java 8 如何在多个属性上获取不同的列表

斯图尔特·马克斯（Stuart Marks）的回答几乎是一字不差的：

import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.concurrent.ConcurrentHashMap;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.stream.Collectors;

class Class {

  public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
    Map<Object, Boolean> seen = new ConcurrentHashMap<>();
    return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
  }

  private static List<Pojo> getList() {
    return Arrays.asList(
      new Pojo("123", 100),
      new Pojo("123", 100),
      new Pojo("123", 100),
      new Pojo("456", 200)
    );
  }

  public static void main(String[] args) {

    System.out.println(getList().stream()
      // extract a key for each Pojo in here. 
      // concatenating name and price together works as an example
      .filter(distinctByKey(p -> p.getName() + p.getPrice()))
      .collect(Collectors.toList()));
  }

}

class Pojo {
  private final String name;
  private final Integer price;

  public Pojo(final String name, final Integer price) {
    this.name = name;
    this.price = price;
  }

  public String getName() {
    return name;
  }

  public Integer getPrice() {
    return price;
  }

  @Override
  public String toString() {
    final StringBuilder sb = new StringBuilder("Pojo{");
    sb.append("name='").append(name).append('\'');
    sb.append(", price=").append(price);
    sb.append('}');
    return sb.toString();
  }
}

此主要方法产生：

[Pojo{name='123'， price=100}， Pojo{name='456'， price=200}]

编辑

根据尤金的提示，定价。int

注意：如果你想充实它，你可以使用更有趣的东西作为键：

import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.concurrent.ConcurrentHashMap;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.stream.Collectors;

class Class {

  public static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
    Map<Object, Boolean> seen = new ConcurrentHashMap<>();
    return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
  }

  private static List<Pojo> getList() {
    return Arrays.asList(
      new Pojo("123", 100),
      new Pojo("123", 100),
      new Pojo("123", 100),
      new Pojo("456", 200)
    );
  }

  private static class NameAndPricePojoKey {
    final String name;
    final int price;

    public NameAndPricePojoKey(final Pojo pojo) {
      this.name = pojo.getName();
      this.price = pojo.getPrice();
    }

    @Override
    public boolean equals(final Object o) {
      if (this == o) return true;
      if (o == null || getClass() != o.getClass()) return false;

      final NameAndPricePojoKey that = (NameAndPricePojoKey) o;

      if (price != that.price) return false;
      return name != null ? name.equals(that.name) : that.name == null;

    }

    @Override
    public int hashCode() {
      int result = name != null ? name.hashCode() : 0;
      result = 31 * result + price;
      return result;
    }
  }

  public static void main(String[] args) {

    System.out.println(getList().stream()
      // extract a key for each Pojo in here. 
      .filter(distinctByKey(NameAndPricePojoKey::new))
      .collect(Collectors.toList()));
  }

}

class Pojo {
  private String name;
  private Integer price;
  private Object otherField1;
  private Object otherField2;

  public Pojo(final String name, final Integer price) {
    this.name = name;
    this.price = price;
  }

  public String getName() {
    return name;
  }

  public void setName(final String name) {
    this.name = name;
  }

  public Integer getPrice() {
    return price;
  }

  public void setPrice(final Integer price) {
    this.price = price;
  }

  public Object getOtherField1() {
    return otherField1;
  }

  public void setOtherField1(final Object otherField1) {
    this.otherField1 = otherField1;
  }

  public Object getOtherField2() {
    return otherField2;
  }

  public void setOtherField2(final Object otherField2) {
    this.otherField2 = otherField2;
  }

  @Override
  public boolean equals(final Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;

    final Pojo pojo = (Pojo) o;

    if (name != null ? !name.equals(pojo.name) : pojo.name != null) return false;
    if (price != null ? !price.equals(pojo.price) : pojo.price != null) return false;
    if (otherField1 != null ? !otherField1.equals(pojo.otherField1) : pojo.otherField1 != null) return false;
    return otherField2 != null ? otherField2.equals(pojo.otherField2) : pojo.otherField2 == null;

  }

  @Override
  public int hashCode() {
    int result = name != null ? name.hashCode() : 0;
    result = 31 * result + (price != null ? price.hashCode() : 0);
    result = 31 * result + (otherField1 != null ? otherField1.hashCode() : 0);
    result = 31 * result + (otherField2 != null ? otherField2.hashCode() : 0);
    return result;
  }

  @Override
  public String toString() {
    final StringBuilder sb = new StringBuilder("Pojo{");
    sb.append("name='").append(name).append('\'');
    sb.append(", price=").append(price);
    sb.append(", otherField1=").append(otherField1);
    sb.append(", otherField2=").append(otherField2);
    sb.append('}');
    return sb.toString();
  }
}

我会选择这样的东西，它相当简单和灵活，并以您的示例为基础：

public static <T> List<T> distinctList(List<T> list, Function<? super T, ?>... keyExtractors) {

    return list
        .stream()
        .filter(distinctByKeys(keyExtractors))
        .collect(Collectors.toList());
}

private static <T> Predicate<T> distinctByKeys(Function<? super T, ?>... keyExtractors) {

    final Map<List<?>, Boolean> seen = new ConcurrentHashMap<>();

    return t -> {

        final List<?> keys = Arrays.stream(keyExtractors)
            .map(ke -> ke.apply(t))
            .collect(Collectors.toList());

        return seen.putIfAbsent(keys, Boolean.TRUE) == null;

    };

}

然后可以通过以下方式调用它：

final List<Xyz> distinct = distinctList(list, Xyz::getName, Xyz::getPrice)