Getting the highest value of a column in MongoDB

Question

I've been for some help on getting the highest value on a column for a mongo document. I can sort it and get the top/bottom, but I'm pretty sure there is a better way to do it.

I tried the following (and different combinations):

transactions.find("id" => x).max({"sellprice" => 0})

But it keeps throwing errors. What's a good way to do it besides sorting and getting the top/bottom?

Thank you!

Supertax

You should include the errors it's throwing. – Tiddlywinks

Answer 1

max() does not work the way you would expect it to in SQL for Mongo. This is perhaps going to change in future versions but as of now, max,min are to be used with indexed keys primarily internally for sharding.

see http://www.mongodb.org/display/DOCS/min+and+max+Query+Specifiers

Unfortunately for now the only way to get the max value is to sort the collection desc on that value and take the first.

transactions.find("id" => x).sort({"sellprice" => -1}).limit(1).first()

Satrap

Answer 2

Sorting might be overkill. You can just do a group by

db.messages.group(
           {key: { created_at:true },
            cond: { active:1 },
            reduce: function(obj,prev) { if(prev.cmax<obj.created_at) prev.cmax = obj.created_at; },
            initial: { cmax: **any one value** }
            });

Monto

I wouldn't say it's an overkill considering this quote from the manual "When a $sort immediately precedes a $limit in the pipeline, the $sort operation only maintains the top n results as it progresses, where n is the specified limit" – Trudietrudnak

Answer 3

db.collectionName.aggregate(
  {
    $group : 
    {
      _id  : "",
      last : 
      {
        $max : "$sellprice"
      }
    }
  }
)

Progress

Answer 4

Example mongodb shell code for computing aggregates.

see mongodb manual entry for group (many applications) :: http://docs.mongodb.org/manual/reference/aggregation/group/#stage._S_group

In the below, replace the $vars with your collection key and target variable.

db.activity.aggregate( 
  { $group : {
      _id:"$your_collection_key", 
      min: {$min : "$your_target_variable"}, 
      max: {$max : "$your_target_variable"}
    }
  } 
)

Spindling

Answer 5

Use aggregate():

db.transactions.aggregate([
  {$match: {id: x}},
  {$sort: {sellprice:-1}},
  {$limit: 1},
  {$project: {sellprice: 1}}
]);

Athabaska

Answer 6

It will work as per your requirement.

transactions.find("id" => x).sort({"sellprice" => -1}).limit(1).first()

Goodtempered

What's the point of this answer? Repeating the accepted answer? – Schwarz

Answer 7

If the column's indexed then a sort should be OK, assuming Mongo just uses the index to get an ordered collection. Otherwise it's more efficient to iterate over the collection, keeping note of the largest value seen. e.g.

max = nil
coll.find("id" => x).each do |doc| 
    if max == nil or doc['sellprice'] > max then
        max = doc['sellprice'] 
    end
end

(Apologies if my Ruby's a bit ropey, I haven't used it for a long time - but the general approach should be clear from the code.)

Painless

Consider to iterate with MapReduce !!! This why MongoDB is cool, else use flat files ;-) – Pentheam

Answer 8

Assuming I was using the Ruby driver (I saw a mongodb-ruby tag on the bottom), I'd do something like the following if I wanted to get the maximum _id (assuming my _id is sortable). In my implementation, my _id was an integer.

result = my_collection.find({}, :sort => ['_id', :desc]).limit(1)

To get the minimum _id in the collection, just change :desc to :asc

Betrothed

Wouldn't find_one() be better than using limit()? result = my_collection.find_one({}, :sort => ['_id', :desc]) – Footle

Answer 9

Following query does the same thing: db.student.find({}, {'_id':1}).sort({_id:-1}).limit(1)

For me, this produced following result: { "_id" : NumberLong(10934) }

Spires