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); } }
