Groupby with TimeGrouper 'backwards'

Question

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 =         


        

Answer 1

Since the question now focuses on grouping by week, you can simply:

rdf.resample('W-{}'.format(rdf.index[-1].strftime('%a')), closed='right', label='right').sum()

You can use loffset to get it to work - at least for most periods (using .resample()):

for i in range(2, 7):
    print(i)
    print(rdf.resample('{}D'.format(i), closed='right', loffset='{}D'.format(i)).sum())

2
             a
2016-06-01  24
2016-06-03  48
2016-06-05  48
2016-06-07  48
3
             a
2016-06-01  24
2016-06-04  72
2016-06-07  72
4
             a
2016-06-01  24
2016-06-05  96
2016-06-09  48
5
              a
2016-06-01   24
2016-06-06  120
2016-06-11   24
6
              a
2016-06-01   24
2016-06-07  144

However, you could also create custom groupings that calculate the correct values without TimeGrouper like so:

days = rdf.index.to_series().dt.day.unique()[::-1]
for n in range(2, 7):
    chunks = [days[i:i + n] for i in range(0, len(days), n)][::-1]
    grp = pd.Series({k: v for d in [zip(chunk, [idx] * len(chunk)) for idx, chunk in enumerate(chunks)] for k, v in d})
    rdf.groupby(rdf.index.to_series().dt.day.map(grp))['a'].sum()

 2
groups
0    24
1    48
2    48
3    48
Name: a, dtype: int64

 3
groups
0    24
1    72
2    72
Name: a, dtype: int64

 4
groups
0    72
1    96
Name: a, dtype: int64

 5
groups
0     48
1    120
Name: a, dtype: int64

 6
groups
0     24
1    144
Name: a, dtype: int64