How to Group mongodb - mapReduce output?

Question

i have a query regarding the mapReduce framework in mongodb, so i have a result of key value pair from mapReduce function , now i want to run the query on this output of mapReduce.

So i am using mapReduce to find out the stats of user like this

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()

which give me output like this

{ "_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") } }
......

Now i want to run query and group the users on the basis of count in different categories like for user with count less than 5 in one group , 5-10 in another, etc

and want output something like this

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

roanjain · Accepted Answer · 2016-02-19 20:57:48Z

Try this,

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 } } }]);

you will get you desired result

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

user3700158 · Accepted Answer · 2016-02-16 10:43:15Z

0

I wrote a query using your data in aggregation as per my knowledge, there may be better way to solve this problem.

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"}}}])

insert a, b, c into another collection

edited Feb 16, 2016 at 10:43

answered Feb 16, 2016 at 10:38

user3700158

12 bronze badges

2 Comments

user29578 Over a year ago

actually your solution suggests that first i need to insert data in test collection, also the queries are limited to sample data i have shown, but in real i want to have all such categories of 5 records interval.

user3700158 Over a year ago

I have taken some test collection for testing, replace it with your "order_total" collection.I will also find if any better approach is there to solve this problem.

Community · Accepted Answer · 2017-05-23 11:50:34Z

0

You could try to group the output data after mapreduce to every 5 interval count through aggregate like below

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

Also maybe here is one related question here.

edited May 23, 2017 at 11:50

CommunityBot

11 silver badge

answered Feb 16, 2016 at 11:37

zangw

49.1k24 gold badges214 silver badges249 bronze badges

Collectives™ on Stack Overflow

How to Group mongodb - mapReduce output?

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related