MongoDB 聚合：计算不同的字段
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MongoDB Aggregation: Counting distinct fields
提问by user1438162
I am trying to write an aggregation to identify accounts that use multiple payment sources. Typical data would be.
我正在尝试编写一个聚合来识别使用多个付款来源的帐户。典型的数据是。
{
account:"abc",
vendor:"amazon",
}
...
{
account:"abc",
vendor:"overstock",
}
Now, I'd like to produce a list of accounts similar to this
现在，我想生成一个与此类似的帐户列表
{
account:"abc",
vendorCount:2
}
How would I write this in Mongo's aggregation framework
我将如何在 Mongo 的聚合框架中编写此内容
回答by user1438162
I figured this out by using the $addToSet and $unwind operators.
我通过使用 $addToSet 和 $unwind 运算符来解决这个问题。
Mongodb Aggregation count array/set size
db.collection.aggregate([
{
$group: { _id: { account: '$account' }, vendors: { $addToSet: '$vendor'} }
},
{
$unwind:"$vendors"
},
{
$group: { _id: "$_id", vendorCount: { $sum:1} }
}
]);
Hope it helps someone
希望它可以帮助某人
回答by Mayur Kataria
I think its better if you execute query like following which will avoid unwind
我认为如果您执行如下查询会更好，这将避免放松
db.t2.insert({_id:1,account:"abc",vendor:"amazon"});
db.t2.insert({_id:2,account:"abc",vendor:"overstock"});
db.t2.aggregate(
{ $group : { _id : { "account" : "$account", "vendor" : "$vendor" }, number : { $sum : 1 } } },
{ $group : { _id : "$_id.account", number : { $sum : 1 } } }
);
Which will show you following result which is expected.
这将向您显示以下预期结果。
{ "_id" : "abc", "number" : 2 }
回答by Rahul
I do not see why somebody would have to use $group twice
我不明白为什么有人必须使用 $group 两次
db.t2.aggregate([ { $group: {"_id":"$account" , "number":{$sum:1}} } ])
This will work perfectly fine.
这将工作得很好。
回答by Hett
You can use sets
您可以使用 sets
db.test.aggregate([
{$group: {
_id: "$account",
uniqueVendors: {$addToSet: "$vendor"}
}},
{$project: {
_id: 1,
vendorsCount: {$size: "$uniqueVendors"}
}}
]);
回答by Anuj Gupta
This approach doesn't make use of $unwind and other extra operations. Plus, this won't affect anything if new things are added into the aggregation. There's a flaw in the accepted answer. If you have other accumulated fields in the $group, it would cause issues in the $unwind stage of the accepted answer.
这种方法不使用 $unwind 和其他额外的操作。另外，如果将新内容添加到聚合中，这不会影响任何事情。接受的答案存在缺陷。如果您在 $group 中有其他累积字段，则会在已接受答案的 $unwind 阶段引起问题。
db.collection.aggregate([{
"$group": {
"_id": "$account",
"vendors": {"$addToSet": "$vendor"}
}
},
{
"$addFields": {
"vendorCount": {
"$size": "$vendors"
}
}
}])
回答by Amitesh
To identify accounts that use multiple payment sources:
要识别使用多个付款来源的帐户：
- Use grouping
- Use a match to filter count
- 使用分组
- 使用匹配来过滤计数
db.payment_collection.aggregate([ { $group: {"_id":"$account" , "number":{$sum:1}} }, { "$match": { "number": { "$gt": 1 } } } ])
db.payment_collection.aggregate([ { $group: {"_id":"$account" , "number":{$sum:1}} }, { "$match": { "number": { "$gt" : 1 } } } ])
回答by Abilash Raghu
db.UserModule.aggregate(
{ $group : { _id : { "companyauthemail" : "$companyauthemail", "email" : "$email" }, number : { $sum : 1 } } },
{ $group : { _id : "$_id.companyauthemail", number : { $sum : 1 } } }
);
回答by user3448195
An example
一个例子
db.collection.distinct("example.item").forEach( function(docs) {
print(docs + "==>>" + db.collection.count({"example.item":docs}))
});
