我想合并两个聚合操作。第一个操作返回,例如:
{ "_id" : "Colors", "count" : 12 }
{ "_id" : "Animals", "count" : 6 }
第二个操作返回,例如:
{ "_id" : "Red", "count" : 10 }
{ "_id" : "Blue", "count" : 9 }
{ "_id" : "Green", "count" : 9 }
{ "_id" : "White", "count" : 7 }
{ "_id" : "Yellow", "count" : 7 }
{ "_id" : "Orange", "count" : 7 }
{ "_id" : "Black", "count" : 5 }
{ "_id" : "Goose", "count" : 4 }
{ "_id" : "Chicken", "count" : 3 }
{ "_id" : "Grey", "count" : 3 }
{ "_id" : "Cat", "count" : 3 }
{ "_id" : "Rabbit", "count" : 3 }
{ "_id" : "Duck", "count" : 3 }
{ "_id" : "Turkey", "count" : 2 }
{ "_id" : "Elephant", "count" : 2 }
{ "_id" : "Shark", "count" : 2 }
{ "_id" : "Fish", "count" : 2 }
{ "_id" : "Tiger", "count" : 2 }
{ "_id" : "Purple", "count" : 1 }
{ "_id" : "Pink", "count" : 1 }
如何结合这 2 个操作来实现以下目标?
{ "_id" : "Colors", "count" : 12, "items" :
[
{ "_id" : "Red", "count" : 10 },
{ "_id" : "Blue", "count" : 9 },
{ "_id" : "Green", "count" : 9 },
{ "_id" : "White", "count" : 7 },
{ "_id" : "Yellow", "count" : 7 },
{ "_id" : "Orange", "count" : 7 },
{ "_id" : "Black", "count" : 5 },
{ "_id" : "Grey", "count" : 3 },
{ "_id" : "Purple", "count" : 1 },
{ "_id" : "Pink", "count" : 1 }
]
},
{ "_id" : "Animals", "count" : 6, "items" :
[
{ "_id" : "Goose", "count" : 4 },
{ "_id" : "Chicken", "count" : 3 },
{ "_id" : "Cat", "count" : 3 },
{ "_id" : "Rabbit", "count" : 3 },
{ "_id" : "Duck", "count" : 3 },
{ "_id" : "Turkey", "count" : 2 },
{ "_id" : "Elephant", "count" : 2 },
{ "_id" : "Shark", "count" : 2 },
{ "_id" : "Fish", "count" : 2 },
{ "_id" : "Tiger", "count" : 2 }
]
}
架构
var ListSchema = new Schema({
created: {
type: Date,
default: Date.now
},
title: {
type: String,
default: '',
trim: true,
required: 'Title cannot be blank'
},
items: {
type: Array,
default: [String],
trim: true
},
creator: {
type: Schema.ObjectId,
ref: 'User'
}
});
操作一
db.lists.aggregate(
[
{ $group: { _id: "$title", count: { $sum: 1 } } },
{ $sort: { count: -1 } }
]
)
操作2
db.lists.aggregate(
[
{ $unwind: "$items" },
{ $group: { _id: "$items", count: { $sum: 1 } } },
{ $sort: { count: -1 } }
]
)
最佳答案
这实际上取决于您在响应中所追求的结果类型。您询问的内容似乎表明您正在寻找结果中的“方面计数”,但稍后我会谈到这一点。
作为基本结果,这种方法没有任何问题:
Thing.aggregate(
[
{ "$group": {
"_id": {
"type": "$type", "name": "$name"
},
"count": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.type",
"count": { "$sum": "$count" },
"names": {
"$push": { "name": "$_id.name", "count": "$count" }
}
}}
],
function(err,results) {
console.log(JSON.stringify(results, undefined, 2));
callback(err);
}
)
这应该会给你这样的结果:
[
{
"_id": "colours",
"count": 50102,
"names": [
{ "name": "Green", "count": 9906 },
{ "name": "Yellow", "count": 10093 },
{ "name": "Red", "count": 10083 },
{ "name": "Orange", "count": 9997 },
{ "name": "Blue", "count": 10023 }
]
},
{
"_id": "animals",
"count": 49898,
"names": [
{ "name": "Tiger", "count": 9710 },
{ "name": "Lion", "count": 10058 },
{ "name": "Elephant", "count": 10069 },
{ "name": "Monkey", "count": 9963 },
{ "name": "Bear", "count": 10098 }
]
}
]
这里最基本的方法是简单地 $group分两个阶段,第一阶段将键组合聚合到最低(最细粒度)分组级别,然后再次处理 $group 以基本上“加起来”最高级别的总数(最小粒度)分组级别,因此也将较低的结果添加到项目数组中。
但这并不像在“方面计数”中那样“分离”,因此这样做变得有点复杂,也有点疯狂。但首先是示例:
Thing.aggregate(
[
{ "$group": {
"_id": {
"type": "$type",
"name": "$name"
},
"count": { "$sum": 1 }
}},
{ "$group": {
"_id": "$_id.type",
"count": { "$sum": "$count" },
"names": {
"$push": { "name": "$_id.name", "count": "$count" }
}
}},
{ "$group": {
"_id": null,
"types": {
"$push": {
"type": "$_id", "count": "$count"
}
},
"names": { "$push": "$names" }
}},
{ "$unwind": "$names" },
{ "$unwind": "$names" },
{ "$group": {
"_id": "$types",
"names": { "$push": "$names" }
}},
{ "$project": {
"_id": 0,
"facets": {
"types": "$_id",
"names": "$names",
},
"data": { "$literal": [] }
}}
],
function(err,results) {
console.log(JSON.stringify(results[0], undefined, 2));
callback(err);
}
);
这将产生这样的输出:
{
"facets": {
"types": [
{ "type": "colours", "count": 50102 },
{ "type": "animals", "count": 49898 }
],
"names": [
{ "name": "Green", "count": 9906 },
{ "name": "Yellow", "count": 10093 },
{ "name": "Red", "count": 10083 },
{ "name": "Orange", "count": 9997 },
{ "name": "Blue", "count": 10023 },
{ "name": "Tiger", "count": 9710 },
{ "name": "Lion", "count": 10058 },
{ "name": "Elephant", "count": 10069 },
{ "name": "Monkey", "count": 9963 },
{ "name": "Bear", "count": 10098 }
]
},
"data": []
}
但显而易见的是,虽然“可能”,但在管道中进行的这种“杂耍”以产生这种输出格式并不是真正有效的。与第一个示例相比,这里有很多开销只是简单地将结果拆分为它们自己的数组响应并且独立于分组键。随着要生成的“方面”越来越多,这明显变得更加复杂。
此外,正如此处输出中所暗示的,人们通常对“方面计数”的要求是,除了聚合方面之外,结果“数据”也包含在响应(可能已分页)中。因此,进一步的并发症应该在这里显而易见:
{ "$group": {
"_id": null,
(...)
此类操作的要求基本上是将每条数据“填充”到单个对象中。在大多数情况下,当然在您想要结果中的实际数据(在此示例中使用 100,000)的情况下,遵循这种方法变得完全不切实际,并且几乎肯定会超过 16MB 的 BSON 文档大小限制。
在这种情况下,如果您想在响应中生成结果和该数据的“方面”,那么这里最好的方法是将每个聚合和输出页面作为单独的查询操作运行,并“流式传输”输出JSON(或其他格式)返回给接收客户端。
作为一个独立的例子:
var async = require('async'),
mongoose = require('mongoose'),
Schema = mongoose.Schema;
mongoose.connect('mongodb://localhost/things');
var data = {
"colours": [
"Red","Blue","Green","Yellow","Orange"
],
"animals": [
"Lion","Tiger","Bear","Elephant","Monkey"
]
},
dataKeys = Object.keys(data);
var thingSchema = new Schema({
"name": String,
"type": String
});
var Thing = mongoose.model( 'Thing', thingSchema );
var writer = process.stdout;
mongoose.connection.on("open",function(err) {
if (err) throw err;
async.series(
[
function(callback) {
process.stderr.write("removing\n");
Thing.remove({},callback);
},
function(callback) {
process.stderr.write("inserting\n");
var bulk = Thing.collection.initializeUnorderedBulkOp(),
count = 0;
async.whilst(
function() { return count < 100000; },
function(callback) {
var keyLen = dataKeys.length,
keyIndex = Math.floor(Math.random(keyLen)*keyLen),
type = dataKeys[keyIndex],
types = data[type],
typeLen = types.length,
nameIndex = Math.floor(Math.random(typeLen)*typeLen),
name = types[nameIndex];
var obj = { "type": type, "name": name };
bulk.insert(obj);
count++;
if ( count % 1000 == 0 ) {
process.stderr.write('insert count: ' + count + "\n");
bulk.execute(function(err,resp) {
bulk = Thing.collection.initializeUnorderedBulkOp();
callback(err);
});
} else {
callback();
}
},
callback
);
},
function(callback) {
writer.write("{ \n \"page\": 1,\n \"pageSize\": 25,\n")
writer.write(" \"facets\": {\n"); // open object response
var stream = Thing.collection.aggregate(
[
{ "$group": {
"_id": "$name",
"count": { "$sum": 1 }
}}
],
{
"cursor": {
"batchSize": 1000
}
}
);
var counter = 0;
stream.on("data",function(data) {
stream.pause();
if ( counter == 0 ) {
writer.write(" \"names\": [\n");
} else {
writer.write(",\n");
}
data = { "name": data._id, "count": data.count };
writer.write(" " + JSON.stringify(data));
counter++;
stream.resume();
});
stream.on("end",function() {
writer.write("\n ],\n");
var stream = Thing.collection.aggregate(
[
{ "$group": {
"_id": "$type",
"count": { "$sum": 1 }
}}
],
{
"cursor": {
"batchSize": 1000
}
}
);
var counter = 0;
stream.on("data",function(data) {
stream.pause();
if ( counter == 0 ) {
writer.write(" \"types\": [\n");
} else {
writer.write(",\n");
}
data = { "name": data._id, "count": data.count };
writer.write(" " + JSON.stringify(data));
counter++;
stream.resume();
});
stream.on("end",function() {
writer.write("\n ]\n },\n");
var stream = Thing.find({}).limit(25).stream();
var counter = 0;
stream.on("data",function(data) {
stream.pause();
if ( counter == 0 ) {
writer.write(" \"data\": [\n");
} else {
writer.write(",\n");
}
writer.write(" " + JSON.stringify(data));
counter++;
stream.resume();
});
stream.on("end",function() {
writer.write("\n ]\n}\n");
callback();
});
});
});
}
],
function(err) {
if (err) throw err;
process.exit();
}
);
});
输出如下:
{
"page": 1,
"pageSize": 25,
"facets": {
"names": [
{"name":"Red","count":10007},
{"name":"Tiger","count":10012},
{"name":"Yellow","count":10119},
{"name":"Monkey","count":9970},
{"name":"Elephant","count":10046},
{"name":"Bear","count":10082},
{"name":"Orange","count":9982},
{"name":"Green","count":10005},
{"name":"Blue","count":9884},
{"name":"Lion","count":9893}
],
"types": [
{"name":"colours","count":49997},
{"name":"animals","count":50003}
]
},
"data": [
{"_id":"55bf141f3edc150b6abdcc02","type":"animals","name":"Lion"},
{"_id":"55bf141f3edc150b6abdc81b","type":"colours","name":"Blue"},
{"_id":"55bf141f3edc150b6abdc81c","type":"colours","name":"Orange"},
{"_id":"55bf141f3edc150b6abdc81d","type":"animals","name":"Bear"},
{"_id":"55bf141f3edc150b6abdc81e","type":"animals","name":"Elephant"},
{"_id":"55bf141f3edc150b6abdc81f","type":"colours","name":"Orange"},
{"_id":"55bf141f3edc150b6abdc820","type":"colours","name":"Green"},
{"_id":"55bf141f3edc150b6abdc821","type":"animals","name":"Lion"},
{"_id":"55bf141f3edc150b6abdc822","type":"animals","name":"Monkey"},
{"_id":"55bf141f3edc150b6abdc823","type":"colours","name":"Yellow"},
{"_id":"55bf141f3edc150b6abdc824","type":"colours","name":"Yellow"},
{"_id":"55bf141f3edc150b6abdc825","type":"colours","name":"Orange"},
{"_id":"55bf141f3edc150b6abdc826","type":"animals","name":"Monkey"},
{"_id":"55bf141f3edc150b6abdc827","type":"colours","name":"Blue"},
{"_id":"55bf141f3edc150b6abdc828","type":"animals","name":"Tiger"},
{"_id":"55bf141f3edc150b6abdc829","type":"colours","name":"Red"},
{"_id":"55bf141f3edc150b6abdc82a","type":"animals","name":"Monkey"},
{"_id":"55bf141f3edc150b6abdc82b","type":"animals","name":"Elephant"},
{"_id":"55bf141f3edc150b6abdc82c","type":"animals","name":"Tiger"},
{"_id":"55bf141f3edc150b6abdc82d","type":"animals","name":"Bear"},
{"_id":"55bf141f3edc150b6abdc82e","type":"colours","name":"Yellow"},
{"_id":"55bf141f3edc150b6abdc82f","type":"animals","name":"Lion"},
{"_id":"55bf141f3edc150b6abdc830","type":"animals","name":"Elephant"},
{"_id":"55bf141f3edc150b6abdc831","type":"colours","name":"Orange"},
{"_id":"55bf141f3edc150b6abdc832","type":"animals","name":"Elephant"}
]
}
这里有一些注意事项,特别是 Mongoose .aggregate()并不真正直接支持标准节点流接口(interface)。 .cursor() 提供了一个 .each() 方法在聚合方法上,但是 core API method 隐含的“流”这里提供了更多的控制权,所以 .collection 方法在这里获取底层 driver object是可取的。希望 future 的 Mongoose 版本会考虑这一点。
因此,如果您的最终目标是这样一个“方面计数”以及此处演示的结果,那么每个聚合和结果都最适合以演示的方式“流式传输”。如果没有它,聚合就会变得过于复杂,并且很可能会超过 BSON 限制,就像在这种情况下这样做一样。
关于mongodb - 在单个结果中组合聚合操作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31777590/
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