Question

我上课

class Person {  
    public List<BaseballPlayer> baseballPlayers;    
    public List<MmaFighter> mmaFighters;    
    public List<RugbyPlayer> rugbyPlayers;  
}

在每个对象播放器中，其id都有String属性。我正在尝试收集列表中的所有ID

List<String> baseballPlayersIds = person.baseballPlayers.stream()
.map(s -> s.getId()).collect(Collectors.toList());  

List<String> mmaFightersIds = person.mmaFighters.stream()
.map(s -> s.getId()).collect(Collectors.toList());  

List<String> rugbyPlayersIds = person.rugbyPlayers.stream()
.map(s -> s.getId()).collect(Collectors.toList());

baseballPlayersIds.addAll(mmaFightersIds);
baseballPlayersIds.addAll(rugbyPlayersIds);

现在我正在尝试通过使用Stream.concat（）简化事情并改善逻辑。

Stream<List<BaseballPlayer>> baseballPlayersIdsStream =  Stream.of(person.baseballPlayers);    
Stream<List<MmaFighter>> mmaFightersIdsStream = Stream.of(person.mmaFighters);
Stream<List<RugbyPlayer>> rugbyPlayersIdsStream = Stream.of(person.rugbyPlayers);       

Stream<List<? extends Object>> personStream = Stream.concat(baseballPlayersIdsStream, Stream.concat(mmaFightersIdsStream, rugbyPlayersIdsStream));

但是我不知道我是否应该对这3个流中出现的新流类型使用泛型？还尝试为所有3个类创建父类，以在流diamand中代替Object使用。此personStream值得怀疑。

Answer 1

尝试一下。

<input type="text" />

或

List<String> allIds = Stream.of(
    person.baseballPlayers.stream().map(p -> p.getId()),
    person.mmaFighters.stream().map(p -> p.getId()),
    person.rugbyPlayers.stream().map(p -> p.getId()))
    .flatMap(s -> s)
    .collect(Collectors.toList());

Answer 2

如果您在所有类中实现一个通用接口：

interface Team {
    String getId();
}

您可以使用以下代码：

List<BaseballPlayer> baseballPlayers = ...;
List<MmaFighter> mmaFighters = ...;
List<RugbyPlayer> rugbyPlayers = ...;

List<String> identfiers = Stream.of(baseballPlayers, mmaFighters, rugbyPlayers)
        .flatMap(Collection::stream)
        .map(Team::getId)
        .collect(Collectors.toList());

在concat 3从不同对象的列表流传输时获取List <string>

2 个答案: