CODEWITHSUNDEEP
pythonpandas
Pablo
Pablo

Reputation: 1003

How to construct a TimeGrouper with a DateOffset?

I'm trying to programatically construct a pandas TimeGrouper. A quick look at the code shows that the freq parameter of the TimeGrouper's __init__ method gets converted into a DateOffset by the to_offset() function. Furthermore, to_offset() checks whether its parameter is an instance of DateOffset and if true, returns it.

So, this code should work:

import pandas as pd
period = 'minute'
value = 10
time_grouper = pd.TimeGrouper(pd.DateOffset(**{period:value}))

However, I get the following exception:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\site-packages\pandas\tseries\resample.py", line 45, in __init__
    rule = self.freq.rule_code
  File "C:\Python27\lib\site-packages\pandas\tseries\offsets.py", line 217, in rule_code
    raise NotImplementedError
NotImplementedError

I'm using version 0.12.0 of pandas.

What am I missing to get working instances of DateOffset and TimeGrouper?

Upvotes: 2

Views: 1216

Answers (1)

Jeff
Jeff

Reputation: 128948

You don't need to construct an offset explicity (though you can if you want), simply passing '10T' is enough (this is what to_offset does, converts it to an offset object). Furthermore, you rarely need to explicity construct aTimeGrouper(and in 0.14.0, releasing shorty), you don't need to at all. You generally justresample``

In [5]: pd.offsets.Minute('10')
Out[5]: <10 * Minutes>

In [6]: pd.TimeGrouper(freq='10T')
Out[6]: <pandas.tseries.resample.TimeGrouper at 0x3faab90>

In [7]: pd.TimeGrouper(freq='10T').freq
Out[7]: <10 * Minutes>

In [8]: pd.TimeGrouper(freq=pd.offsets.Minute('10')).freq
Out[8]: <10 * Minutes>

You could also do this:

In [1]: values = { 'minute' : 10, 'hour' : 5 }

In [3]: [ getattr(pd.offsets,k.capitalize())(v) for k,v in values.items() ]
Out[3]: [<5 * Hours>, <10 * Minutes>]

Or this (by specifying a tuple)

In [3]: pd.TimeGrouper(freq=(5,'Min')).freq
Out[3]: <5 * Minutes>

In [4]: pd.TimeGrouper(freq=(10,'H')).freq
Out[4]: <10 * Hours>

Upvotes: 1

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