Python重新采样无法索引的时间序列数据

时间：2018-01-22 10:18:21

标签： python pandas pandas-groupby

这个问题的目的是知道每秒（计数）中发生了多少交易以及交易的总交易量（总和）。

我有时间序列数据无法编入索引（因为有多个条目具有相同的时间戳 - 可以在同一毫秒内获得许多交易）因此使用resample，如此处所述不能工作

另一种方法是首先按时间分组here（后来每秒重新采样）。问题是分组将只对分组项目产生一个基本算术（我只能求和/平均/标准等），而在这个数据中，我需要将'tradeVolume'列按总和分组'ask1'按平均值分组。

所以我的问题是： 1.如何group by每列不同的算术 如果不可能，还有其他方法可以将毫秒数据重新采样为秒而没有日期时间索引。

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

1 个答案:

答案 0 :(得分：2)

首先，你需要有一个秒的列（自纪元以来），然后{@ 1}}使用该列，然后对你想要的列进行聚合。

您希望将时间戳降低到一秒精度，并使用它进行分组。然后应用聚合来获得所需的均值/ sum / std 

md-backdrop

我修改了数据以确保其实际上有不同的秒数，

.md-select-menu-container {
    z-index: 900; }

md-backdrop.md-select-backdrop {
    z-index: 899; }

和输出

groupby

脚注

OP说原版（下图）速度更快，所以我跑了一些时间

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})

我在两个数据集上测试了一个，大小是第一个数据集的10倍，结果

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

因此，最好使用In [53]: groups.agg({'ask1': np.mean, 'tradeVolume': np.sum}) Out[53]: ask1 tradeVolume seconds 1511164809 12869.0 10 1511164810 12869.0 10 版本，因为这是最快的并且可以扩展到最佳版本。

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