问题描述
好的，我有列表< Person> 。每个人都有列表< String> ，这是该人拥有的电话号码列表。
Ok, so I have a List<Person>. and each Person has a List<String> that is a list of phone numbers that that person owns.
所以这是基本结构：
public class Person {
private String name;
private List<String> phoneNumbers;
// constructors and accessor methods
}
我想创建一个 Map< String，Person> ，其中密钥是该人拥有的每个电话号码，值是实际的人。
I would like to create a Map<String, Person> where the key is each phone number that the person owns, and the value is the actual person.
所以要更好地解释。如果我有列表< Person> ：
So to explain better. If I had this List<Person>:
Person bob = new Person("Bob");
bob.getPhoneNumbers().add("555-1738");
bob.getPhoneNumbers().add("555-1218");
Person john = new Person("John");
john.getPhoneNumbers().add("518-3718");
john.getPhoneNumbers().add("518-3115");
john.getPhoneNumbers().add("519-1987");
List<Person> list = new ArrayList<>();
list.add(bob);
list.add(john);
我调用了这个方法。它会给我以下 Map< String，Person>
and I invoked this method. It would give me the following Map<String, Person>
Map<String, Person> map = new HashMap<>();
map.put("555-1738", bob);
map.put("555-1218", bob);
map.put("518-3718", john);
map.put("518-3115", john);
map.put("519-1987", john);
我想知道如何使用流来实现这个目标 API，因为我已经知道如何使用来执行循环等。
I would like to know how to achieve this using the stream API, as I already know how I would do it using for loops and such.
推荐答案
如果您有名为人的人员名单和一个名为 PhoneNumberAndPerson 的人（你）可以使用通用元组或对代替）
If you have list of persons called persons and a class called PhoneNumberAndPerson (you could use a generic Tuple or Pair instead)
以下是步骤：
对于每个人，请记下该人的每个电话号码。对于每个电话号码，创建 PhoneNumberAndPerson 的新实例，并将其添加到列表中。使用flatMap制作所有这些较小列表的单个列表。要从此列表中生成 Map ，您需要提供一个函数来从 PhoneNumberAndPerson 中提取密钥，并提供另一个要提取的函数一个人来自同一个 PhoneNumberAndPerson 。
For each person, take each phone number of that person. For each of those phone numbers, create a new instance of PhoneNumberAndPerson and add that to a list. Use flatMap to make one single list of all these smaller lists. To make a Map out of this list you supply one function to extract a key from a PhoneNumberAndPerson and another function to extract a Person from that same PhoneNumberAndPerson.
persons.stream()
.flatMap(person -> person.getPhoneNumbers().stream().map(phoneNumber -> new PhoneNumberAndPerson(phoneNumber, person)))
.collect(Collectors.toMap(pp -> pp.getPhoneNumber(), pp -> pp.getPerson()));
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