如何分組mongodb - mapReduce輸出？

Question

我有關於mongodb中的mapReduce框架的查詢，所以我有mapReduce函數的鍵值對的結果，現在我想在mapReduce的這個輸出上運行查詢。

所以我使用mapReduce來查找這樣的用戶統計信息

db.order.mapReduce(function() { emit (this.customer,{count:1,orderDate:this.orderDate.interval_start}) },
function(key,values){ 
    var sum =0 ; var lastOrderDate;  
    values.forEach(function(value) {
     if(value['orderDate']){ 
        lastOrderDate=value['orderDate'];
    }  
    sum+=value['count'];
}); 
    return {count:sum,lastOrderDate:lastOrderDate}; 
},
{ query:{status:"DELIVERED"},out:"order_total"}).find()

這給了我這樣的輸出

{ "_id" : ObjectId("5443765ae4b05294c8944d5b"), "value" : { "count" : 1, "orderDate" : ISODate("2014-10-18T18:30:00Z") } }
{ "_id" : ObjectId("54561911e4b07a0a501276af"), "value" : { "count" : 2, "lastOrderDate" : ISODate("2015-03-14T18:30:00Z") } }
{ "_id" : ObjectId("54561b9ce4b07a0a501276b1"), "value" : { "count" : 1, "orderDate" : ISODate("2014-11-01T18:30:00Z") } }
{ "_id" : ObjectId("5458712ee4b07a0a501276c2"), "value" : { "count" : 2, "lastOrderDate" : ISODate("2014-11-03T18:30:00Z") } }
{ "_id" : ObjectId("545f64e7e4b07a0a501276db"), "value" : { "count" : 15, "lastOrderDate" : ISODate("2015-06-04T18:30:00Z") } }
{ "_id" : ObjectId("54690771e4b0070527c657ed"), "value" : { "count" : 6, "lastOrderDate" : ISODate("2015-06-03T18:30:00Z") } }
{ "_id" : ObjectId("54696c64e4b07f3c07010b4a"), "value" : { "count" : 1, "orderDate" : ISODate("2014-11-18T18:30:00Z") } }
{ "_id" : ObjectId("546980d1e4b07f3c07010b4d"), "value" : { "count" : 4, "lastOrderDate" : ISODate("2015-03-24T18:30:00Z") } }
{ "_id" : ObjectId("54699ac4e4b07f3c07010b51"), "value" : { "count" : 30, "lastOrderDate" : ISODate("2015-05-23T18:30:00Z") } }
{ "_id" : ObjectId("54699d0be4b07f3c07010b55"), "value" : { "count" : 1, "orderDate" : ISODate("2014-11-16T18:30:00Z") } }
{ "_id" : ObjectId("5469a1dce4b07f3c07010b59"), "value" : { "count" : 2, "lastOrderDate" : ISODate("2015-04-29T18:30:00Z") } }
{ "_id" : ObjectId("5469a96ce4b07f3c07010b5e"), "value" : { "count" : 1, "orderDate" : ISODate("2014-11-16T18:30:00Z") } }
{ "_id" : ObjectId("5469c1ece4b07f3c07010b64"), "value" : { "count" : 9, "lastOrderDate" : ISODate("2015-04-15T18:30:00Z") } }
{ "_id" : ObjectId("5469f422e4b0ce7d5ee021ad"), "value" : { "count" : 5, "lastOrderDate" : ISODate("2015-06-01T18:30:00Z") } }
......

現在我想運行查詢並根據不同類別的計數對用戶進行分組，例如一組中小於5，另一組中小於5的用戶等

並希望輸出這樣的東西

{userLessThan5: 9 }
{user5to10: 2 }
{user10to15: 1 }
{user15to20: 0 }
  ....

Answer 1

嘗試這個，

db.order.mapReduce(function() { emit (this.customer,{count:1,orderDate:this.orderDate.interval_start}) },
function(key,values){ 
var category; // add this new field
var sum =0 ; var lastOrderDate;  
values.forEach(function(value) {
 if(value['orderDate']){ 
    lastOrderDate=value['orderDate'];
}  
sum+=value['count'];
}); 
// at this point you are already aware in which category your records lies , just add a new field to mark it
 if(sum < 5){ category: userLessThan5};
 if(sum >= 5 && sum <=10){ category: user5to10};
 if(sum <= 10 && sum >= 15){ category: user10to15};
 if(sum <= 15 && sum >=20){ category: user15to20};
  ....
return {count:sum,lastOrderDate:lastOrderDate,category:category}; 
},
{ query:{status:"DELIVERED"},out:"order_total"}).find()
 db.order_total.aggregate([{ $group: { "_id": "$value.category", "users": { $sum: 1 } } }]);

你會得到你想要的結果

{userLessThan5: 9 }
{user5to10: 2 }
{user10to15: 1 }
{user15to20: 0 }
 ....

Answer 2

根據我的知識，我在聚合中使用您的數據編寫了一個查詢，可能有更好的方法來解決這個問題。

var a=db.test.aggregate([{$match:{"value.count":{$lt:5}}},
              { $group: { _id:"$value.count",total:{"$sum":1}}},
             {$group:{_id:"less than 5",total:{$sum:"$total"}}}])              

var b=db.test.aggregate([{$match:{"value.count":{$lt:10,$gt:5}}},
            { $group: { _id:"$value.count",total:{"$sum":1}}},
            {$group:{_id:"between 5 and 10",total:{$sum:"$total"}}}])

var c=db.test.aggregate([{$match:{"value.count":{$lt:15,$gt:10}}},
       { $group: { _id:"$value.count",total:{"$sum":1}}},
       {$group:{_id:"between 10 and 15",total:{$sum:"$total"}}}])

將a，b，c插入另一個集合中

Answer 3

你可以通過嘗試組映射精簡后的輸出數據每5間隔計數aggregate如下圖所示

db.data.aggregate([
    { "$group": {
        "_id": {
            "$subtract": [
                { "$subtract": [ "$value.count", 0 ] },
                { "$mod": [ 
                    { "$subtract": [ "$value.count", 0 ] },
                    5
                ]}
            ]
        },
        "count": { "$sum": 1 }
    }}
])

也許這里有一個相關的問題 。