使用stream的Collectors.toMap()方法常见的问题及解决

2023-03-06 11:03:17 方法 解决 常见

使用stream的Collectors.toMap()方法常见问题

java8开始的流式编程很大程度上简化了我们的代码,提高了开发效率。

我们经常会使用到stream的Collectors.toMap()来将List转换Map

在使用过程中有两个小坑需要注意

1、java.lang.IllegalStateException: Duplicate key

2、java.lang.NullPointerException

第一个是由于在List转Map过程中Map集合的key重复导致的;

第二个是由于在List转Map过程中Map集合的value有null导致的(当存在value值为空时,使用Collectors.toMap()会报NPE,因为底层调用了Map的merge方法,而map方法规定了此处的vlue不能为null,从而抛出空指针异常);

解决方案

1、Collectors.toMap(dto ->key值 , dto -> dto,(v1,v2) -> v1)

在后面添加(v1,v2)->v1 指定选取第一个值 当key值重复的时候,根据情况而定选取第一个还是第二个)

2、自定义一个Map来接收,不使用Collectors.toMap()

第一种情况示例:

import com.Google.common.collect.Lists;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import lombok.Data;
 
public class Test {
 
    private static List<User> userList = Lists.newArrayList();
 
    @Data
    public static class User {
        private String userCode;
        private String userName;
    }
 
    
    public static void initData() {
        User user1 = new User();
        user1.setUserCode("10001");
        user1.setUserName("张三");
 
        User user2 = new User();
        user2.setUserCode("10002");
        user2.setUserName("李四");
 
        User user3 = new User();
        user3.setUserCode("10002");
        user3.setUserName("王五");
 
        userList.add(user1);
        userList.add(user2);
        userList.add(user3);
    }
 
    public static void main(String[] args) {
        initData();
        //反例
      //  Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName));
 
        //正例,在后面添加(u1,u2)->u1 指定选取第一个值 当key值重复的时候,根据情况而定选取第一个还是第二个
        Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName, (u1, u2) -> u1));
 
        System.out.println(userMap);
    }
}

第二种情况示例:

import com.google.common.collect.Lists;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import lombok.Data;
 
public class Test {
 
    private static List<User> userList = Lists.newArrayList();
 
    @Data
    public static class User {
        private String userCode;
        private String userName;
    }
 
    
    public static void initData() {
        User user1 = new User();
        user1.setUserCode("10001");
        user1.setUserName("张三");
 
        User user2 = new User();
        user2.setUserCode("10002");
        user2.setUserName("李四");
 
        User user3 = new User();
        user3.setUserCode("10003");
        user3.setUserName(null);
 
        userList.add(user1);
        userList.add(user2);
        userList.add(user3);
    }
 
    public static void main(String[] args) {
        initData();
        //反例
       // Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName));
 
        //正例 (如果对转换后的顺序有要求,这里还可以使用LinkedHashMap)
        Map<String, String> userMap = userList.stream().collect(HashMap::new, (map, user) -> map.put(user.getUserCode(), user.getUserName()), HashMap::putAll);
 
        System.out.println(userMap);
    }
 
}

Stream ToMap(Collectors.toMap) 实践

Requirements

List TO Map

List Stream 转换 Map时向collect()方法中传递Collector对象,对象由Collectors.toMap()方法返回。

如下实现List转换为Map

List<GroupBrandCateBO> list = new ArrayList<>(
      Arrays.asList(
              new GroupBrandCateBO("v1", "g1", "b1"),
              new GroupBrandCateBO("v1", "g1", "b1"),
              new GroupBrandCateBO("v3", "g3", "b3")
      )
);
Map<String, String> map = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal, LinkedHashMap::new)); 
System.out.println(map.getClass());
Map<String, String> map0 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal));
System.out.println(map0.getClass());
System.out.println(map0.toString());
Map<String, String> map1 = list.stream().collect(Collectors.toMap(GroupBrandCateBO::getVersion, GroupBrandCateBO::getGroupCode));
System.out.println(map1.toString());

Console
class java.util.LinkedHashMap
class java.util.HashMap
{v1=g1, v3=g3}
Exception in thread “main” java.lang.IllegalStateException: Duplicate key g1
at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)

问题分析

toMap()函数重载:

  • 未指定合并函数mergeFunction情况下,传入throwingMerger()返回BinaryOperator对象,当出现key重复时,调用合并函数!
  • 未指定Supplier实例情况下,默认生成HashMap实例。
public static <T, K, U>
Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                Function<? super T, ? extends U> valueMapper) {
    return toMap(keyMapper, valueMapper, throwingMerger(), HashMap::new);
}

public static <T, K, U>
Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                Function<? super T, ? extends U> valueMapper,
                                BinaryOperator<U> mergeFunction) {
    return toMap(keyMapper, valueMapper, mergeFunction, HashMap::new);
}

public static <T, K, U, M extends Map<K, U>>
Collector<T, ?, M> toMap(Function<? super T, ? extends K> keyMapper,
                            Function<? super T, ? extends U> valueMapper,
                            BinaryOperator<U> mergeFunction,
                            Supplier<M> mapSupplier) {
    BiConsumer<M, T> accumulator
            = (map, element) -> map.merge(keyMapper.apply(element),
                                          valueMapper.apply(element), mergeFunction);
    return new CollectorImpl<>(mapSupplier, accumulator, mapMerger(mergeFunction), CH_ID);
}

private static <T> BinaryOperator<T> throwingMerger() {
    return (u,v) -> { throw new IllegalStateException(String.fORMat("Duplicate key %s", u)); };
}

补充

关于合并函数

List<GroupBrandCateBO> list = new ArrayList<>(
       Arrays.asList(
               new GroupBrandCateBO("v1", "g1", "b1"),
               new GroupBrandCateBO("v1", "g2", "b2"),
               new GroupBrandCateBO("v1", "g2", "b2"),
               new GroupBrandCateBO("v3", "g3", "b3")
       )
);
Map<String, String> map00 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> currVal));
Map<String, String> map01 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal + currVal));
System.out.println(map00.toString());
System.out.println(map01.toString());

Console
{v1=g2, v3=g3}
{v1=g1g2g2, v3=g3}

传入Lambda表达式将转化为BinaryOperator<U> mergeFunction对象,合并处理value,非Key!!!

比如:

(oldVal, currVal) -> currVal) // key相同时当前值替换原始值
(oldVal, currVal) -> oldVal + currVal //key相同时保留原始值和当前值

总结

以上为个人经验,希望能给大家一个参考,也希望大家多多支持。

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