数据库中的 Pymongo 计算

Question

我的目标：

计算：

"open" * 1.1

在pymongo输出数据之前，因为我认为它会更快

数据如下所示：

{
    "_id": "000016.XSHE_2020-01-02-09-32",
    "order_book_id": "000016.XSHE",
    "datetime": {
        "$date": "2020-01-02T09:32:00.000Z"
    },
    "num_trades": 0,
    "low": 4.3131,
    "close": 4.3228,
    "high": 4.3326,
    "total_turnover": 2826041,
    "volume": 639124,
    "open": 4.3228
}

{
    "_id": "000016.XSHE_2020-01-02-09-33",
    "order_book_id": "000016.XSHE",
    "datetime": {
        "$date": "2020-01-02T09:33:00.000Z"
    },
    "num_trades": 0,
    "low": 4.3131,
    "close": 4.3228,
    "high": 4.3326,
    "total_turnover": 1087488,
    "volume": 246300,
    "open": 4.3131
}....

我正在尝试以下代码，但 none 有效：

tdcursor = MinutesDB.find({"order_book_id": "000016.XSHE", "datetime": {"$gte": recent11TradingDays[-1]},
                           "uplimit":{"$multiply":{"open": 1.1}}},
                          {"_id": 0, "order_book_id":1, "datetime": 1, "open": 1, "close": 1})
datadf = pd.DataFrame(tdcursor)

或

multiply = MinutesDB.aggregate([
    {"$group":
    {"order_book_id": "000016.XSHE",
    "uplimit": {"$multiply": ["$open", 1.1]}
    }
}
])
datadf = pd.DataFrame(multiply)

上面的代码给出如下错误：

pymongo.errors.OperationFailure: The $in accumulator is a unary operator, full error: {'ok': 0.0, 'errmsg': 'The $in accumulator is a unary operator', 'code': 40237, 'codeName': 'Location40237'}

感谢您的帮助！

Answer 1

这里有一个聚合的解决方案

db.collection.aggregate([
  {
    "$match": {
      "order_book_id": "000016.XSHE",
      
    }
  },
  {
    "$project": {
      "uplimit": {
        "$multiply": [
          "$open",
          1.1
        ]
      }
    }
  }
])

由于您只想查询匹配的文档：order_book_id": "000016.XSHE，您将需要 $match聚合阶段

然后$project使用$multiply

试一试here

数据库中的 Pymongo 计算

Pymongo calculation within database

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