Java8之Stream之List转Map有哪些坑

Duplicate key 问题

当 key 值重复时会有这个问题,异常如下

Exception in thread "main" java.lang.IllegalStateException: Duplicate key 小C
at java.util.stream.Collectors.lambda$throwingMerger$0(Unknown Source)
at java.util.HashMap.merge(Unknown Source)
at java.util.stream.Collectors.lambda$toMap$58(Unknown Source)
at java.util.stream.ReduceOps$3ReducingSink.accept(Unknown Source)
at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(Unknown Source)
at java.util.stream.AbstractPipeline.copyInto(Unknown Source)
at java.util.stream.AbstractPipeline.wrapAndCopyInto(Unknown Source)
at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(Unknown Source)
at java.util.stream.AbstractPipeline.evaluate(Unknown Source)
at java.util.stream.ReferencePipeline.collect(Unknown Source)
at JavaBase.lamda.List2Map.main(List2Map.java:47)

Duplicate key 解决办法一:遇到重复的key就使用后者替换

// 后面的值代替之前的值
Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(value1 , value2)-> value2 ));

Duplicate key 解决办法二:重复时将前面的value和后面的value拼接起来

// 重复时将前面的value 和后面的value拼接起来
Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(value1 , value2)-> value1+","+value2 ));

Duplicate key 解决办法三:重复时将重复key的数据组成集合

// 重复时将重复key的数据组成集合
Map<String, List<String>> map = list.stream().collect(Collectors.toMap(Person::getId, p -> {
List<String> getNameList = new ArrayList<>();
getNameList.add(p.getName());
return getNameList;
}, (List<String> value1, List<String> value2) -> {
value1.addAll(value2);
return value1;
}));

NullPointerException 问题

Exception in thread "main" java.lang.NullPointerException
at java.util.HashMap.merge(Unknown Source)
at java.util.stream.Collectors.lambda$toMap$58(Unknown Source)
at java.util.stream.ReduceOps$3ReducingSink.accept(Unknown Source)
at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(Unknown Source)
at java.util.stream.AbstractPipeline.copyInto(Unknown Source)
at java.util.stream.AbstractPipeline.wrapAndCopyInto(Unknown Source)
at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(Unknown Source)
at java.util.stream.AbstractPipeline.evaluate(Unknown Source)
at java.util.stream.ReferencePipeline.collect(Unknown Source)
at JavaBase.lamda.List2Map.main(List2Map.java:47)

解决办法

Map<String, String> map = (Map<String, String>) list.stream().collect(HashMap::new,(k, v) ->k.put(v.getId(),v.getName()),HashMap::putAll);

Map类集合Key/Value能否存储null值情况表格

集合类

Key

Value

Super

说明

ConcurrentHashMap

不允许为null

不允许为null

AbstractMap

分段锁技术

Hashtable

不允许为null

不允许为null

Dictionary

线程安全

HashMap

允许为null

允许为null

AbstractMap

线程不安全

TreeMap

不允许为null

允许为null

AbstractMap

线程不安全

完整测试代码

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

class Person {
public String getId() {
return id;
}

public void setId(String id) {
this.id = id;
}

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}
private String id;
private String name;
public Person(String id, String name) {
this.id = id;
this.name = name;
}
}

public class List2Map {

public static void main(String[] args) {
// 声明一个List集合
List<Person> list = new ArrayList();
list.add(new Person("1001", "小A"));
list.add(new Person("1002", "小B"));
list.add(new Person("1003", "小C"));
// list.add(new Person("1003", "小D"));
list.add(new Person("1004", null));
// list.add(new Person(null, "小D"));
// 将list转换map
Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName));
// 后面的值代替之前的值
// Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(value1 , value2)-> value2 ));
// 重复时将前面的value 和后面的value拼接起来
// Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName,(value1 , value2)-> value1+","+value2 ));
// 重复时将重复key的数据组成集合
/* Map<String, List<String>> map = list.stream().collect(Collectors.toMap(Person::getId, p -> {
List<String> getNameList = new ArrayList<>();
getNameList.add(p.getName());
return getNameList;
}, (List<String> value1, List<String> value2) -> {
value1.addAll(value2);
return value1;
}));*/

// Map<String, String> map = (Map<String, String>) list.stream().collect(HashMap::new,(k, v) ->k.put(v.getId(),v.getName()),HashMap::putAll);
System.out.println(map);
}
}

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