在MongoDB中按条件分组 [英] Group By Condition in MongoDB
问题描述
我在MongoDB中有一系列文档(检查事件),如下所示:
I have a series of documents (check events) in MongoDB that look like this:
{
"_id" : ObjectId("5397a78ab87523acb46f56"),
"inspector_id" : ObjectId("5397997a02b8751dc5a5e8b1"),
"status" : 'defect',
"utc_timestamp" : ISODate("2014-06-11T00:49:14.109Z")
}
{
"_id" : ObjectId("5397a78ab87523acb46f57"),
"inspector_id" : ObjectId("5397997a02b8751dc5a5e8b2"),
"status" : 'ok',
"utc_timestamp" : ISODate("2014-06-11T00:49:14.109Z")
}
我需要得到一个看起来像这样的结果集:
I need to get a result set that looks like this:
[
{
"date" : "2014-06-11",
"defect_rate" : '.92'
},
{
"date" : "2014-06-11",
"defect_rate" : '.84'
},
]
换句话说,我需要获取每天的平均缺陷率.这可能吗?
In other words, I need to get the average defect rate per day. Is this possible?
推荐答案
您想要的是聚合框架:
db.collection.aggregate([
{ "$group": {
"_id": {
"year": { "$year": "$utc_timestamp" },
"month": { "$month": "$utc_timestamp" },
"day": { "$dayOfMonth": "$utc_timestamp" },
},
"defects": {
"$sum": { "$cond": [
{ "$eq": [ "$status", "defect" ] },
1,
0
]}
},
"totalCount": { "$sum": 1 }
}},
{ "$project": {
"defect_rate": {
"$cond": [
{ "$eq": [ "$defects", 0 ] },
0,
{ "$divide": [ "$defects", "$totalCount" ] }
]
}
}}
])
因此,您首先需要使用日期聚合运算符进行分组,并获得totalCount特定日期的商品数量.在这里使用 $cond
运算符可以确定是否状态"实际上是否是缺陷,其结果是有条件的 $sum
,其中仅计算缺陷"值.
So first you group on the day using the date aggregation operators and get the totalCount of items on the given day. The use of the $cond
operator here determines whether the "status" is actually a defect or not and the result is a conditional $sum
where only the "defect" values are counted.
将这些内容每天分组后,您只需 $divide
结果,并通过 $cond
进行另一次检查确保您没有被零除.
Once those are grouped per day you simply $divide
the result, with another check with $cond
to make sure you are not dividing by zero.
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