我正在处理Java 8流,我想知道我是否可以用奇特的方式解决这个问题。通过流和平面图传递对象
这就是我的场景: 假设我有一个派对的列表,并且在每个元素中都有成员的名字。我想遍历列表并使用名称和属于哪个方来创建一个新列表。
我的第一种方法是:
@Test
public void test(){
Party firstParties = new Party("firstParty",Lists.newArrayList("Member 1","Member 2","Member 3"));
Party secondParty = new Party("secondParty",Lists.newArrayList("Member 4","Member 5","Member 6"));
List<Party> listOfParties = Lists.newArrayList();
listOfParties.add(firstParty);
listOfParties.add(secondParty);
List<Elector> electors = new ArrayList<>();
listOfParties.stream().forEach(party ->
party.getMembers().forEach(memberName ->
electors.add(new Elector(memberName,party.name))
)
);
}
class Party {
List<String> members = Lists.newArrayList();
String name = "";
public Party(String name, List<String> members) {
this.members = members;
this.name = name;
}
public List<String> getMembers() {
return members;
}
}
class Elector{
public Elector(String electorName,String partyName) {
}
}
在我的第二个方法我试过用映射的flatmap的操作:
@Test
public void test(){
Party firstParty = new Party("firstParty",Lists.newArrayList("Member 1","Member 2","Member 3"));
Party secondParty = new Party("secondParty",Lists.newArrayList("Member 4","Member 5","Member 6"));
List<Party> listOfParties = Lists.newArrayList();
listOfParties.add(firstParty);
listOfParties.add(secondParty);
List<Elector> people = listOfParties.stream().map(party -> party.getMembers())
.flatMap(members -> members.stream())
.map(membersName -> new Elector(membersName, party.name)) #Here is my problem variable map doesn't exist
.collect(Collectors.toList());
}
的问题是我无法访问党物体内地图操作。 所以问题再次是我可以做更多功能的方式? (如第二种方法)
谢谢!
是的,黑客将平面映射到一个元组流中,该元组将包含成员及其派对。就像'.flatMap(party - > party.getMembers()。stream()。map(member - > new Tuple <>(party,member))'。因为没有内置的元组类,所以你可以你自己或(ab)使用'AbstractMap.SimpleEntry' ... – Tunaki
或者'Pair.of(party,member)'from commons-lang如果你已经拥有这个依赖关系。 –
@Tunaki:你想太复杂了... – Holger