如何解决使用流按多个属性分组和排序:Java 8
我有 MainEntity
的列表public class MainEntity {
private String keyword;
private double cost;
private String company;
}
我有 CompanyEntity
public class CompanyEntity {
private double cost;
private String company;
}
我正在尝试将列表转换成Map<String,List<CompanyEntity>>
,其中键将为keyword
,而List<CompanyEntity>
将具有所有成本的平均值,并且也将其排序。我正在尝试在流和Java 8中实现。
对于输入的特定关键字,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()));
}
但是我试图为所有关键字创建一个地图。是否可能或再次遍历整个列表有意义?
当前keyowordMap
的类型为Map<String,MainEntity>
,我通过迭代MainEntity
的列表来做到这一点,但是我想要Map<String,List<MainEntity>>
。
解决方法
首先,制作一个keyWordMap
Map<String,List<MainEntity>> keyWordMap =
mainEntityList
.stream()
.collect(Collectors.groupingBy(MainEntity::getKeyword));
然后迭代地图,对于每个关键字,您可以直接获取按平均值排序的CompanyEntity
列表,并使用map()
转换数据并收集为List,然后放入{{1 }}
result
或者您想一次性完成
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);
}
,
在最初误解了该问题之后,另一个答案完全改变了答案,并以良好的StackOverflow精神吸引了第一个赞,因此现在被接受并被最高赞。但这在代码中显示了正在发生的事情的更多步骤:
这应该为您带来结果:
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)])}
您也许可以跳过一些步骤,只需快速输入即可。当然,您可以内联东西以使其看起来更短/更干净,但是这种格式可以显示正在发生的事情。
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