Java-stream过滤列表？

Question

如何在java 8流中使用过滤器过滤出复杂的对象列表。 假设我有这样的课程

public class InfoLite{
    private String right;
    List<Others> options;
    private String new;
}
//getter,setter and constructor

public class Others{    
    private String workId;
    private String work;
}
//getter,setter and constructor

我有一个与此类似的信息对象

List<InfoLite> info = new ArrayList<>();
List<Others> options = new ArrayList<>();
options.add(new Others(1,"game"))
options.add(new Others(2,"unit"))
options.add(new Others(3,"dest"))
List<Others> options2 = new ArrayList<>();
options.add(new Others(1,"console"))
options.add(new Others(2,"unit"))
List<Others> options3 = new ArrayList<>();
options.add(new Others(1,"zan"))
options.add(new Others(2,"zebvra"))
List<Others> options4 = new ArrayList<>();
options.add(new Others(1,"zan"))
options.add(new Others(2,"lon"))
options.add(new Others(3,"car"))
info.add(new Test("game,unit",options,"Florida"));
info.add(new Test("console",options2,"Florida"));
info.add(new Test("zebvra",options3,"Florida"));
info.add(new Test("lon",options4,"Florida"));

在这里我如何过滤info 列表，以便我可以获得这样的新列表

List<InfoLite> newInfo = new ArrayList<>();

List<Others> options = new ArrayList<>();
options.add(new Others(1,"game"))
options.add(new Others(2,"unit"))

List<Others> options2 = new ArrayList<>();
options.add(new Others(1,"console"))

List<Others> options3 = new ArrayList<>();
options.add(new Others(2,"zebvra"))

List<Others> options4 = new ArrayList<>();
options.add(new Others(2,"lon"))

info.add(new Test("game,unit",options,"Florida"));
info.add(new Test("console",options2,"Florida"));
info.add(new Test("zebvra",options3,"Florida"));
info.add(new Test("lon",options4,"Florida"));

这里Others 对象基本上被过滤了，即只过滤Others 列表中与info 中的right 匹配值的对象。如果info 中的right 是逗号分隔的，则拆分right 并签入每个Others 列表对象并返回值。

我尝试的到这里，但是我很困惑下一步该做什么。

info.stream()
            .filter(option -> Arrays.stream(option.getRight().split(",")).collect(Collectors.toList()).contains(option.getOptions().stream().filter(Others :: getWork))))``

Answer 1

假设，您打算打印与right 值InfoLite 中的字符串之一匹配的所有Others 实例。

如果是这种情况，下面的 sn-p 应该会有所帮助

System.out.println(info.stream()
                       .flatMap(infoLite -> infoLite.getOptions()
                                                    .stream()
                                                    .filter(option -> Arrays.stream(infoLite.getRight()
                                                                                            .split(","))
                                                                            .collect(Collectors.toSet())
                                                                            .contains(option.getWork())))
                       .collect(Collectors.toSet()));

输出：

[Others{workId=1, work='console'}, Others{workId=2, work='zebvra'}, Others{workId=1, work='game'}, Others{workId=2, work='unit'}, Others{workId=2, work='lon'}]

---

注意：我在请求程序中做了一些更改。

• 在下面的sn-p

  List<Others> options2 = new ArrayList<>();
  options.add(new Others(1, "console"));
  

  应该是options2.add...
• 假设Test 是InfoLite 的子类
• 为Others 引入了equals() 和hashcode()。

  @Override
  public boolean equals(Object o) {
      if (this == o) return true;
      if (o == null || getClass() != o.getClass()) return false;
      Others others = (Others) o;
      return workId == others.workId &&
            Objects.equals(work, others.work);
  }
  
  @Override
  public int hashCode() {
      return Objects.hash(workId, work);
  }
  

---

更新 要检索完整的InfoList 实例，修改后的 sn-p 是

info.forEach(infoLite -> {
    Set<String> rightSet = Arrays.stream(infoLite.getRight()
                                                 .split(","))
                                 .collect(Collectors.toSet());
    infoLite.getOptions()
            .removeIf(option -> !rightSet.contains(option.getWork()));
});

info.forEach(System.out::println);

它产生以下输出

InfoLite{right='game,unit', newString='Florida', options=[Others{workId=1, work='game'}, Others{workId=2, work='unit'}]}
InfoLite{right='console', newString='Florida', options=[Others{workId=1, work='console'}]}
InfoLite{right='zebvra', newString='Florida', options=[Others{workId=2, work='zebvra'}]}
InfoLite{right='lon', newString='Florida', options=[Others{workId=2, work='lon'}]}