$lastN (aggregation accumulator)

Null and Missing ValuesNull和缺少值

• $lastN does not filter out null values.不会筛选出空值。
• $lastN converts missing values to null.将缺少的值转换为null。

Consider the following aggregation that returns the last five documents from a group:考虑以下从组中返回最后五个文档的聚合：

db.aggregate( [
   {
      $documents: [
         { playerId: "PlayerA", gameId: "G1", score: 1 },
         { playerId: "PlayerB", gameId: "G1", score: 2 },
         { playerId: "PlayerC", gameId: "G1", score: 3 },
         { playerId: "PlayerD", gameId: "G1"},
         { playerId: "PlayerE", gameId: "G1", score: null }
      ]
   },
   {
      $group:
      {
         _id: "$gameId",
         lastFiveScores:
            {
               $lastN:
                  {
                     input: "$score",
                     n: 5
                  }
            }
      }
   }
] )

In this example:在此示例中：

• $documents creates the literal documents that contain player scores.创建包含玩家分数的文字文档。
• $group groups the documents by gameId. 按gameId对文档进行分组。This example has only one gameId, G1.这个例子只有一个gameId，G1。
• PlayerD has a missing score and PlayerE has a null score. 缺少score，而PlayerE的score为空。These values are both considered as null.这些值都被视为null。
• The lastFiveScores field is specified using input : "$score" and returned as an array.lastFiveScores字段是使用input : "$score"指定的，并作为数组返回。
• Since there is no sort criteria the last 5 score fields are returned.由于没有排序条件，因此返回最后5个score字段。

[
   {
      _id: "G1",
      lastFiveScores: [ 1, 2, 3, null, null ]
   }
]

Comparison of $lastN and $bottomN$lastN和$bottomN的比较

Both $lastN and $bottomN accumulators can accomplish similar results.$lastN和$bottomN累加器都可以实现类似的结果。

In general:一般来说：

• If the documents coming into $group are already ordered, you should use $lastN.如果进入$group的文档已经订购，则应使用$lastN。
• If you're sorting and selecting the bottom n elements then you can use $bottomN to accomplish both tasks with one accumulator.如果要排序并选择底部n个元素，那么可以使用$bottomN用一个累加器完成这两个任务。
• $lastN can be used as an aggregation expression, $bottomN cannot.$lastN可以用作聚合表达式，$bottomN不能。

Window Function and Aggregation Expression Support窗口函数和聚合表达式支持

$lastN is supported as an aggregation expression.$lastN支持作为聚合表达式。

For details on aggregation expression usage see Using $lastN as an Aggregation Expression.有关聚合表达式用法的详细信息，请参阅使用$lastN作为聚合表达式。

$lastN is supported as a window operator.支持作为窗口运算符。

Memory Limit Considerations内存限制注意事项

Aggregation pipelines which call $lastN are subject to the 100 MB limit. 调用$lastN的聚合管道受100 MB限制。If this limit is exceeded for an individual group, the aggregation fails with an error.如果单个组超过了此限制，则聚合将失败并返回错误。

Find the Last Three Player Scores for a Single Game查找单个游戏的最后三个玩家得分

You can use the $lastN accumulator to find the last three scores in a single game.您可以使用$lastN累加器查找单个游戏中的最后三个分数。

db.gamescores.aggregate( [
   {
      $match : { gameId : "G1" }
   },
   {
      $group:
         {
            _id: "$gameId",
            lastThreeScores:
               {
                  $lastN:
                  {
                     input: ["$playerId", "$score"],
                     n:3
                  }
               }
         }
   }
] )

The example pipeline:示例管道：

• Uses $match to filter the results on a single gameId. 使用$match筛选单个gameId上的结果。In this case, G1.在本例中为G1。
• Uses $group to group the results by gameId. 使用$group按gameId对结果进行分组。In this case, G1.在本例中为G1。
• Specifies the fields that are output from $lastN with output : ["$playerId"," $score"].使用output : ["$playerId"," $score"]指定从$lastN输出的字段。
• Uses $lastN to return the last three documents for the G1 game with n : 3.使用$lastN返回编号为n:3的G1游戏的最后三个文档。

The operation returns the following results:该操作返回以下结果：

[
   {
      _id: "G1",
      lastThreeScores: [ [ "PlayerB", 33 ], [ "PlayerC", 99 ], [ "PlayerD", 1 ] ]
   }
]

Finding the Last Three Player Scores Across Multiple Games在多场比赛中查找最后三名球员的得分

You can use the $lastN accumulator to find the last n input fields in each game.您可以使用$lastN累加器查找每个游戏中最后n个输入字段。

db.gamescores.aggregate( [
      {
         $group:
         {
         _id: "$gameId", playerId:
            {
               $lastN:
                  {
                     input: [ "$playerId","$score" ],
                     n: 3
                  }
            }
         }
      }
] )

The example pipeline:示例管道：

• Uses $group to group the results by gameId.使用$group按gameId对结果进行分组。
• Uses $lastN to return the last three documents for each game with n: 3.使用$lastN返回n:3的每个游戏的最后三个文档。
• Specifies the fields that are input for $lastN with input : ["$playerId", "$score"].使用input : ["$playerId", "$score"]指定$lastN的输入字段。

The operation returns the following results:该操作返回以下结果：

[
   {
      _id: 'G2',
      playerId: [ [ 'PlayerB', 14 ], [ 'PlayerC', 66 ], [ 'PlayerD', 80 ] ]
   },
   {
      _id: 'G1',
      playerId: [ [ 'PlayerB', 33 ], [ 'PlayerC', 99 ], [ 'PlayerD', 1 ] ]
   }
]

Using $sort With $lastN将$sort与$lastN一起使用

Using a $sort stage earlier in the pipeline can influence the results of the $lastN accumulator.在管道的前面使用$sort阶段可能会影响$lastN累加器的结果。

In this example:在此示例中：

• {$sort : { score : -1 } } sorts the highest scores to the back of each group.将最高分数排序到每组后面。
• lastN returns the three lowest scores from the back of each group.返回每组后面的三个最低分数。

db.gamescores.aggregate( [
      { $sort : { score : -1 } },
      {
         $group:
         { _id: "$gameId", playerId:
            {
               $lastN:
                  {
                     input: [ "$playerId","$score" ],
                     n: 3
                  }
            }
         }
      }
   ] )

The operation returns the following results:该操作返回以下结果：

[
   {
      _id: 'G2',
      playerId: [ [ 'PlayerC', 66 ], [ 'PlayerB', 14 ], [ 'PlayerA', 10 ] ]
   },
   {
      _id: 'G1',
      playerId: [ [ 'PlayerB', 33 ], [ 'PlayerA', 31 ], [ 'PlayerD', 1 ] ]
   }
]

Computing n Based on the Group Key for $group基于$Group的组密钥计算n

You can also assign the value of n dynamically. 还可以动态指定n的值。In this example, the $cond expression is used on the gameId field.在本例中，$cond表达式用于gameId字段。

db.gamescores.aggregate([
   {
      $group:
      {
         _id: {"gameId": "$gameId"},
         gamescores:
            {
               $lastN:
                  {
                     input: "$score",
                     n: { $cond: { if: {$eq: ["$gameId","G2"] }, then: 1, else: 3 } }
                  }
            }
      }
   }
] )

The example pipeline:示例管道：

• Uses $group to group the results by gameId.使用$group按gameId对结果进行分组。
• Specifies the fields that input for $lastN with input : "$score".使用input : "$score"指定$lastN的输入的字段。
• If the gameId is G2 then n is 1, otherwise n is 3.如果gameId为G2，则n为1，否则n为3。

The operation returns the following results:该操作返回以下结果：

[
   { _id: { gameId: "G2" }, gamescores: [ 10 ] },
   { _id: { gameId: "G1" }, gamescores: [ 33, 31, 1 ] }
]

Using $lastN as an Aggregation Expression使用$lastN作为聚合表达式

You can also use $lastN as an aggregation expression.还可以使用$lastN作为聚合表达式。

In this example:在此示例中：

• $documents creates the literal document that contains an array of values.创建包含值数组的文本文档。
• $project is used to return the output of $lastN.用于返回$lastN的输出。
• _id is omited from the output with _id : 0.在_id:0的输出中省略。
• $lastN uses the input array of [10, 20, 30, 40].使用输入数组[10, 20, 30, 40]。
• The last three elements of the array are returned for the input document.为输入文档返回数组的最后三个元素。

db.aggregate( [
   {
      $documents: [
         { array: [10, 20, 30, 40] } ]
   },
   { $project: {
      lastThreeElements:{
                           $lastN:
                           {
                              input: "$array",
                              n: 3
                           }
                        }
               }
   }
] )

The operation returns the following results:该操作返回以下结果：

[ { lastThreeElements: [ 20, 30, 40 ] } ]