如何轉換列表<object>至 Map<k, v> 使用 java stream<div id="text_translate"><p> 我想將List&lt;ObjectInList&gt;轉換為Map&lt;K, V&gt;</p><pre> class ObjectInList { List&lt;Long&gt; listWithLong; Map&lt;String, Object&gt; dataMap; // there is 'id' key, and i want to use this id as key in map to be converted }</pre><p> 新的 map 格式如下</p><pre>String type; // this type is value of dataMap. List&lt;Long&gt; contents</pre><p> List&lt;Object&gt;中的每個 Object 都可以有重復的類型</p><p>例如</p><pre>///////// 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] }</pre></div></k,></object>

Question

我想將List<ObjectInList>轉換為Map<K, V>

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
}

新的 map 格式如下

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

List<Object>中的每個 Object 都可以有重復的類型

例如

///////// 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]
}

Answer 1

您可以使用groupingBy按type分組，使用flatMapping來展平Long數據列表並收集為單個列表。

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

Answer 2

我不確定你為什么需要Map<String, Object> dataMap; 當它應該只有一個值時。 為簡單起見，我已將您的ObjectInList class 修改為

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

要獲得分組列表，我們可以這樣做 -

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

解釋：

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采用keyMapper ，它是ObjectInList::getType （根據類型進行分組）， valueMaper 是ObjectInList::getListWithLong ，因為我們有重復的鍵，我們需要提供一個mergeFunction作為(oldList, newList) -> {oldList.addAll(newList);return oldList;}

從文檔 -

合並 function，用於解決與同一鍵關聯的值之間的沖突，如提供給 {@link Map#merge(Object, Object, BiFunction)