Map<String, String> phoneBook = people.stream()
.collect(toMap(Person::getName,
Person::getAddress));
我得到java.lang.IllegalStateException:当找到一个重复的元素时，重复键。
是否有可能忽略这种例外添加值到地图?
当有重复的键时，应该忽略重复的键继续执行。
当前回答
感觉toMap经常工作，但并不总是java流的一个黑暗的弱点。好像他们应该把它叫做“ouniquemap”之类的…
最简单的方法是使用收集器。groupingBy而不是collections . tomap。
默认情况下它会返回一个List类型的输出，但是碰撞问题已经解决了，这也许是你想要的存在的倍数。
Map<String, List<Person>> phoneBook = people.stream()
.collect(groupingBy((x) -> x.name));
如果一个Set类型的地址集合与特定的名称相关联，groupingBy也可以做到这一点:
Map<String, Set<String>> phoneBook = people.stream()
.collect(groupingBy((x) -> x.name, mapping((x) -> x.address, toSet())));
另一种方法是“开始”一个哈希或一个集…并仔细跟踪以确保键在输出流中不会重复。啊。这里有一个例子，碰巧在这种情况下幸存下来…
其他回答
对于其他遇到此问题但没有在映射流中重复键的人，请确保keyMapper函数不返回空值。
跟踪这个是非常烦人的，因为当它处理第二个元素时，Exception会说“重复键1”，而1实际上是条目的值而不是键。
在我的情况下，我的keyMapper函数试图在不同的映射中查找值，但由于字符串中的错别字返回空值。
final Map<String, String> doop = new HashMap<>();
doop.put("a", "1");
doop.put("b", "2");
final Map<String, String> lookup = new HashMap<>();
doop.put("c", "e");
doop.put("d", "f");
doop.entrySet().stream().collect(Collectors.toMap(e -> lookup.get(e.getKey()), e -> e.getValue()));
假设你有people对象列表
Map<String, String> phoneBook=people.stream()
.collect(toMap(Person::getName, Person::getAddress));
现在你需要两个步骤:
1)
people =removeDuplicate(people);
2)
Map<String, String> phoneBook=people.stream()
.collect(toMap(Person::getName, Person::getAddress));
下面是删除重复的方法
public static List removeDuplicate(Collection<Person> list) {
if(list ==null || list.isEmpty()){
return null;
}
Object removedDuplicateList =
list.stream()
.distinct()
.collect(Collectors.toList());
return (List) removedDuplicateList;
}
在这里添加完整的示例
package com.example.khan.vaquar;
import java.util.Arrays;
import java.util.Collection;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
public class RemovedDuplicate {
public static void main(String[] args) {
Person vaquar = new Person(1, "Vaquar", "Khan");
Person zidan = new Person(2, "Zidan", "Khan");
Person zerina = new Person(3, "Zerina", "Khan");
// Add some random persons
Collection<Person> duplicateList = Arrays.asList(vaquar, zidan, zerina, vaquar, zidan, vaquar);
//
System.out.println("Before removed duplicate list" + duplicateList);
//
Collection<Person> nonDuplicateList = removeDuplicate(duplicateList);
//
System.out.println("");
System.out.println("After removed duplicate list" + nonDuplicateList);
;
// 1) solution Working code
Map<Object, Object> k = nonDuplicateList.stream().distinct()
.collect(Collectors.toMap(s1 -> s1.getId(), s1 -> s1));
System.out.println("");
System.out.println("Result 1 using method_______________________________________________");
System.out.println("k" + k);
System.out.println("_____________________________________________________________________");
// 2) solution using inline distinct()
Map<Object, Object> k1 = duplicateList.stream().distinct()
.collect(Collectors.toMap(s1 -> s1.getId(), s1 -> s1));
System.out.println("");
System.out.println("Result 2 using inline_______________________________________________");
System.out.println("k1" + k1);
System.out.println("_____________________________________________________________________");
//breacking code
System.out.println("");
System.out.println("Throwing exception _______________________________________________");
Map<Object, Object> k2 = duplicateList.stream()
.collect(Collectors.toMap(s1 -> s1.getId(), s1 -> s1));
System.out.println("");
System.out.println("k2" + k2);
System.out.println("_____________________________________________________________________");
}
public static List removeDuplicate(Collection<Person> list) {
if (list == null || list.isEmpty()) {
return null;
}
Object removedDuplicateList = list.stream().distinct().collect(Collectors.toList());
return (List) removedDuplicateList;
}
}
// Model class
class Person {
public Person(Integer id, String fname, String lname) {
super();
this.id = id;
this.fname = fname;
this.lname = lname;
}
private Integer id;
private String fname;
private String lname;
// Getters and Setters
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getFname() {
return fname;
}
public void setFname(String fname) {
this.fname = fname;
}
public String getLname() {
return lname;
}
public void setLname(String lname) {
this.lname = lname;
}
@Override
public String toString() {
return "Person [id=" + id + ", fname=" + fname + ", lname=" + lname + "]";
}
}
结果:
Before removed duplicate list[Person [id=1, fname=Vaquar, lname=Khan], Person [id=2, fname=Zidan, lname=Khan], Person [id=3, fname=Zerina, lname=Khan], Person [id=1, fname=Vaquar, lname=Khan], Person [id=2, fname=Zidan, lname=Khan], Person [id=1, fname=Vaquar, lname=Khan]]
After removed duplicate list[Person [id=1, fname=Vaquar, lname=Khan], Person [id=2, fname=Zidan, lname=Khan], Person [id=3, fname=Zerina, lname=Khan]]
Result 1 using method_______________________________________________
k{1=Person [id=1, fname=Vaquar, lname=Khan], 2=Person [id=2, fname=Zidan, lname=Khan], 3=Person [id=3, fname=Zerina, lname=Khan]}
_____________________________________________________________________
Result 2 using inline_______________________________________________
k1{1=Person [id=1, fname=Vaquar, lname=Khan], 2=Person [id=2, fname=Zidan, lname=Khan], 3=Person [id=3, fname=Zerina, lname=Khan]}
_____________________________________________________________________
Throwing exception _______________________________________________
Exception in thread "main" java.lang.IllegalStateException: Duplicate key Person [id=1, fname=Vaquar, lname=Khan]
at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)
at java.util.HashMap.merge(HashMap.java:1253)
at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)
at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
at java.util.Spliterators$ArraySpliterator.forEachRemaining(Spliterators.java:948)
at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:481)
at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:471)
at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)
at com.example.khan.vaquar.RemovedDuplicate.main(RemovedDuplicate.java:48)
这可以通过使用collector的mergeFunction参数实现。toMap(keyMapper, valueMapper, mergeFunction):
Map<String, String> phoneBook =
people.stream()
.collect(Collectors.toMap(
Person::getName,
Person::getAddress,
(address1, address2) -> {
System.out.println("duplicate key found!");
return address1;
}
));
mergeFunction是一个对与同一个键相关联的两个值进行操作的函数。address1对应于收集元素时遇到的第一个地址，address2对应于遇到的第二个地址:这个lambda只是告诉保留第一个地址而忽略第二个地址。
正如在JavaDocs中所说:
如果映射的键包含重复项(根据 Object.equals(Object))时，当异常时抛出IllegalStateException 执行收集操作。如果映射的键可能有 重复，使用toMap(Function keyMapper, Function valueMapper, BinaryOperator mergeFunction)代替。
所以你应该使用toMap(Function keyMapper, Function valueMapper, BinaryOperator mergeFunction)来代替。只需提供一个合并函数，它将确定将哪个副本放入映射中。
例如，如果你不关心是哪个，打电话就可以了
Map<String, String> phoneBook = people.stream().collect(
Collectors.toMap(Person::getName, Person::getAddress, (a1, a2) -> a1));
我有同样的情况，发现最简单的解决方案(假设你只是想覆盖重复键的映射值)是:
Map<String, String> phoneBook =
people.stream()
.collect(Collectors.toMap(Person::getName,
Person::getAddress,
(key1, key2)-> key2));
