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));

来自alaster的答案对我帮助很大，但如果有人试图对数据进行分组，我想添加有意义的信息。

例如，如果您有两个订单，每个订单的代码相同，但产品数量不同，并且您希望将数量相加，您可以执行以下操作:

List<Order> listQuantidade = new ArrayList<>();
listOrders.add(new Order("COD_1", 1L));
listOrders.add(new Order("COD_1", 5L));
listOrders.add(new Order("COD_1", 3L));
listOrders.add(new Order("COD_2", 3L));
listOrders.add(new Order("COD_3", 4L));

listOrders.collect(Collectors.toMap(Order::getCode, 
                                    o -> o.getQuantity(), 
                                    (o1, o2) -> o1 + o2));

结果:

{COD_3=4, COD_2=3, COD_1=9}

或者，在javadocs中，你可以组合地址:

 Map<String, String> phoneBook
     people.stream().collect(toMap(Person::getName,
                                   Person::getAddress,
                                   (s, a) -> s + ", " + a));

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)

我在分组对象时遇到过这样的问题，我总是用一个简单的方法解决它们:使用java.util.Set执行一个自定义过滤器，以删除重复的对象，无论你选择的属性如下所示

Set<String> uniqueNames = new HashSet<>();
Map<String, String> phoneBook = people
                  .stream()
                  .filter(person -> person != null && !uniqueNames.add(person.getName()))
                  .collect(toMap(Person::getName, Person::getAddress));

希望这对有同样问题的人有所帮助!

按对象分组

Map<Integer, Data> dataMap = dataList.stream().collect(Collectors.toMap(Data::getId, data-> data, (data1, data2)-> {LOG.info("Duplicate Group For :" + data2.getId());return data1;}));

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()));

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…

为了完整起见，这里介绍了如何将重复项“减少”到只有一个。

如果你同意最后一个:

  Map<String, Person> phoneBook = people.stream()
          .collect(groupingBy(x -> x.name, reducing(null, identity(), (first, last) -> last)));

如果你只想要第一个:

  Map<String, Person> phoneBook = people.stream()
          .collect(groupingBy(x -> x.name, reducing(null, identity(), (first, last) -> first != null ? first : last)));

如果你想要last but“address as String”(不使用identity()作为参数)。

  Map<String, String> phoneBook = people.stream()
          .collect(groupingBy(x -> x.name, reducing(null, x -> x.address, (first, last) -> last)));

源

因此，本质上，groupingBy与还原收集器配对，开始表现得非常类似于toMap收集器，有一些类似于它的mergeFunction…结果是一样的……

可以使用lambda函数:比较从key(…)到键字符串

List<Blog> blogsNoDuplicates = blogs.stream()
            .collect(toMap(b->key(b), b->b, (b1, b2) -> b1))  // b.getAuthor() <~>key(b) as Key criteria for Duplicate elimination
            .values().stream().collect(Collectors.toList());

static String key(Blog b){
    return b.getTitle()+b.getAuthor(); // make key as criteria of distinction
}

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));