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$lastN

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  • Definition
  • Aggregation Accumulator
  • Syntax
  • Behavior
  • Restrictions
  • Examples
  • Array Operator
  • Syntax
  • Behavior
  • Example

New in version 5.2.

$lastN can be used as an aggregation accumulator or array operator. As an aggregation accumulator, it an aggregation of the last n elements within a group. As an array operator, it returns the specified number of elements from the end of an array.

$lastN

When $lasttN is used as an aggregation accumulator, the elements returned are meaningful only if they are in a specified sort order. If the group contains fewer than n elements, $lastN returns all elements in the group.

{
$lastN:
{
input: <expression>,
n: <expression>
}
}
  • input specifies the field(s) from the document to take the last n of. Input can be any expression.

  • n has to be a positive integral expression that is either a constant or depends on the _id value for $group. For details see group key example.

  • $lastN does not filter out null values.

  • $lastN converts missing values to null.

Consider the following aggregation that returns the last five documents from a group:

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:

  • $documents creates the literal documents that contain player scores.

  • $group groups the documents by gameId. This example has only one gameId, G1.

  • PlayerD has a missing score and PlayerE has a null score. These values are both considered as null.

  • The lastFiveScores field is specified using input : "$score" and returned as an array.

  • Since there is no sort criteria the last 5 score fields are returned.

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

Both $lastN and $bottomN accumulators can accomplish similar results.

In general:

  • If the documents coming into $group are already ordered, you should use $lastN.

  • If you're sorting and selecting the bottom n elements then you can use $bottomN to accomplish both tasks with one accumulator.

  • $lastN can be used as an aggregation expression, $bottomN cannot.

$lastN is supported as an aggregation expression.

$lastN is supported as a window operator.

Consider a gamescores collection with the following documents:

db.gamescores.insertMany([
{ playerId: "PlayerA", gameId: "G1", score: 31 },
{ playerId: "PlayerB", gameId: "G1", score: 33 },
{ playerId: "PlayerC", gameId: "G1", score: 99 },
{ playerId: "PlayerD", gameId: "G1", score: 1 },
{ playerId: "PlayerA", gameId: "G2", score: 10 },
{ playerId: "PlayerB", gameId: "G2", score: 14 },
{ playerId: "PlayerC", gameId: "G2", score: 66 },
{ playerId: "PlayerD", gameId: "G2", score: 80 }
])

You can use the $lastN accumulator to find the last three scores in a single game.

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

The example pipeline:

  • Uses $match to filter the results on a single gameId. In this case, G1.

  • Uses $group to group the results by gameId. In this case, G1.

  • Specifies the fields that are output from $lastN with output : ["$playerId"," $score"].

  • Uses $lastN to return the last three documents for the G1 game with n : 3.

The operation returns the following results:

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

You can use the $lastN accumulator to find the last n input fields in each game.

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

The example pipeline:

  • Uses $group to group the results by gameId.

  • Uses $lastN to return the last three documents for each game with n: 3.

  • Specifies the fields that are input for $lastN with input : ["$playerId", "$score"].

The operation returns the following results:

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

Using a $sort stage earlier in the pipeline can influence the results of the $lastN accumulator.

In this example:

  • {$sort : { score : -1 } } sorts the highest scores to the back of each group.

  • lastN returns the three lowest scores from the back of each group.

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

The operation returns the following results:

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

You can also assign the value of n dynamically. In this example, the $cond expression is used on the gameId field.

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

The example pipeline:

  • Uses $group to group the results by gameId.

  • Specifies the fields that input for $lastN with input : "$score".

  • If the gameId is G2 then n is 1, otherwise n is 3.

The operation returns the following results:

[
{ _id: { gameId: "G1" }, gamescores: [ 33, 99, 1 ] },
{ _id: { gameId: "G2" }, gamescores: [ 80 ] }
]

You can also use $lastN as an aggregation expression.

In this example:

  • $documents creates the literal document that contains an array of values.

  • $project is used to return the output of $lastN.

  • _id is omited from the output with _id : 0.

  • $lastN uses the input array of [10, 20, 30, 40].

  • The last three elements of the array are returned for the input document.

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

The operation returns the following results:

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

$lastN has the following syntax:

{ $lastN: { n: <expression>, input: <expression> } }
Field
Description
n
An expression that resolves to a positive integer. The integer specifies the number of array elements that $lastN returns.
input
An expression that resolves to the array from which to return n elements.
  • $lastN returns elements in the same order they appear in the input array.

  • $lastN does not filter out null values in the input array.

  • You cannot specify a value of n less than 1.

  • If the specified n is greater than or equal to the number of elements in the input array, $lastN returns the input array.

  • If input resolves to a non-array value, the aggregation operation errors.

The collection games has the following documents:

db.games.insertMany([
{ "playerId" : 1, "score" : [ 1, 2, 3 ] },
{ "playerId" : 2, "score" : [ 12, 90, 7, 89, 8 ] },
{ "playerId" : 3, "score" : [ null ] },
{ "playerId" : 4, "score" : [ ] },
{ "playerId" : 5, "score" : [ 1293, null, 3489, 9 ]},
{ "playerId" : 6, "score" : [ "12.1", 2, NumberLong("2090845886852"), 23 ]}
])

The following example uses the $lastN operator to retrieve the last three scores for each player. The scores are returned in the new field lastScores created by $addFields.

db.games.aggregate([
{ $addFields: { lastScores: { $lastN: { n: 3, input: "$score" } } } }
])

The operation returns the following results:

[{
"playerId": 1,
"score": [ 1, 2, 3 ],
"lastScores": [ 1, 2, 3 ]
},
{
"playerId": 2,
"score": [ 12, 90, 7, 89, 8 ],
"lastScores": [ 7, 89, 8 ]
},
{
"playerId": 3,
"score": [ null ],
"lastScores": [ null ]
},
{
"playerId": 4,
"score": [ ],
"lastScores": [ ]
},
{
"playerId": 5,
"score": [ 1293, null, 3489, 9 ],
"lastScores": [ null, 3489, 9 ]
},
{
"playerId": 6,
"score": [ "12.1", 2, NumberLong("2090845886852"), 23 ],
"lastScores": [ 2, NumberLong("2090845886852"), 23 ]
}]

Tip

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$last (aggregation)