使用流按多个属性分组和排序: Java 8

Question

我有MainEntity的名单

public class MainEntity {
private String keyword;
private double cost;
private String company;
}

我有CompanyEntity

public class CompanyEntity {
    private double cost;
    private String company;
    }

我正在尝试将我的列表转换为Map<String,List<CompanyEntity>>，其中的键将是keyword，List<CompanyEntity>将有所有成本的平均值，并进行排序。我正试图在流和Java 8中完成它。

对于一个特定的关键字作为输入，我正在这样做。

List<MainEntity> entityList = keyWordMap.get(entity.getKeyword());
        entityList.add(entity);
        keyWordMap.put(entity.getKeyword(), entityList);

Map<String, Double> average = (keyWordMap.get(keyword)).stream()
            .collect(groupingBy(MainEntity::getCompany,
                    Collectors.averagingDouble(MainEntity::getCtr)));
    result.put(keyword, new ArrayList<>());

    for (Map.Entry<String, Double> entity : average.entrySet()) {
        result.get(keyword).add(new CompanyEntity(entity.getKey(), entity.getValue()));
    }

但我试着为所有关键词创建一个地图。是否有可能再次迭代整个列表？目前，keyowordMap是Map<String,MainEntity>类型的，我是通过迭代MainEntity列表来实现的，但是我想要Map<String,List<MainEntity>>。

Answer 1

首先，制作一个keyWordMap

Map<String, List<MainEntity>> keyWordMap = 
            mainEntityList
            .stream()
            .collect(Collectors.groupingBy(MainEntity::getKeyword));

然后迭代映射，对于每个关键字，可以根据平均值直接获取CompanyEntity排序列表，并使用map()转换数据并将其作为列表收集，然后放入result中。

Map<String,List<CompanyEntity>> result = ....
for (Map.Entry<String, List<MainEntity> entry : keyWordMap.entrySet()) {
    List<CompanyEntity> list = entry.getValue().stream()
                .collect(groupingBy(MainEntity::getCompany,
                        Collectors.averagingDouble(MainEntity::getCtr)))
                .entrySet()
                .stream()
                .sorted(Comparator.comparing(e -> e.getValue()))
                .map(e -> new CompanyEntity(e.getKey(), e.getValue()))
                .collect(Collectors.toList());
    result.put(entry.getKey(), list);
}

或者你想一蹴而就

Map<String,List<CompanyEntity>> mapData = 
           mainEntityList
            .stream()
            .collect(Collectors.groupingBy(MainEntity::getKeyWord,
                     Collectors.groupingBy(MainEntity::getCtr,
                        Collectors.averagingDouble(MainEntity::getCtr))))
            .entrySet()
            .stream()
            .collect(Collectors.toMap(m -> m.getKey(),
                 m -> m.entrySet()
                       .stream()
                       .sorted(Comparator.comparing(e -> e.getValue()))
                       .map(e -> new CompanyEntity(e.getKey(), e.getValue()))
                       .collect(Collectors.toList())));

Answer 2

另一个答案在最初误解了问题后完全改变了它的答案，并且在良好的StackOverflow精神中它吸引了第一个向上的选票，所以现在被接受并且被最高的支持。但是在代码中还有几个步骤显示了正在发生的事情：

这会让你得到这样的结果：

import java.io.IOException;
import java.util.AbstractMap.SimpleEntry;
import java.util.Comparator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.util.stream.Stream;
import lombok.Value;

public class CompanyEntityStackOverflowQuestion {

    public static void main(String[] args) throws IOException {

        //setup test data
        MainEntity one = new MainEntity("key1", 10D, "company1");
        MainEntity two = new MainEntity("key2", 5D, "company2");
        MainEntity three = new MainEntity("key1", 7D, "company3");
        MainEntity four = new MainEntity("key2", 3D, "company4");
        List<MainEntity> mainEntityList = List.of(one, two, three, four);

        //group list by keyword
        Map<String, List<MainEntity>> mainEntityByKeyword = mainEntityList.stream()
            .collect(Collectors.groupingBy(MainEntity::getKeyword));

        //map to companyEntity object
        Stream<SimpleEntry<String, List<CompanyEntity>>> mapped = mainEntityByKeyword.entrySet().stream()
            .map(entry -> new SimpleEntry<>(entry.getKey(), entry.getValue().stream().map(
                getCompanyListFunction()).collect(Collectors.toList())));

        //sort and calculate average
        Stream<SimpleEntry<String, CompanyEntityListWithStats>> mappedToListWithStats = mapped
            .map(entry -> new SimpleEntry<>(entry.getKey(),
                new CompanyEntityListWithStats(entry.getValue().stream().mapToDouble(company -> company.cost).average().orElse(0D), //or use Collectors.averagingDouble(company -> company.cost))
                    sortList(entry.getValue()))));

        //collect back to map
        Map<String, CompanyEntityListWithStats> collect = mappedToListWithStats
            .collect(Collectors.toMap(Entry::getKey, Entry::getValue));

        //show result
        System.out.println(collect);
    }

    //sort by cost
    private static List<CompanyEntity> sortList(List<CompanyEntity> list) {
        list.sort(Comparator.comparing(company -> company.cost));
        return list;
    }

    //map MainEntity to CompanyEntity
    private static Function<MainEntity, CompanyEntity> getCompanyListFunction() {
        return mainEntity -> new CompanyEntity(mainEntity.cost, mainEntity.company);
    }

    @Value
    public static class MainEntity {

        public String keyword;
        public double cost;
        public String company;
    }

    @Value
    public static class CompanyEntity {

        public double cost;
        public String company;
    }

    @Value
    public static class CompanyEntityListWithStats {

        public double average;
        public List<CompanyEntity> companyList;
    }


}

输出：{key1=CompanyEntityStackOverflowQuestion.CompanyEntityListWithStats(average=8.5, companyList=[CompanyEntityStackOverflowQuestion.CompanyEntity(cost=7.0, company=company3), CompanyEntityStackOverflowQuestion.CompanyEntity(cost=10.0, company=company1)]), key2=CompanyEntityStackOverflowQuestion.CompanyEntityListWithStats(average=4.0, companyList=[CompanyEntityStackOverflowQuestion.CompanyEntity(cost=3.0, company=company4), CompanyEntityStackOverflowQuestion.CompanyEntity(cost=5.0, company=company2)])}

您可能可以跳过一些步骤，这只是快速打印出来。当然，你可以用内嵌的东西来让它看起来更短/更干净，但是这种格式显示了正在发生的事情。