Question

我正试图在整个集合中获得数组元素的总体数量。

输入集合

 db.test.insert(    { "_id" : 1, "item" : "A", "hx" : [ { "label" : "new"    , "price" : 10.99 , "category" : "P1"} , 
                                                { "label" : "active" , "price" : 12.99 , "category" : "P2"} ] } )

 db.test.insert(    { "_id" : 2, "item" : "B", "hx" : [ { "label" : "new"    , "price" : 10.99 , "category" : "P2"} , 
                                                { "label" : "active" , "price" :  8.99 , "category" : "P3"} ] } )

 db.test.insert(    { "_id" : 3, "item" : "C", "hx" : [ { "label" : "new"    , "price" : 10.99 , "category" : "P1"} , 
                                                { "label" : "active" , "price" : 15.99 , "category" : "P4"} ] } )

我尝试了各种

 a = db.test.aggregate(
    [ { 
          $group: { 
                   _id : "$cat", 
                   cat : { $addToSet : "$hx.category" } 
                  } 
      },  
      { $unwind : "$hx.category" } , 
      { $group : { _id : "$cat", count: { $sum : 1 } } }
    ])

到达

  db.res.insert({ "category" : "P1" , "count" : 2 })
  db.res.insert({ "category" : "P2" , "count" : 2 })
  db.res.insert({ "category" : "P3" , "count" : 1 })
  db.res.insert({ "category" : "P4" , "count" : 1 })

但我得到的结果是

 > a
 { "result" : [ ], "ok" : 1 }

最终我想在pymongo中运行它：非常感谢任何关于pymongo实现的提示。

Answer 1

您可以尝试以下聚合：

>>> from bson.son import SON
>>> pipeline = [
...     {"$unwind": "$hx"},
...     {"$group": {"_id": "$hx.category", "count": {"$sum": 1}}},
...     {"$project": SON([("count", -1), ("_id", -1)])}
... ]
>>> list(db.test.aggregate(pipeline))
[{u'count': 1, u'_id': u'P3'}, {u'count': 2, u'_id': u'P2'}, {u'count': 1, u'_id': u'P4'}, {u'count': 2, u'_id': u'P1'}]

Answer 2

我假设您的类别字段是具有唯一条目的集合。

db.test.aggregate([ 
    {$unwind: "$hx"},
    {$group: {_id: "$hx.category", count:{$sum:1}}}
])

Answer 3

看起来像Map-Reduce

的任务

mongodb：计算独特的数组元素

3 个答案: