pandas -按日期分组日内时间序列 [英] Pandas - grouping intra day timeseries by date

问题描述

我有一个连续几天的日内日志返回值，我想将其降采样为每日ohlc.我可以做类似的事情

I have an intra day series of log returns over multiple days that I would like to downsample to daily ohlc. I can do something like

hi = series.resample('B', how=lambda x: np.max(np.cumsum()))
low = series.resample('B', how=lambda x: np.min(np.cumsum())) 

但是在每次通话中计算总和似乎效率低下.有没有一种方法可以先计算出总和，然后对数据应用"ohcl"?

But it seems inefficient to compute cumsum on each call. Is there a way to first compute the cumsums and then apply 'ohcl' to the data?

1999-08-09 12:30:00-04:00   -0.000486
1999-08-09 12:31:00-04:00   -0.000606
1999-08-09 12:32:00-04:00   -0.000120
1999-08-09 12:33:00-04:00   -0.000037
1999-08-09 12:34:00-04:00   -0.000337
1999-08-09 12:35:00-04:00    0.000100
1999-08-09 12:36:00-04:00    0.000219
1999-08-09 12:37:00-04:00    0.000285
1999-08-09 12:38:00-04:00   -0.000981
1999-08-09 12:39:00-04:00   -0.000487
1999-08-09 12:40:00-04:00    0.000476
1999-08-09 12:41:00-04:00    0.000362
1999-08-09 12:42:00-04:00   -0.000038
1999-08-09 12:43:00-04:00   -0.000310
1999-08-09 12:44:00-04:00   -0.000337
...
1999-09-28 06:45:00-04:00    0.000000
1999-09-28 06:46:00-04:00    0.000000
1999-09-28 06:47:00-04:00    0.000000
1999-09-28 06:48:00-04:00    0.000102
1999-09-28 06:49:00-04:00   -0.000068
1999-09-28 06:50:00-04:00    0.000136
1999-09-28 06:51:00-04:00    0.000566
1999-09-28 06:52:00-04:00    0.000469
1999-09-28 06:53:00-04:00    0.000000
1999-09-28 06:54:00-04:00    0.000000
1999-09-28 06:55:00-04:00    0.000000
1999-09-28 06:56:00-04:00    0.000000
1999-09-28 06:57:00-04:00    0.000000
1999-09-28 06:58:00-04:00    0.000000
1999-09-28 06:59:00-04:00    0.000000

推荐答案

df.groupby([df.index.year, df.index.month, df.index.day]).transform(np.cumsum).resample('B', how='ohlc')

我认为这可能是我想要的，但我必须进行测试.

I think this might be what I want but I have to test.

在zelazny7的回复之后:

After zelazny7's repsonse:

df.groupby(pd.TimeGrouper('D')).transform(np.cumsum).resample('D', how='ohlc')

有效，并且比我以前的解决方案更有效.

works and is also more efficient than my previous solution.

更新:

由于 pandas v0.21.0 .

使用 pd.Grouper() 相反:

df.groupby(pd.Grouper(freq='D')).transform(np.cumsum).resample('D', how='ohlc')

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