筛选列表值的仅一个单个值的列表

标签 list java-8 java-stream

我想做分而治之来解决这个问题

在我的问题 Generate List with Combination of Subset of List, Java 中显示之前,我有这门课

我有一个名为 Competor 的类,具有类型、名称和功率。

public class Competitor {
  private final int type;
  private final String name;
  private final int power;

  public Competitor(int type, String name, int power) {
    this.type = type;
    this.name = name;
    this.power = power;
  }

  public int getType() {
    return type;
  }

  public String getName() {
    return name;
  }

  public int getPower() {
    return power;
  }

  @Override
  public String toString() {
    return "Competitor{" + "type=" + type + ", name=" + name + ", power=" + power + '}';
  }

}

现在我填写 list
public class Game {
  public static void main(String... args) {
    List<Competitor> listCompetitors = new ArrayList<>();
    listCompetitors.add(new Competitor(1, "Cat 00", 93));
    listCompetitors.add(new Competitor(1, "Cat 10", 11));
    listCompetitors.add(new Competitor(1, "Cat 23", 20));

    listCompetitors.add(new Competitor(2, "Dog 61", 54));
    listCompetitors.add(new Competitor(2, "Dog 18", 40));
    listCompetitors.add(new Competitor(2, "Dog 45", 71));
    listCompetitors.add(new Competitor(2, "Dog 30", 68));

    listCompetitors.add(new Competitor(3, "Pig 90", 90));
    listCompetitors.add(new Competitor(3, "Pig 78", 32));

    listCompetitors.add(new Competitor(4, "Cow 99", 90));

    I want to obtain a List with values -> 1, 2, 3, and 4
  }
}

我试图用类型值创建一个列表:
    // List with only types
    List<Integer> listTypes = listCompetitors.stream().sorted(
        Comparator.comparing(Competitor::getType)
    ).collect(
        Collectors.toMap(Competitor::getType, Competitor::getType)
    ); // Does n't compile!



    // List with only types
    List<Integer> listTypes = listCompetitors.stream().sorted(
        Comparator.comparing(Competitor::getType)
    ).collect(
        Collectors.groupingBy(Competitor::getType)
    ); // Does n't compile!

如何创建仅包含单个项目的列表 type来自列表listCompetitors ?

最佳答案

您只需要不同的值,因为您必须使用 Set .这是它的外观。

Set<Integer> typeList = listCompetitors.stream()
    .map(Competitor::getType)
    .collect(Collectors.toSet());

关于筛选列表值的仅一个单个值的列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56100895/

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