Top documents per bucket - mongodb

Question

I would like to get the documents with the N highest fields for each of N categories. For example, the posts with the 3 highest scores from each of the past 3 months. So each month would have 3 posts that "won" for that month.
Here is what my work so far has gotten, simplified.
// simplified
db.posts.aggregate([
{$bucket: {
groupBy: "$createdAt",
boundaries: [
ISODate('2019-06-01'),
ISODate('2019-07-01'),
ISODate('2019-08-01')
],
default: "Other",
output: {
posts: {
$push: {
// ===
// This gets all the posts, bucketed by month
score: '$score',
title: '$title'
// ===
}
}
}
}},
{$match: {_id: {$ne: "Other"}}}
])
I attempted to use the $slice operator in between the // ===s, but go an error (below).
postResults: {
$each: [{
score: '$score',
title: '$title'
}],
$sort: {score: -1},
$slice: 2
}
An object representing an expression must have exactly one field: { $each: [ { score: \"$score\", title: \"$title\" } ], $sort: { baseScore: -1.0 }, $slice: 2.0 }

Answer 1

$slice you're trying to use is dedicated for update operations. To get top N posts you need to run $unwind, then $sort and $group to get ordered array. As a last step you can use $slice (aggregation), try:
db.posts.aggregate([
{$bucket: {
groupBy: "$createdAt",
boundaries: [
ISODate('2019-06-01'),
ISODate('2019-07-08'),
ISODate('2019-08-01')
],
default: "Other",
output: {
posts: {
$push: {
score: '$score',
title: '$title'
}
}
}
}},
{ $match: {_id: {$ne: "Other"}}},
{ $unwind: "$posts" },
{ $sort: { "posts.score": -1 } },
{ $group: { _id: "$_id", posts: { $push: { "score": "$posts.score", "title": "$posts.title" } } } },
{ $project: { _id: 1, posts: { $slice: [ "$posts", 3 ] } } }
])