mongodb 使用 mongo 聚合框架按数组的特定元素分组
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Group by specific element of array with mongo aggregation framework
提问by Russell
Is it possible to use the aggregation framework to group by a specific element of an array?
是否可以使用聚合框架按数组的特定元素进行分组？
Such that with documents like this:
这样的文件是这样的：
{
name: 'Russell',
favourite_foods: [
{ name: 'Pizza', type: 'Four Cheeses' },
{ name: 'Burger', type: 'Veggie'}
],
height: 6
}
I could get a distinct list of top favourite foods (ie. foods at index 0) along with the height of the tallest person who's top favourite food that is?
我可以得到一份最喜欢的食物（即索引为 0 的食物）的明确列表，以及最喜欢的食物中最高的人的身高？
Something like this (although it doesn't work as the array index access dot notation doesn't seem to work in the aggregation framework):
像这样的东西（尽管它不起作用，因为数组索引访问点表示法在聚合框架中似乎不起作用）：
db.people.aggregate([
{ $group : { _id: "$favourite_foods.0.name", max_height: { $max : "$height" } } }
])
回答by Asya Kamsky
Seems like you are relying on the favorite food for each person being first in the array. If so, there is an aggregation framework operator you can take advantage of.
似乎您依靠每个人最喜欢的食物排在第一位。如果是这样，您可以利用一个聚合框架运算符。
Here is the pipeline you can use:
这是您可以使用的管道：
db.people.aggregate(
[
{
"$unwind" : "$favourite_foods"
},
{
"$group" : {
"_id" : {
"name" : "$name",
"height" : "$height"
},
"faveFood" : {
"$first" : "$favourite_foods"
}
}
},
{
"$group" : {
"_id" : "$faveFood.name",
"height" : {
"$max" : "$_id.height"
}
}
}
])
On this sample dataset:
在此示例数据集上：
> db.people.find().pretty()
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 6
}
{
"_id" : ObjectId("5088950bd4197aa2b9490742"),
"name" : "Lucy",
"favourite_foods" : [
{
"name" : "Pasta",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 5.5
}
{
"_id" : ObjectId("5088951dd4197aa2b9490743"),
"name" : "Landy",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 5
}
{
"_id" : ObjectId("50889541d4197aa2b9490744"),
"name" : "Augie",
"favourite_foods" : [
{
"name" : "Sushi",
"type" : "Four Cheeses"
},
{
"name" : "Pizza",
"type" : "Veggie"
}
],
"height" : 6.2
}
You get these results:
你会得到这些结果：
{
"result" : [
{
"_id" : "Pasta",
"height" : 5.5
},
{
"_id" : "Pizza",
"height" : 6
},
{
"_id" : "Sushi",
"height" : 6.2
}
],
"ok" : 1
}
回答by JohnnyHK
Looks like it isn't currently possible to extract a specific element from an array in aggregation: https://jira.mongodb.org/browse/SERVER-4589
看起来目前无法从聚合数组中提取特定元素：https: //jira.mongodb.org/browse/SERVER-4589
回答by Jenna
I think you can make use of the $project and $unwind operators (let me know if this isn't what you're trying to accomplish):
我认为您可以使用 $project 和 $unwind 运算符（如果这不是您想要完成的，请告诉我）：
> db.people.aggregate(
{$unwind: "$favourite_foods"},
{$project: {food : "$favourite_foods", height: 1}},
{$group : { _id: "$food", max_height: { $max : "$height" } } })
{
"result" : [
{
"_id" : {
"name" : "Burger",
"type" : "Veggie"
},
"max_height" : 6
},
{
"_id" : {
"name" : "Pizza",
"type" : "Four Cheeses"
},
"max_height" : 6
}
],
"ok" : 1
}
回答by Eddy
JUST add more information about the result after using "$wind":
只需在使用后添加有关结果的更多信息"$wind"：
DOCUMENT :
文档 ：
> db.people.find().pretty()
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : [
{
"name" : "Pizza",
"type" : "Four Cheeses"
},
{
"name" : "Burger",
"type" : "Veggie"
}
],
"height" : 6
},
...
AGGREAGATION :
聚合：
db.people.aggregate([{
$unwind: "$favourite_foods"
}]);
RESULT :
结果 ：
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" :{
"name" : "Pizza",
"type" : "Four Cheeses"
},
"height" : 6
},
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "Russell",
"favourite_foods" : {
"name" : "Burger",
"type" : "Veggie"
},
"height" : 6
}
In Addition:
If there are more than two array fields in one collection record,
we can use "$project"stage to specify the array field.
另外：
如果一个集合记录中有两个以上的数组字段，我们可以使用“$project”阶段来指定数组字段。
db.people.aggregate([
{
$project:{
"favourite_foods": 1
}
},
{
$unwind: "$favourite_foods"
}
]);
