Groupby 与 TimeGrouper '向后' [英] Groupby with TimeGrouper 'backwards'

问题描述

我有一个包含时间序列的 DataFrame:

I have a DataFrame containing a time series:

rng = pd.date_range('2016-06-01', periods=24*7, freq='H')
ones = pd.Series([1]*24*7, rng)
rdf = pd.DataFrame({'a': ones})

最后一个条目是 2016-06-07 23:00:00.我现在想按两天分组，基本上是这样的:

Last entry is 2016-06-07 23:00:00. I now want to group this by, say two days, basically like so:

rdf.groupby(pd.TimeGrouper('2D')).sum()

但是，我想从最后一个数据点开始向后分组，所以不要得到这个结果:

However, I want to group starting from my last data point backwards, so instead of getting this result:

            a
2016-06-01  48
2016-06-03  48
2016-06-05  48
2016-06-07  24

我更希望这样:

            a
2016-06-01  24
2016-06-03  48
2016-06-05  48
2016-06-07  48

当按'3D'分组时:

            a
2016-06-01  24
2016-06-04  72
2016-06-07  72

按'4D'分组时的预期结果是:

Expected outcome when grouping by '4D' is:

            a
2016-06-03  72
2016-06-07  96

我无法通过我能想到的 closed、label 等的每种组合来实现这一点.

I am not able to get this with every combination of closed, label etc. I could think of.

我怎样才能做到这一点?

How can I achieve this?

推荐答案

由于我主要想按 7 天(也就是一周)分组，所以我现在使用这种方法来找到所需的垃圾箱:

Since I primarily want to group by 7 days, aka one week, I am using this method now to come to the desired bins:

from pandas.tseries.offsets import Week

# Let's not make full weeks
hours = 24*6*4
rng = pd.date_range('2016-06-01', periods=hours, freq='H')

# Set week start to whatever the last weekday of the range is
print("Last day is %s" % rng[-1])
freq = Week(weekday=rng[-1].weekday())

ones = pd.Series([1]*hours, rng)
rdf = pd.DataFrame({'a': ones})
rdf.groupby(pd.TimeGrouper(freq=freq, closed='right', label='right')).sum()

这给了我想要的输出

2016-06-25  96
2016-07-02  168
2016-07-09  168

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