MongoDB aggregate array of objects together by object id and count occurences - mongodb

Question

I'm trying to figure out what I'm doing wrong, I have collected the following, "Subset of data", "Desired output"
This is how my data objects look
[{
"survey_answers": [
{
"id": "9ca01568e8dbb247", // As they are, this is the key to groupBy
"option_answer": 5, // Represent the index of the choosen option
"type": "OPINION_SCALE" // Opinion scales are 0-10 (meaning elleven options)
},
{
"id": "ba37125ec32b2a99",
"option_answer": 3,
"type": "LABELED_QUESTIONS" // Labeled questions are 0-x (they can change it from survey to survey)
}
],
"survey_id": "test"
},
{
"survey_answers": [
{
"id": "9ca01568e8dbb247",
"option_answer": 0,
"type": "OPINION_SCALE"
},
{
"id": "ba37125ec32b2a99",
"option_answer": 3,
"type": "LABELED_QUESTIONS"
}
],
"survey_id": "test"
}]
My desired output is:
[
{
id: '9ca01568e8dbb247'
results: [
{ _id: 5, count: 1 },
{ _id: 0, count: 1 }
]
},
{
id: 'ba37125ec32b2a99'
results: [
{ _id: 3, count: 2 }
]
}
]
Active query
Model.aggregate([
{
$match: {
'survey_id': survey_id
}
},
{
$unwind: "$survey_answers"
},
{
$group: {
_id: "$survey_answers.option_answer",
count: {
$sum: 1
}
}
}
])
Current output
[
{
"_id": 0,
"count": 1
},
{
"_id": 3,
"count": 2
},
{
"_id": 5,
"count": 1
}
]

Answer 1

I added your records to my db. Post that I tried your commands one by one.
$unwind results you similar to -
> db.survey.aggregate({$unwind: "$survey_answers"})
{ "_id" : ObjectId("5c3859e459875873b5e6ee3c"), "survey_answers" : { "id" : "9ca01568e8dbb247", "option_answer" : 5, "type" : "OPINION_SCALE" }, "survey_id" : "test" }
{ "_id" : ObjectId("5c3859e459875873b5e6ee3c"), "survey_answers" : { "id" : "ba37125ec32b2a99", "option_answer" : 3, "type" : "LABELED_QUESTIONS" }, "survey_id" : "test" }
{ "_id" : ObjectId("5c3859e459875873b5e6ee3d"), "survey_answers" : { "id" : "9ca01568e8dbb247", "option_answer" : 0, "type" : "OPINION_SCALE" }, "survey_id" : "test" }
{ "_id" : ObjectId("5c3859e459875873b5e6ee3d"), "survey_answers" : { "id" : "ba37125ec32b2a99", "option_answer" : 3, "type" : "LABELED_QUESTIONS" }, "survey_id" : "test" }
I am not adding code for match since that is okay in your query as well
The grouping would be -
> db.survey.aggregate({$unwind: "$survey_answers"},{$group: { _id: { 'optionAnswer': "$survey_answers.option_answer", 'id':"$survey_answers.id"}, count: { $sum: 1}}})
{ "_id" : { "optionAnswer" : 0, "id" : "9ca01568e8dbb247" }, "count" : 1 }
{ "_id" : { "optionAnswer" : 3, "id" : "ba37125ec32b2a99" }, "count" : 2 }
{ "_id" : { "optionAnswer" : 5, "id" : "9ca01568e8dbb247" }, "count" : 1 }
You can group on $survey_answers.id to bring it into projection.
The projection is what you're missing in your query -
> db.survey.aggregate({$unwind: "$survey_answers"},{$group: { _id: { 'optionAnswer': "$survey_answers.option_answer", 'id':'$survey_answers.id'}, count: { $sum: 1}}}, {$project : {answer: '$_id.optionAnswer', id: '$_id.id', count: '$count', _id:0}})
{ "answer" : 0, "id" : "9ca01568e8dbb247", "count" : 1 }
{ "answer" : 3, "id" : "ba37125ec32b2a99", "count" : 2 }
{ "answer" : 5, "id" : "9ca01568e8dbb247", "count" : 1 }
Further you can add a group on id and add results to a set. And your final query would be -
db.survey.aggregate(
{$unwind: "$survey_answers"},
{$group: {
_id: { 'optionAnswer': "$survey_answers.option_answer", 'id':'$survey_answers.id'},
count: { $sum: 1}
}},
{$project : {
answer: '$_id.optionAnswer',
id: '$_id.id',
count: '$count',
_id:0
}},
{$group: {
_id:{id:"$id"},
results: { $addToSet: {answer: "$answer", count: '$count'} }
}},
{$project : {
id: '$_id.id',
answer: '$results',
_id:0
}})
Hope this helps.