    private String lastName;
    public LastNameModel(String idNumber, String lastName) {
        this.idNumber = idNumber;
        this.lastName = lastName;
    }
    public String getIdNumber() { return idNumber; }
    public String getLastName() { return lastName; }
}

public class MatchDemo {
    public static void main(String[] args) {
        List<FirstNameModel> randomFirstNameList = List.of(
                new FirstNameModel("5631ab", "Bob"),
                new FirstNameModel("ca790a", "George"),
                new FirstNameModel("j8f1sa", "Alice")
        );
        List<LastNameModel> randomLastNameList = List.of(
                new LastNameModel("j8f1sa", "Smith"),
                new LastNameModel("5631ab", "Johnson"),
                new LastNameModel("ca790a", "Takai")
        );

        // 1. 构建LastName集合的ID索引，指定重复key的合并策略避免报错
        Map<String, LastNameModel> lastNameIndex = randomLastNameList.stream()
                .collect(Collectors.toMap(
                        LastNameModel::getIdNumber,
                        Function.identity(),
                        (oldVal, newVal) -> newVal
                ));
        // 2. 流式关联生成目标Map，过滤无匹配项
        Map<FirstNameModel, LastNameModel> matchMap = randomFirstNameList.stream()
                .filter(first -> lastNameIndex.containsKey(first.getIdNumber()))
                .collect(Collectors.toMap(
                        Function.identity(),
                        first -> lastNameIndex.get(first.getIdNumber())
                ));

        // 遍历输出
        matchMap.forEach((key, value) -> {
            System.out.printf("Name for id %s is %s %s%n",
                    key.getIdNumber(), key.getFirstName(), value.getLastName());
        });
    }
}

• 如果需要保留ID不匹配的项，移除filter步骤，取值时用lastNameIndex.getOrDefault(first.getIdNumber(), null)即可。

内容的提问来源于stack exchange，提问作者Bowerick