Python:数据帧分组后timedelta的平均值

Question

我知道，这是一个非常流行的问题，有很多关于平均时间增量和日期时间参数的话题。不幸的是，我被我的方法卡住了，我想请您协助完成以下任务：

作为行的结果：

time_to_rent = user_payments[user_payments.rentComplete].groupby(['rentId','creditCardId','rentComplete'], as_index=False).agg({'createdAt': np.min, 'updatedAt': np.max})

我得到了数据帧(dict格式(：

time_to_rent = {'rentId': {0: 43.0, 1: 87.0, 2: 140.0, 3: 454.0, 4: 1458.0}, 'creditCardId': {0: 40, 1: 40, 2: 40, 3: 40, 4: 40}, 'rentComplete': {0: True, 1: True, 2: True, 3: True, 4: False}, 'createdAt': {0: Timestamp('2020-08-24 16:13:11.850216'), 1: Timestamp('2020-09-10 10:47:31.748628'), 2: Timestamp('2020-09-13 15:29:06.077622'), 3: Timestamp('2020-09-24 08:08:39.852348'), 4: Timestamp('2020-10-19 08:54:09.891518')}, 'updatedAt': {0: Timestamp('2020-08-24 20:26:31.805939'), 1: Timestamp('2020-09-10 20:05:18.759421'), 2: Timestamp('2020-09-13 18:38:10.044112'), 3: Timestamp('2020-09-24 08:53:22.512533'), 4: Timestamp('2020-10-19 09:24:03.982986')}, 'rent_time': {0: Timedelta('0 days 04:13:19.955723'), 1: Timedelta('0 days 09:17:47.010793'), 2: Timedelta('0 days 03:09:03.966490'), 3: Timedelta('0 days 00:44:42.660185'), 4: Timedelta('0 days 00:29:54.091468')}}

然后我再添加一列：

time_to_rent['rent_time'] = time_to_rent['updatedAt'] - time_to_rent['createdAt']

我想将time_to_rent分组为"；creditCardId"；并且具有"0"的平均值；rent_ time"；柱

该代码返回错误：

average_per_user = time_to_rent.groupby('creditCardId').agg({'rent_time' : np.mean})

这是返回的错误：

~anaconda3libsite-packagespandascoregroupbygroupby.py in _cython_agg_general(self, how, alt, numeric_only, min_count)
906 
907         if len(output) == 0:
--> 908             raise DataError("No numeric types to aggregate")
909 
910         return self._wrap_aggregated_output(output)
DataError: No numeric types to aggregate

不确定为什么len(输出(等于0。。。

Answer 1

如果我理解正确，您需要从df.rentComplete中排除False值(并用)结束整个过程(。在Pandas:中，使用布尔列进行过滤很简单

average_per_user = time_to_rent[time_to_rent.rentComplete] 
.groupby('creditCardId').agg({'rent_time' : np.mean})

Answer 2

试试这个：

average_per_user = time_to_rent.groupby('creditCardId').mean()['rent_time']