I am very new to MongoDB and have been trying to create a new field within my collection that is calculated using existing data.
Is there a way to add the field myRating to the movies collection?
Here is what I came up with.
db.movies.aggregate([
{$unwind: "$genres"},
{$project:{_id:0, title:1, genres:1,
durationScore: {$cond: {if: {$gte: ["$runtime", 90]}, then: 10, else: 5}},
yearScore: {$cond: {if: {$gte: ["$year", 1990]}, then: 10, else: 5}},
genreScore: {$switch:{branches:[
{
case: {$eq :["$genres", "Action"]}, "then": 30 ,
},
{
case: {$eq :["$genres", "Western"]}, "then": 20 ,
},
{
case: {$eq :["$genres", "Comedy"]}, "then": 5 ,
},
{
case: {$eq :["$genres", "Drama"]}, "then": 15 ,
},
],
default: 10
}},
directorScore: {$switch:{branches:[
{
case: {$eq :["$director", "Quentin Tarantino"]}, "then": 20 ,
},
{
case: {$eq :["$director", "Martin Scorsese"]}, "then": 20 ,
},
],
default: 10
}}
}},
{$addFields: { myRating: { $sum: [ "$yearScore", "$durationScore", "$genreScore", "$directorScore" ]}}},
])
Sample of Data.
{
"_id": {
"$oid": "60502686eb0d3e3e849677ef"
},
"title": "Once Upon a Time in the West",
"year": 1968,
"rated": "PG-13",
"runtime": 175,
"countries": [
"Italy",
"USA",
"Spain"
],
"genres": [
"Western"
],
"director": "Sergio Leone",
"writers": [
"Sergio Donati",
"Sergio Leone",
"Dario Argento",
"Bernardo Bertolucci",
"Sergio Leone"
],
"actors": [
"Claudia Cardinale",
"Henry Fonda",
"Jason Robards",
"Charles Bronson"
],
"plot": "Epic story of a mysterious stranger with a harmonica who joins forces with a notorious desperado to protect a beautiful widow from a ruthless assassin working for the railroad.",
"poster": "http://ia.media-imdb.com/images/M/MV5BMTEyODQzNDkzNjVeQTJeQWpwZ15BbWU4MDgyODk1NDEx._V1_SX300.jpg",
"imdb": {
"id": "tt0064116",
"rating": 8.6,
"votes": 201283
},
"tomato": {
"meter": 98,
"image": "certified",
"rating": 9,
"reviews": 54,
"fresh": 53,
"consensus": "A landmark Sergio Leone spaghetti western masterpiece featuring a classic Morricone score.",
"userMeter": 95,
"userRating": 4.3,
"userReviews": 64006
},
"metacritic": 80,
"awards": {
"wins": 4,
"nominations": 5,
"text": "4 wins & 5 nominations."
},
"type": "movie"
}
I would suggest you keep _id field in $project stage.
Without considering performance, simply iterating through the aggregate result and $set myRating field through updateOne using the _id field.
db.movies.aggregate([
...
{$project:{_id:1, title:1, genres:1,
...
]).forEach(result = > {
db.movies.updateOne(
{_id : result._id},
{$set : {myRating : {result.myRating}}
})
})
Starting in MongoDB 4.2, you can use the aggregation pipeline for update operations. Try this query:
db.movies.updateOne(
{ "_id": ObjectId("60502686eb0d3e3e849677ef") },
[
{
$set: {
myRating: {
$let: {
vars: {
durationScore: { $cond: { if: { $gte: ["$runtime", 90] }, then: 10, else: 5 } },
yearScore: { $cond: { if: { $gte: ["$year", 1990] }, then: 10, else: 5 } },
genreScore: {
$switch: {
branches: [
{ case: { $in: ["Action", "$genres"] }, "then": 30 },
{ case: { $in: ["Western", "$genres"] }, "then": 20 },
{ case: { $in: ["Comedy", "$genres"] }, "then": 5 },
{ case: { $in: ["Drama", "$genres"] }, "then": 15 }
],
default: 10
}
},
directorScore: {
$switch: {
branches: [
{ case: { $eq: ["$director", "Quentin Tarantino"] }, "then": 20 },
{ case: { $eq: ["$director", "Martin Scorsese"] }, "then": 20 }
],
default: 10
}
}
},
in: { $sum: ["$$yearScore", "$$durationScore", "$$genreScore", "$$directorScore"] }
}
}
}
}
]
);
Related
I'm working on simple program that counts total number of special units through n number of players.
I have documents similar to this (simplified), where array rosterUnits could be of length 0 to 7. There is a total of 7 special units. I need to know how many of each unit players have in roster.
{
{
_id: ObjectId(...),
member: {
rosterUnits: [ "Unit1", "Unit2", "Unit3", "Unit4"]
}
},
{
_id: ObjectId(...),
member: {
rosterUnits: [ "Unit1", "Unit3"]
}
},
...
}
Expected result would be something like this:
{
_id: ...
result: [
{
name: "Unit1"
count: 2
},
{
name: "Unit2"
count: 1
},
{
name: "Unit3"
count: 2
},
...
{
name: "Unit7"
count: 0
}
]
}
How do I achieve this using aggregate pipeline?
EDIT (2/7/2023)
Excuse me everyone, I thought I provided enough details here but...
Documents are very big and pipeline until this stage is very long.
I wanted to spare you the trouble with the documents
I have guild with up to 50 players. I search for guild then $unwind members of guild and $lookup into members to get member.rosterUnit(s).
This is a full query I came up with:
db.getCollection('guilds').aggregate([
{ $match: { 'profile.id': 'jrl9Q-_CRDGdMyNjTQH1rQ' } },
//{ $match: { 'profile.id': { $in : ['jrl9Q-_CRDGdMyNjTQH1rQ', 'Tv_j9nhRTgufvH7C7oUYAA']} } },
{ $project: { member: 1, profile: 1 } },
{ $unwind: "$member" },
{
$lookup: {
from: "players",
localField: "member.playerId",
foreignField: "playerId",
pipeline: [
{
$project: {
profileStat: 1,
rosterUnit: {
$let: {
vars: { gls: ["JABBATHEHUTT:SEVEN_STAR", "JEDIMASTERKENOBI:SEVEN_STAR", "GRANDMASTERLUKE:SEVEN_STAR", "LORDVADER:SEVEN_STAR", "GLREY:SEVEN_STAR", "SITHPALPATINE:SEVEN_STAR", "SUPREMELEADERKYLOREN:SEVEN_STAR"], },
in: {
$reduce: {
input: "$rosterUnit",
initialValue: [],
in: {
$cond: {
if: { $gt: [{ $indexOfArray: ["$$gls", "$$this.definitionId"] }, -1] },
then: { $concatArrays: ["$$value", [{ definitionId: "$$this.definitionId", count: 1 }]] },
else: { $concatArrays: ["$$value", []] }
}
},
}
}
}
}
}
}
],
as: "member"
}
},
{
$addFields: {
member: { $arrayElemAt: ["$member", 0] },
gpStats: {
$let: {
vars: { member: { $arrayElemAt: ["$member", 0] } },
in: {
$reduce: {
input: "$$member.profileStat",
initialValue: {},
in: {
characterGp: {
$arrayElemAt: [
"$$member.profileStat.value",
{
$indexOfArray: [
"$$member.profileStat.nameKey",
"STAT_CHARACTER_GALACTIC_POWER_ACQUIRED_NAME"
]
}
]
},
shipGp: {
$arrayElemAt: [
"$$member.profileStat.value",
{
$indexOfArray: [
"$$member.profileStat.nameKey",
"STAT_SHIP_GALACTIC_POWER_ACQUIRED_NAME"
]
}
]
}
}
}
}
}
}
}
},
{
$group: {
_id: "$profile.id",
guildName: { $first: "$profile.name" },
memberCount: { $first: "$profile.memberCount" },
guildGp: { $first: "$profile.guildGalacticPower" },
totalGp: { $sum: { $sum: [{ $toInt: "$gpStats.characterGp" }, { $toInt: "$gpStats.shipGp" }] } },
avgTotalGp: { $avg: { $sum: [{ $toInt: "$gpStats.characterGp" }, { $toInt: "$gpStats.shipGp" }] } },
characterGp: { $sum: { $toInt: "$gpStats.characterGp" } },
shipGp: { $sum: { $toInt: "$gpStats.shipGp" } },
}
}
])
I want to add new field in group with desired result from above.
If I do $unwind on member.rosterUnit how do I go back to member grouping?
(Excuse me once again, this is my first question)
Use $unwind to deconstruct the rosterUnits array into separate documents.
Then use $group to group the documents by the rosterUnits values and calculate the count for each unit.
Then use $project to format the output to include only the name and count fields.
db.collection.aggregate([
{
$unwind: "$member.rosterUnits"
},
{
$group: {
_id: "$member.rosterUnits",
count: { $sum: 1 }
}
},
{
$project: {
_id: 0,
name: "$_id",
count: "$count"
}
}
])
Yes I think that the best way of do that is using aggregations.
I'm sure there is a better way to do it.
But here is the solution, I hope it works for you friend.
Basically we are going to use a "$group" aggregation and within it using an operator "$cond" and "$in" we are going to validate case by case if the searched element is found. In the case that it is so, we will add one and if the element is not found, zero.
I recommend you download mongodb compass to try it
Aggregation:
[{
$group: {
_id: null,
Unit1: {
$sum: {
$cond: [
{
$in: [
'Unit1',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit2: {
$sum: {
$cond: [
{
$in: [
'Unit2',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit3: {
$sum: {
$cond: [
{
$in: [
'Unit3',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit4: {
$sum: {
$cond: [
{
$in: [
'Unit4',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit5: {
$sum: {
$cond: [
{
$in: [
'Unit5',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit6: {
$sum: {
$cond: [
{
$in: [
'Unit6',
'$member.rosterUnits'
]
},
1,
0
]
}
},
Unit7: {
$sum: {
$cond: [
{
$in: [
'Unit7',
'$member.rosterUnits'
]
},
1,
0
]
}
}
}
}, {
$project: {
_id: 0
}
}]
Query
because you want to count values that might not exists, you can make the groups manualy, and do conditional count
after the group you can do extra tranformation(if you really need the expected outpute exactly like that). Object to array, and map to give the field names(name,count)
Playmongo
aggregate(
[{"$unwind": "$member.rosterUnits"},
{"$group":
{"_id": null,
"Unit1":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit1"]}, 1, 0]}},
"Unit2":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit2"]}, 1, 0]}},
"Unit3":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit3"]}, 1, 0]}},
"Unit4":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit4"]}, 1, 0]}},
"Unit5":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit5"]}, 1, 0]}},
"Unit6":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit6"]}, 1, 0]}},
"Unit7":
{"$sum":
{"$cond": [{"$eq": ["$member.rosterUnits", "Unit7"]}, 1, 0]}}}},
{"$unset": ["_id"]},
{"$project":
{"result":
{"$map":
{"input": {"$objectToArray": "$$ROOT"},
"in": {"name": "$$this.k", "count": "$$this.v"}}}}}])
How can I update a value in a document based on applying functions to another field (which is in a different embedded document)?
With the sample data below, I want to
get the col field for the farm having id 12
multiply that by 0.025
add the current value of the statistic.crypt field
ensure the value is a double by converting it with $toDouble
store the result back into statistic.crypt
data:
{
"_id": {
"$oid": "6128c238c144326c57444227"
},
"statistic": {
"balance": 112570,
"diamond": 14,
"exp": 862.5,
"lvl": 76,
"mn_exp": 2.5,
"lvl_mn_exp": 15,
"coll_ms": 8047,
"all_exp": 67057.8,
"rating": 0,
"crypt": 0
},
"inventory": {
"farm": [{
"id": 12,
"col": 100,
"currency": "diamond",
"cost": 2,
"date": "2021-09-02 18:58:39"
}, {
"id": 14,
"col": 1,
"currency": "diamond",
"cost": 2,
"date": "2021-09-02 16:57:08"
}],
"items": []
},
...
}
My initial attempt is:
self.collection
.update_many({"inventory.farm.id": 12}, [{
"$set": {
"test": {
'$toDouble': {
"$sum": [
{'$multiply':["$inventory.farm.$[].col", 0.025]},
'$test'
]
}
} }
},])
This does not work as it applies to test rather than statistic.crypt, and I cannot figure out how to modify it to apply to statistic.crypt.
A field can be updated based on another in the following stages:
add a field containing the farm
set statistic.crypt to the result of the mathematical expression (applied to the newly embedded farm)
remove extra fields
In code:
self.collection.update_many({"inventory.farm.id": 12 }, [
{
$addFields: {
hh: {
$filter: {
input: "$inventory.farm",
as: "z",
cond: { $eq: ["$$z.id", 12] },
},
},
},
},
{
$set: {
"statistic.crypt": {
$toDouble: {
$sum: [
{
$multiply: [{ $first: "$hh.col" }, 0.025],
},
"statistic.crypt",
],
},
},
},
},
{
$project: {
id_pr: 1,
id_server: 1,
role: 1,
warns: 1,
id_clan: 1,
statistic: 1,
design: 1,
date: 1,
inventory: 1,
voice: 1,
},
},)
I have a collection called shows, with documents as:
{
"url": "http://www.tvmaze.com/shows/167/24",
"name": "24",
"genres": [
"Drama",
"Action"
],
"runtime": 60
},
{
"url": "http://www.tvmaze.com/shows/4/arrow",
"name": "Arrow",
"genres": [
"Drama",
"Action",
"Science-Fiction"
],
"runtime": 60
}
I wanted to search shows with genre 'Action' and project the result array as
{
"url": "http://www.tvmaze.com/shows/167/24",
"name": "24",
"genres": [
"Action" // I want only the matched item in
//my result array
],
"runtime": 60
} , //same for the second doc as well
If I use
db.shows.find({genres:'Action'}, {'genres.$': 1});
It works but the same does not work in aggregate method with $project
Shows.aggregate([
{
$match: { 'genres': 'Action'}
},
{
$project: {
_id: 0,
url: 1,
name: 1,
runtime: 1,
'genres.$': 1
}
}
]);
this is the error I get on this aggregate query
Invalid $project :: caused by :: FieldPath field names may not start with '$'."
db.collection.aggregate([
{
$match: {
"genres": {
$regex: "/^action/",
$options: "im"
}
}
},
{
$project: {
_id: 0,
url: 1,
name: 1,
runtime: 1,
genres: {
$filter: {
input: "$genres",
as: "genre",
cond: {
$regexMatch: {
input: "$$genre",
regex: "/^action/",
options: "im"
}
}
}
}
}
}
])
Here is how I solved it, Thanks #turivishal for the help
Help to "flatten" (to pull nested fields at same level as document's fields) a mongodb document in a query
//this is "anagrafiche" collection
[{
"name": "tizio"
,"surname": "semproni"
,"birthday": "01/02/1923"
,"home": {
"road": "via"
,"roadname": "bianca"
,"roadN": 12
,"city": "rome"
,"country": "italy"
}
},
{
"name": "caio"
,"surname": "giulio"
,"birthday": "02/03/1932"
,"home": {
"road": "via"
,"roadname": "rossa"
,"roadN": 21
,"city": "milan"
,"country": "italy"
}
},
{
"name": "mario"
,"surname": "rossi"
// birthday is not present for this document
,"home": {
"road": "via"
,"roadname": "della pace"
,"roadN": 120
,"city": "rome"
,"country": "italy"
}
}
]
my query:
db.anagrafiche.aggregate([ {$match {"home.city": "rome"}}
{$project:{"name": 1, "surname":1, <an expression to flatten the address>, "birthday": 1, "_id":0}}
]
);
expected result:
{
,"name": "tizio"
,"surname": "semproni"
,"address": "via bianca 12 rome"
,"birthday": 01/02/1923
},{
,"name": "mario"
,"surname": "rossi"
,"address": "via della pace 120 rome"
,"birthday": NULL
}
You can use $objectToArray to get nested document keys and values and then use $reduce along with $concat to concatenate values dynamically:
db.collection.aggregate([
{
$project: {
_id: 0,
name: 1,
surname: 1,
birthday: 1,
address: {
$reduce: {
input: { $objectToArray: "$home" },
initialValue: "",
in: {
$concat: [
"$$value",
{ $cond: [ { $eq: [ "$$value", "" ] }, "", " " ] },
{ $toString: "$$this.v" }
]
}
}
}
}
}
])
Mongo Playground
I have this 'Sales' collection and a sample of it looks like this:
[
{cusID: 'a412q39x',
cusCountry: 'MEX',
itemPurchased: 'Toy_A'
},
{cusID: 'r760e11s',
cusCountry: 'USA',
itemPurchased: 'Toy_B'
},
{cusID: 'g723f01z',
cusCountry: 'USA',
itemPurchased: 'Toy_C'
},
{cusID: 'h277p01c',
cusCountry: 'CAN',
itemPurchased: 'Toy_B'
}
]
This is the result I am hoping to achieve.
[
{item: 'Toy_A',
USA: 4,
MEX: 2,
CAN: 1,
BRA: 0
},
{item: 'Toy_B',
USA: 3,
MEX: 0,
CAN: 2,
BRA: 1
}
]
I tried:
{
$group:{_id:{toy:'$itemPurchased', country: $cusCountry'},'cnt':{'$sum': 1}}
}
The result was not what I wanted.
[
{
_id.toy: 'Toy_A',
_id.country: 'BRA',
cnt: 43
},
{
_id.toy: 'Toy_A',
_id.country: 'USA',
cnt: 102
},
{
_id.toy: 'Toy_A',
_id.country: 'JPN',
cnt: 72
},
{
_id.toy: 'Toy_B',
_id.country: 'CAN',
cnt: 32
}
]
I have also experimented with $facet but to no avail. Mongo gurus, please enlighten. Thanks in advance.
Try as below:
db.collection.aggregate([
{
$group: { _id: '$itemPurchased' , items: { $push: { "country" : "$cusCountry", "count": { $sum :1} } } }
},
{
"$project": {
"countryCounts": {
"$arrayToObject": {
"$map": {
"input": "$items",
"as": "item",
"in": {
"k": "$$item.country",
"v": "$$item.count",
}
}
}
}
}
},
{ $replaceRoot: { newRoot: { "$mergeObjects":[ "$countryCounts" , { "item":"$_id"} ] } } }
])
You will get the result like below:
{
"USA" : 1,
"item" : "Toy_C"
},
/* 2 */
{
"USA" : 1,
"CAN" : 1,
"item" : "Toy_B"
},
/* 3 */
{
"MEX" : 1,
"item" : "Toy_A"
}