Question

如何从对象列表中获取不同的（基于两个属性的不同）列表。例如，让我们有属性名称和价格的对象列表。现在我如何获得具有不同名称或价格的清单假设

list<xyz> l1 = getlist(); // getlist will return the list.

现在让l1具有以下属性（名称，价格）： -
n1，p1
n1，p2
n2，p1
n2，p3

现在在过滤器之后，列表应该是 -
n1，p1
n2，p3

我试着像这样解决 -

public List<xyz> getFilteredList(List<xyz> l1) {

        return l1
                .stream()
                .filter(distinctByKey(xyz::getName))
                .filter(distinctByKey(xyz::getPrice))
                .collect(Collectors.toList());
    }

    private 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;
    }

现在的问题是，当我对名称进行过滤时，列表返回将是 -
n1，p1
n2，p1

然后它会在价格上运行过滤器返回 -
n1，p1

这不是预期的结果。

Answer 1

几乎逐字逐句Stuart Marks' answer：

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();
  }
}

这个主要方法产生：

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

修改

根据Eugene的提示，价格为int。

注意：如果你想充实它，你可以使用更有趣的东西：

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();
  }
}

Answer 2

以下是基于课程Item的解决方案，该课程定义了name和price：

public class Item {

    public String name;
    public double price;

    Item(String name, double price) {
        this.name = name;
        this.price = price;
    }
}

要求是仅从给定的Item获取List<Item>个name，其中price 和不同ItemWrapper s它们发生的顺序。

我发现这个要求是由类public class ItemWrapper { Item item; ItemWrapper(Item item) { this.item = item; } @Override public boolean equals(Object obj) { if (!(obj instanceof ItemWrapper)) return false; ItemWrapper other = (ItemWrapper) obj; return Objects.equals(item.name, other.item.name) || item.price == other.item.price; } @Override public int hashCode() { return 1; } }区分开来的：

List<Item>

现在我们已经准备好了所有内容来过滤给定的List<Item> items = Arrays.asList( new Item("name-1", 100.00), new Item("name-1", 555.00), new Item("name-2", 100.00), new Item("name-2", 999.99), new Item("name-3", 100.00), new Item("name-4", 555.00), new Item("name-5", 999.99) );项：

items.stream()
     .map(item -> new ItemWrapper(item))
     .distinct()
     .map(wrapper -> wrapper.item)
     .collect(Collectors.toList());
}

如下：

import React, { Component } from 'react';
import {
  AppRegistry,
  StyleSheet,
  Text,
  View,
  Navigator,
  TouchableOpacity,
  Image,
  TouchableHighlight,
  Alert,
  TextInput
} from 'react-native';
import Button from 'react-native-button'
import {Actions} from 'react-native-router-flux'
import Home from './Home'

export class Temp extends Component{
constructor(props) {
  super(props);
   this.state = {
    data: '',
    textinput:'',
  }
   state={
            shouldShow: false
        }
}

    componentDidMount(){
    this._onPressButtonGET();
  } 

      _onPressButtonPOST(){
        fetch("url", {
            method: "POST",  
             headers: { 
                 'Accept': 'application/json',
                 'Content-Type': 'application/json',
             },
            body: JSON.stringify({
                "entryDate":"3/2/2017 2:00 AM", 
                "systol": this.state.textinput,
                "mobileType":"ANDROID",
                "userName":"menutest",

               })})
        .then((response) => response.json())
        .then((responseData) => {
            Alert.alert(
                "Blood pressure data",
                "Blood pressure data - " + JSON.stringify(responseData)
            )
        }).catch((error) => {
        console.error(error);
        })
        .done(); 
    }

    _onPressButtonGET(){
        fetch("url", {
            method: "POST",
             headers: {
                 'Accept': 'application/json',
                 'Content-Type': 'application/json',
             },
            body: JSON.stringify({"mobileType":"ANDROID","userName":"menutest"})})
        .then((response) => response.json())
        .then((responseData) => { 

                this.setState({ data : JSON.stringify(responseData)})

            }) .catch((error) => {
        console.error(error);
      })

       .done();
    }
    render(){
        return(

            <View>
                <TouchableHighlight onPress={this._onPressButtonPOST.bind(this)}>
                    <Text>Add</Text> 
                </TouchableHighlight>

            <TouchableOpacity style= {{left:300,top:-20, }}
 onPress={()=>{ this.setState({ shouldShow: !this.state.shouldShow })}}
><Text>Edit</Text></TouchableOpacity>

{this.state.shouldShow ? <TextInput placeholder='systol' 
            onChangeText={(text) => this.setState({textinput: text})}
           /> : null}

                 <TouchableHighlight onPress={this._onPressButtonGET.bind(this)}>
                    <Text>show</Text>
                   </TouchableHighlight>

                <Text>{this.state.data}</Text>  
            </View>
    );
    }
}


module.exports = Temp;

捕获的项目是：

name = name-1，price = 100.0
name = name-2，price = 999.99
name = name-4，price = 555.0

Answer 3

我会选择这样的东西，这是相当简单和灵活的，并以你的榜样为基础：

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)

java 8如何在多个属性上获取不同的列表

3 个答案:

修改