Question

我正在尝试使用pymongo创建一个脚本来计算这样的几个事务：

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

我成功使用MongoDB直接聚合这些数据，以下是示例结果：

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

但是，我没有使用pymongo聚合这些数据并得到错误

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)

这是我的剧本：

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？提前致谢

Answer 1

查看aggregate语法http://api.mongodb.com/python/current/examples/aggregation.html 您需要使用[]作为参数。

在每日分组中使用Pymongo的聚合

1 个答案: