MongoDB distinct aggregation

Question

I\'m working on a query to find cities with most zips for each state:

db.zips.distinct(\"state\", db.zips.aggregate([ {$group:{_id:{state:\"$state\", city:\"         


        

Answer 1

You can use $addToSet with the aggregation framework to count distinct objects.

For example:

db.collectionName.aggregate([{
    $group: {_id: null, uniqueValues: {$addToSet: "$fieldName"}}
}])

Answer 2

Distinct and the aggregation framework are not inter-operable.

Instead you just want:

db.zips.aggregate([ 
    {$group:{_id:{city:'$city', state:'$state'}, numberOfzipcodes:{$sum:1}}}, 
    {$sort:{numberOfzipcodes:-1}},
    {$group:{_id:'$_id.state', city:{$first:'$_id.city'}, 
              numberOfzipcode:{$first:'$numberOfzipcodes'}}}
]);

Answer 3

SQL Query: (group by & count of distinct)

select city,count(distinct(emailId)) from TransactionDetails group by city;

Equivalent mongo query would look like this:

db.TransactionDetails.aggregate([ 
{$group:{_id:{"CITY" : "$cityName"},uniqueCount: {$addToSet: "$emailId"}}},
{$project:{"CITY":1,uniqueCustomerCount:{$size:"$uniqueCount"}} } 
]);

Answer 4

You can call $setUnion on a single array, which also filters dupes:

{ $project: {Package: 1, deps: {'$setUnion': '$deps.Package'}}}