我正在处理Java 8流,想知道是否可以用一种花哨的方法解决这个问题。
这是我的场景: 假设我有一个政党列表,每个元素里面都有成员的姓名。我想遍历列表并创建一个新的列表,其中包含名字和他们所属的政党。
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) {
}
}
在我的第二个尝试中,我尝试使用maps和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());
}
问题是我无法在map操作中访问party对象。所以问题又来了能否以更加函数式的方式实现呢?(比如第二种方法)
谢谢!
.flatMap(party -> party.getMembers().stream().map(member -> new Tuple<>(party, member))
。由于没有内建的元组类,你可以自己创建一个或(滥用)使用AbstractMap.SimpleEntry
... - TunakiPair.of(party, member)
。 - Jean-François Savard