无法重新索引的Python重新采样时间序列数据

Question

这个问题的目的是要知道每秒发生了多少笔交易（计数）以及总交易量（总和）。

我有无法编制索引的时间序列数据（因为存在多个具有相同时间戳的条目-可以在同一毫秒获得很多交易），因此无法使用此处说明的重新采样 。

另一种方法是首先通过一次做组如图这里 （和以后每秒重新取样）。 问题在于，分组将仅对分组的项目造成一种基本算术（我只能求和/均值/标准等），而在此数据中，我需要将“ tradeVolume”列按总和分组，而将列“ ask1”按均值分组。

所以我的问题是：1.如何对每列使用不同的算法进行group by ，如果不可能的话，还有其他方法可以将毫秒数据重新采样为秒，而没有datetime索引。

谢谢！

时间序列（样本）在这里：

SecurityID,dateTime,ask1,ask1Volume,bid1,bid1Volume,ask2,ask2Volume,bid2,bid2Volume,ask3,ask3Volume,bid3,bid3Volume,tradePrice,tradeVolume,isTrade
2318276,2017-11-20 08:00:09.052240,12869.0,1,12868.0,3,12870.0,19,12867.5,2,12872.5,2,12867.0,1,0.0,0,0
2318276,2017-11-20 08:00:09.052260,12869.0,1,12868.0,3,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12861.0,1,1
2318276,2017-11-20 08:00:09.052260,12869.0,1,12868.0,2,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052270,12869.0,1,12868.0,2,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,1
2318276,2017-11-20 08:00:09.052270,12869.0,1,12868.0,1,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052282,12869.0,1,12868.0,1,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,1
2318276,2017-11-20 08:00:09.052282,12869.0,1,12867.5,2,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052291,12869.0,1,12867.5,2,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,1
2318276,2017-11-20 08:00:09.052291,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,0
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.0,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12865.5,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12865.0,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12864.0,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12861.5,2,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12864.0,1,0
2318276,2017-11-20 08:00:09.052335,12869.0,1,12861.5,2,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,1
2318276,2017-11-20 08:00:09.052335,12869.0,1,12861.5,1,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,0
2318276,2017-11-20 08:00:09.052348,12869.0,1,12861.5,1,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,1
2318276,2017-11-20 08:00:09.052348,12869.0,1,12861.0,1,12870.0,19,12860.0,5,12872.5,2,12859.5,3,12861.5,1,0
2318276,2017-11-20 08:00:09.052357,12869.0,1,12861.0,1,12870.0,19,12860.0,5,12872.5,2,12859.5,3,12861.0,1,1
2318276,2017-11-20 08:00:09.052357,12869.0,1,12860.0,5,12870.0,19,12859.5,3,12872.5,2,12858.0,1,12861.0,1,0

Answer 1

首先，您需要有一秒钟的列（自纪元开始），然后使用该列进行 groupby ，然后对所需的列进行汇总。

您希望将时间戳降低到一秒的精度，并使用该精度进行分组。 然后应用聚合以获得所需的均值/和/ std

df = pd.read_csv('data.csv')
df['dateTime'] = df['dateTime'].astype('datetime64[s]')
groups = df.groupby('dateTime')
groups.agg({'ask1': np.mean, 'tradeVolume': np.sum}) 

我修改了数据以确保其中实际上有不同的秒数，

SecurityID,dateTime,ask1,ask1Volume,bid1,bid1Volume,ask2,ask2Volume,bid2,bid2Volume,ask3,ask3Volume,bid3,bid3Volume,tradePrice,tradeVolume,isTrade
2318276,2017-11-20 08:00:09.052240,12869.0,1,12868.0,3,12870.0,19,12867.5,2,12872.5,2,12867.0,1,0.0,0,0
2318276,2017-11-20 08:00:09.052260,12869.0,1,12868.0,3,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12861.0,1,1
2318276,2017-11-20 08:00:09.052260,12869.0,1,12868.0,2,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052270,12869.0,1,12868.0,2,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,1
2318276,2017-11-20 08:00:09.052270,12869.0,1,12868.0,1,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052282,12869.0,1,12868.0,1,12870.0,19,12867.5,2,12872.5,2,12867.0,1,12868.0,1,1
2318276,2017-11-20 08:00:09.052282,12869.0,1,12867.5,2,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12868.0,1,0
2318276,2017-11-20 08:00:09.052291,12869.0,1,12867.5,2,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,1
2318276,2017-11-20 08:00:09.052291,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,0
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.5,1,1
2318276,2017-11-20 08:00:09.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12867.0,1,1
2318276,2017-11-20 08:00:10.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12865.5,1,1
2318276,2017-11-20 08:00:10.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12865.0,1,1
2318276,2017-11-20 08:00:10.052315,12869.0,1,12867.5,1,12870.0,19,12867.0,1,12872.5,2,12865.5,1,12864.0,1,1
2318276,2017-11-20 08:00:10.052315,12869.0,1,12861.5,2,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12864.0,1,0
2318276,2017-11-20 08:00:10.052335,12869.0,1,12861.5,2,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,1
2318276,2017-11-20 08:00:10.052335,12869.0,1,12861.5,1,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,0
2318276,2017-11-20 08:00:10.052348,12869.0,1,12861.5,1,12870.0,19,12861.0,1,12872.5,2,12860.0,5,12861.5,1,1
2318276,2017-11-20 08:00:10.052348,12869.0,1,12861.0,1,12870.0,19,12860.0,5,12872.5,2,12859.5,3,12861.5,1,0
2318276,2017-11-20 08:00:10.052357,12869.0,1,12861.0,1,12870.0,19,12860.0,5,12872.5,2,12859.5,3,12861.0,1,1
2318276,2017-11-20 08:00:10.052357,12869.0,1,12860.0,5,12870.0,19,12859.5,3,12872.5,2,12858.0,1,12861.0,1,0

和输出

In [53]: groups.agg({'ask1': np.mean, 'tradeVolume': np.sum})
Out[53]: 
               ask1  tradeVolume
seconds                         
1511164809  12869.0           10
1511164810  12869.0           10

---

脚注

OP表示原始版本（如下）速度更快，所以我花了一些时间

def test1(df):
    """This is the fastest and cleanest."""
    df['dateTime'] = df['dateTime'].astype('datetime64[s]')
    groups = df.groupby('dateTime')
    agg = groups.agg({'ask1': np.mean, 'tradeVolume': np.sum})

def test2(df):
    """Totally unnecessary amount of datetime floors."""
    def group_by_second(index_loc):
        return df.loc[index_loc, 'dateTime'].floor('S')
    df['dateTime'] = df['dateTime'].astype('datetime64[ns]')
    groups = df.groupby(group_by_second)
    result = groups.agg({'ask1': np.mean, 'tradeVolume': np.sum})

def test3(df):
    """Original version, but the conversion to/from nanoseconds is unnecessary."""
    df['dateTime'] = df['dateTime'].astype('datetime64[ns]')
    df['seconds'] = df['dateTime'].apply(lambda v: v.value // 1e9)
    groups = df.groupby('dateTime')
    agg = groups.agg({'ask1': np.mean, 'tradeVolume': np.sum})

if __name__ == '__main__':
    import timeit
    print('22 rows')
    df = pd.read_csv('data_small.csv')
    print('test1', timeit.repeat("test1(df.copy())", number=50, globals=globals()))
    print('test2', timeit.repeat("test2(df.copy())", number=50, globals=globals()))
    print('test3', timeit.repeat("test3(df.copy())", number=50, globals=globals()))

    print('220 rows')
    df = pd.read_csv('data.csv')
    print('test1', timeit.repeat("test1(df.copy())", number=50, globals=globals()))
    print('test2', timeit.repeat("test2(df.copy())", number=50, globals=globals()))
    print('test3', timeit.repeat("test3(df.copy())", number=50, globals=globals()))

我在两个数据集上进行了测试，结果是第一个数据集的10倍

22 rows
test1 [0.08138518501073122, 0.07786444900557399, 0.0775048139039427]
test2 [0.2644687460269779, 0.26298125297762454, 0.2618108610622585]
test3 [0.10624988097697496, 0.1028324980288744, 0.10304366517812014]
220 rows
test1 [0.07999306707642972, 0.07842653687112033, 0.07848454895429313]
test2 [1.9794962559826672, 1.966513831866905, 1.9625889619346708]
test3 [0.12691736104898155, 0.12642419710755348, 0.126510804053396]

因此，最好使用.astype('datetime[s]')版本，因为这是最快的，并且可以最佳扩展。