mongodb - mongo group and count with condition -

i'm trying group set of documents , count them based on value. example

{ "_id" : 1, "item" : "abc1", "value" : "1" } { "_id" : 2, "item" : "abc1", "value" : "1" } { "_id" : 3, "item" : "abc1", "value" : "11" } { "_id" : 4, "item" : "abc1", "value" : "12" } { "_id" : 5, "item" : "xyz1", "value" : "2" } 

here group "item" , in return count how many times "value" bigger 10 , how many times smaller. so:

{ "item": "abc1", "countsmaller": 2, "countbigger": 1} { "item": "xyz1", "countsmaller": 1, "countbigger": 0} 

a plain count achieved $aggregate, how can achieve above result?

what need $cond operator of aggregation framework. 1 way want be:

db.foo.aggregate([     {         $project: {             item: 1,             lessthan10: {  // set 1 if value < 10                 $cond: [ { $lt: ["$value", 10 ] }, 1, 0]             },             morethan10: {  // set 1 if value > 10                 $cond: [ { $gt: [ "$value", 10 ] }, 1, 0]             }         }     },     {         $group: {             _id: "$item",             countsmaller: { $sum: "$lessthan10" },             countbigger: { $sum: "$morethan10" }         }     } ]) 

note: have assumed value numeric rather string.

output:

{         "result" : [                 {                         "_id" : "xyz1",                         "countsmaller" : 1,                         "countbigger" : 0                 },                 {                         "_id" : "abc1",                         "countsmaller" : 2,                         "countbigger" : 2                 }         ],         "ok" : 1 }