聚合使用Pymongo在每日分组

我试图用pymongo来算几个交易这样创建一个脚本：聚合使用Pymongo在每日分组

{ 
    "_id" : ObjectId("58437604a966aec46dfa249f"), 
    "transaction_id" : 282932121, 
    "transaction_serial_number" : "GtgT46A", 
    "transaction_date" : ISODate("2015-09-28T00:00:00Z"), 
    "card_type" : "MC", 
    "transaction_nominal" : 3500 
} 
...

我成功聚集直接使用MongoDB的这些数据，这里是抽样结果：

{ 
    "_id" : { 
     "card_type" : "MC", 
     "date" : ISODate("2015-09-28T00:00:00Z") 
    }, 
    "count" : 179011 
}

但是，我没有使用pymongo汇总这些数据，就有了一个错误

Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "script_aggregate.py", line 84, in <module> 
    cursor = db.transaction.aggregate(match, proj1, proj2, group); 
TypeError: aggregate() takes exactly 2 arguments (5 given)

这是我的脚本：

from datetime import datetime 
from pymongo import MongoClient 

client = MongoClient('localhost', 27017) 
db = client['mydb'] 
collection = db['transaction'] 

match={"$match" : { 
       "transaction_date" : { "$gt" : datetime(2013,1,1).isoformat() } 
     } 
}; 

proj1={"$project" : { 
     "_id" : 0, 
     "transaction_date" : 1, 
     "card_type" : 1, 
     "h" : { 
      "$hour" : "$transaction_date" 
     }, 
     "m" : { 
      "$minute" : "$transaction_date" 
     }, 
     "s" : { 
      "$second" : "$transaction_date" 
     }, 
     "ml" : { 
      "$millisecond" : "$transaction_date" 
     } 
    } 
}; 

proj2={"$project" : { 
     "_id" : 0, 
     "card_type" : 1, 
     "transaction_date" : { 
      "$subtract" : [ 
       "$transaction_date", 
       { 
        "$add" : [ 
         "$ml", 
         { 
          "$multiply" : [ 
           "$s", 
           1000 
          ] 
         }, 
         { 
          "$multiply" : [ 
           "$m", 
           60, 
           1000 
          ] 
         }, 
         { 
          "$multiply" : [ 
           "$h", 
           60, 
           60, 
           1000 
          ] 
         } 
        ] 
       } 
      ] 
     } 
    } 
}; 

group={"$group" : { 
     "_id" : { 
      "card_type" : "$card_type", 
      "date" : "$transaction_date" 
     }, 
     "count" : { 
      "$sum" : 1 
     } 
    } 
}; 





cursor = db.transaction.aggregate(match, proj1, proj2, group); 

for document in cursor: 
    print(document);

我该怎么办？在此先感谢

来源

2016-12-28 Misbach Imaduddin