Reputation: 533
Taking the example of the collection like:
db.products.insert([
{"product_name": "Product 1", "filters": [{"name": "brand", "value": "Brand 1"}, {"name": "color", "value": "Color 1"}]},
{"product_name": "Product 2", "filters": [{"name": "brand", "value": "Brand 2"}, {"name": "color", "value": "Color 2"}]},
{"product_name": "Product 3", "filters": [{"name": "brand", "value": "Brand 1"}, {"name": "color", "value": "Color 2"}]},
{"product_name": "Product 4", "filters": [{"name": "brand", "value": "Brand 3"}, {"name": "color", "value": "Color 1"}]},
{"product_name": "Product 5", "filters": [{"name": "brand", "value": "Brand 3"}, {"name": "color", "value": "Color 3"}]}
])
If one has to aggregate on list of dictionaries and get the grouped counts for each of them, like the following:
(brand)
Brand 1 : 2
Brand 2 : 1
Brand 3 : 2
(color)
Color 1 : 2
Color 2 : 2
Color 3 : 3
What might be the relevant way to get this?
Upvotes: 1
Views: 2639
Reputation: 7840
Use this aggregation as below :
db.products.aggregate({
"$unwind": "$filters"
}, {
"$group": {
"_id": {
"value": "$filters.value",
"name": "$filters.name"
},
"count": {
"$sum": 1
}
}
}, {
"$group": {
"_id": "$_id.name",
"data": {
"$push": {
"value": "$_id.value",
"totalCount": "$count"
}
}
}
}).pretty()
Upvotes: 3
Reputation: 2222
You have to unwind your data to get a flat dataset and than group by the filter values.
db.products.aggregate(
{
$unwind: "$filters",
}, {
$group: {
_id: "$filters.value",
product_names: { $push: "$product_name" } ,
count: { $sum: 1 }
}
});
Returns
{
"result" : [
{
"_id" : "Color 3",
"product_names" : [
"Product 5"
],
"count" : 1
},
{
"_id" : "Brand 3",
"product_names" : [
"Product 4",
"Product 5"
],
"count" : 2
},
{
"_id" : "Color 2",
"product_names" : [
"Product 2",
"Product 3"
],
"count" : 2
},
{
"_id" : "Color 1",
"product_names" : [
"Product 1",
"Product 4"
],
"count" : 2
},
{
"_id" : "Brand 2",
"product_names" : [
"Product 2"
],
"count" : 1
},
{
"_id" : "Brand 1",
"product_names" : [
"Product 1",
"Product 3"
],
"count" : 2
}
],
"ok" : 1
}
Upvotes: 1