问题

我想将 `List<ObjectInList>` 转换成 `Map<K, V>`

class ObjectInList {
    List<Long> listWithLong;
    Map<String, Object> dataMap; // 这里有一个 'id' 键，我想要将这个 id 作为转换后的映射中的键
}

新的映射格式如下
String type; // 这个 type 是 dataMap 的值
List<Long> contents

在 List<Object> 中的每个对象可以有重复的 type

例如

///////// 转换前 ////////////
[
 {
    list: [1,2,3],
    dataMap: {
      type: "a",
    }
 },
 {
    list: [4,5,6],
    dataMap: {
      type: "b",
    }
 },
 {
    list: [7,8],
    dataMap: {
      type: "a",
    }
 },
]
///////////// 转换后 //////////
{
  "a": [1,2,3,7,8],
  "b": [4,5,6]
}

<details>
<summary>英文:</summary>

I want to convert `List&lt;ObjectInList&gt;` to `Map&lt;K, V&gt;`

class ObjectInList {
List<Long> listWithLong;
Map<String, Object> dataMap; // there is 'id' key, and i want to use this id as key in map to be converted
}

The new map format is like below

String type; // this type is value of dataMap.
List<Long> contents

each Object in `List<Object>` can have duplicated type

for example

///////// before converted ////////////
[
{
list: [1,2,3],
dataMap: {
type: "a",
}
},
{
list: [4,5,6],
dataMap: {
type: "b",
}
},
{
list: [7,8],
dataMap: {
type: "a",
}
},
]
///////////// after converted //////////
{
"a": [1,2,3,7,8],
"b": [4,5,6]
}

</details>


# 答案1
**得分**: 1

你可以使用`groupingBy`按`type`进行分组，并且使用`flatMapping`来展平`Long`数据的列表，并将其收集为单个列表。

```java
Map<String, List<Long>> res = 
   objectInList
    .stream()
    .collect(Collectors.groupingBy(
                       e -> e.getDataMap().get("type"),
                       Collectors.flatMapping(
                                  e -> e.getListWithLong().stream(),
                                  Collectors.toList())
     ));

Map<String, List<Long>> res = 
   objectInList
    .stream()
    .collect(Collectors.groupingBy(
                       e -> e.getDataMap().get("type"),
                       Collectors.flatMapping(
                                  e -> e.getListWithLong().stream(),
                                  Collectors.toList())
     ));

答案2

得分: 0

我不确定为什么你需要 Map<String, Object> dataMap;，当它本应只有一个值。为了简单起见，我已经修改了你的 ObjectInList 类如下：

class ObjectInList {
  List<Long> listWithLong;
  String type; 
}

要获得分组列表，我们可以这样做：

Map<String, List<Long>> grouped =
    objectInLists.stream()
        .collect(
            Collectors.toMap(
                ObjectInList::getType,
                ObjectInList::getListWithLong,
                (oldList, newList) -> {
                  oldList.addAll(newList);
                  return oldList;
                }));

解释：

toMap 方法 -

public static <T, K, U>
    Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                    Function<? super T, ? extends U> valueMapper,
                                    BinaryOperator<U> mergeFunction)

toMap 接受 keyMapper，这里是 ObjectInList::getType（根据类型进行分组），valueMapper 是 ObjectInList::getListWithLong，由于我们有重复的键，我们需要提供一个 mergeFunction，即 (oldList, newList) -> {oldList.addAll(newList); return oldList;}。

根据文档：

一个合并函数，用于解决与同一键关联的值之间的冲突，如提供给 {@link Map#merge(Object, Object, BiFunction)}。

class ObjectInList {
  List<Long> listWithLong;
  String type; 
}

Map<String, List<Long>> grouped =
    objectInLists.stream()
        .collect(
            Collectors.toMap(
                ObjectInList::getType,
                ObjectInList::getListWithLong,
                (oldList, newList) -> {
                  oldList.addAll(newList);
                  return oldList;
                }));

public static <T, K, U>
    Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                    Function<? super T, ? extends U> valueMapper,
                                    BinaryOperator<U> mergeFunction)