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我有一个地图列表，我想按名称对其进行分组

List<Map> allObjs = new ArrayList<Map>();
Map<String, Object> src = new HashMap();
src.put("name", "asset1");
src.put("description", "desc1" );
src.put("definition", "def1" );

Map<String, Object> src2 = new HashMap();
src2.put("name", "asset1");
src2.put("description", "desc2" );
src2.put("definition", "def2" );
allObjs.add(src);
allObjs.add(src2);

我尝试将它们分组

Map<Object, List<Map>> result = allObjs.stream().collect(Collectors.groupingBy(p -> p.get("name")));

我想要的是

{
 name=asset1,
 description=["desc1","desc2"],
 definition=["def1","def2"]
}

我希望将所有其他键分组在一个列表中但我得到的是

{asset1=[
            {name=asset1, description=desc1, definition=def2},
            {name=asset1, description=desc2, definition=def2}
        ]
}

我如何将除名称之外的所有其他键分组到上面预期的列表中？

4

2 回答 2

1

您可以使用groupBywithmapping来获得所需的结果，只是值只能是单一类型，

Map<String, Set<String>> collect1 = allObjs.stream()
                .flatMap(m -> m.entrySet().stream())
                .collect(groupingBy(Map.Entry::getKey, mapping(Map.Entry::getValue, toSet())));

结果：

{
 name=[asset1],
 description=[desc1,desc2],
 definition=[def1,def2]
}
于 2020-03-30T15:02:03.830 回答
0

groupingBy将根据键对项目进行分组。它不会合并两个地图。这是实现你想要的丑陋方式之一。

给定

List<Map<String, String>> allObjs = new ArrayList<>();
Map<String, String> src = new HashMap();
src.put("name", "asset1");
src.put("description", "desc1" );
src.put("definition", "def1" );

Map<String, String> src2 = new HashMap();
src2.put("name", "asset1");
src2.put("description", "desc2" );
src2.put("definition", "def2" );
allObjs.add(src);
allObjs.add(src2);

一种方法是这样做。创建两个类，代表现有结构和最终输出结构

@Getter
@Setter
@AllArgsConstructor
@NoArgsConstructor
@ToString
public class SomeClass {
    private String name;
    private String description;
    private String definition;
}

@Getter
@Setter
@AllArgsConstructor
@NoArgsConstructor
@ToString
public class NewClass {
    private String name;
    private List<String> description;
    private List<String> definition;
}

然后这段代码应该得到你想要的输出

final List<NewClass> collect = allObjs.stream()
        .map(entry -> new SomeClass(entry.get("name"), entry.get("description"), entry.get("definition")))
        .collect(Collectors.toList())
        .stream()
        .collect(Collectors.groupingBy(SomeClass::getName))
        .entrySet()
        .stream()
        .map(entry -> {
            final List<String> descriptions = entry.getValue().stream().map(SomeClass::getDescription).collect(Collectors.toList());
            final List<String> definitions = entry.getValue().stream().map(SomeClass::getDefinition).collect(Collectors.toList());
            return new NewClass(entry.getKey(), descriptions, definitions);
        }).collect(Collectors.toList());

最终结果将是 NewClass 类型的列表。列表中的每个条目都将具有具有唯一键的所需输出。您需要合并很多空检查，因为它是我们正在处理的 Map。

于 2020-03-30T13:26:50.587 回答