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I use TimeGrouper from pandas.tseries.resample to sum monthly return to 6M as follows:

6m_return = monthly_return.groupby(TimeGrouper(freq='6M')).aggregate(numpy.sum)

where monthly_return is like:

2008-07-01    0.003626
2008-08-01    0.001373
2008-09-01    0.040192
2008-10-01    0.027794
2008-11-01    0.012590
2008-12-01    0.026394
2009-01-01    0.008564
2009-02-01    0.007714
2009-03-01   -0.019727
2009-04-01    0.008888
2009-05-01    0.039801
2009-06-01    0.010042
2009-07-01    0.020971
2009-08-01    0.011926
2009-09-01    0.024998
2009-10-01    0.005213
2009-11-01    0.016804
2009-12-01    0.020724
2010-01-01    0.006322
2010-02-01    0.008971
2010-03-01    0.003911
2010-04-01    0.013928
2010-05-01    0.004640
2010-06-01    0.000744
2010-07-01    0.004697
2010-08-01    0.002553
2010-09-01    0.002770
2010-10-01    0.002834
2010-11-01    0.002157
2010-12-01    0.001034

The 6m_return is like:

2008-07-31    0.003626
2009-01-31    0.116907
2009-07-31    0.067688
2010-01-31    0.085986
2010-07-31    0.036890
2011-01-31    0.015283

However I want to get the 6m_return starting 6m from 7/2008 like the following:

2008-12-31    ...
2009-06-31    ...
2009-12-31    ...
2010-06-31    ...
2010-12-31    ...

Tried the different input options (i.e. loffset) in TimeGrouper but doesn't work. Any suggestion will be really appreciated!

Andy Hayden
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user2019264
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  • You *want* to be able to do something like: `s.resample('2Q-DEC', how='sum')` but six monthlys don't seem well supported (or perhaps there is a bug). loffset just changes the label, not the calculation, so I don't think you want to do that. – Andy Hayden Jan 28 '13 at 20:09
  • @AndyHayden I think it is a strange bug as `s.index[0] + pd.datetools.QuarterEnd(startingMonth=12) * 2` is ``, however `s.resample(pd.datetools.QuarterEnd(startingMonth=12) * 2)` starts with `2008-09-30`. I opened a [ticket](https://github.com/pydata/pandas/issues/2764) – bmu Jan 28 '13 at 20:44

3 Answers3

8

The problem can be solved by adding closed = 'left'

df.groupby(pd.TimeGrouper('6M', closed = 'left')).aggregate(numpy.sum)
Ankur
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Anja
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3

TimeGrouper that is suggested in other answers is deprecated and will be removed from Pandas. It is replaced with Grouper. So a solution to your question using Grouper is:

df.groupby(pd.Grouper(freq='6M', closed='left')).aggregate(numpy.sum)
Primoz
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2

This is a workaround for what seems a bug, but give it a try and see if it works for you.

In [121]: ts = pandas.date_range('7/1/2008', periods=30, freq='MS')

In [122]: df = pandas.DataFrame(pandas.Series(range(len(ts)), index=ts))

In [124]: df[0] += 1

In [125]: df

Out[125]: 
             0
2008-07-01   1
2008-08-01   2
2008-09-01   3
2008-10-01   4
2008-11-01   5
2008-12-01   6
2009-01-01   7
2009-02-01   8
2009-03-01   9
2009-04-01  10
2009-05-01  11
2009-06-01  12
2009-07-01  13
2009-08-01  14
2009-09-01  15
2009-10-01  16
2009-11-01  17
2009-12-01  18
2010-01-01  19
2010-02-01  20
2010-03-01  21
2010-04-01  22
2010-05-01  23
2010-06-01  24
2010-07-01  25
2010-08-01  26
2010-09-01  27
2010-10-01  28
2010-11-01  29
2010-12-01  30

I've used integers to help confirm that the sums are correct. The workaround that seems to work is to add a month to the front of the dataframe to trick the TimeGrouper into doing what you need.

In [127]: df2 = pandas.DataFrame([0], index = [df.index.shift(-1, freq='MS')[0]])

In [129]: df2.append(df).groupby(pandas.TimeGrouper(freq='6M')).aggregate(numpy.sum)[1:]
Out[129]: 
              0
2008-12-31   21
2009-06-30   57
2009-12-31   93
2010-06-30  129
2010-12-31  165

Note the final [1:] is there to trim off the first group.

D. A.
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