我有课
class Person {
public List<BaseballPlayer> baseballPlayers;
public List<MmaFighter> mmaFighters;
public List<RugbyPlayer> rugbyPlayers;
}
在每个对象玩家中,都有一个用于其 id 的 String 属性。我正在尝试收集我所做的列表中的所有 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);
现在我尝试使用 Stream.concat() 来简化事情并改进逻辑
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));
但我不知道我是否应该使用泛型来表示来自 3 个流的新流类型?还尝试为所有 3 个类创建父类以在流钻石中使用而不是 Object。此 personStream 有疑问。
最佳答案
试试这个。
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());
或者
List<String> allIds = Stream.of(
person.baseballPlayers.stream().map(BaseballPlayer::getId),
person.mmaFighters.stream().map(MmaFighter::getId),
person.rugbyPlayers.stream().map(RugbyPlayer::getId))
.flatMap(Function.identity())
.collect(Collectors.toList());
关于java - 从不同对象的列表中连接 3 个流时获取 List<String>,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55909788/