Resampling with custom periods

Question

Is there a \'cookbook\' way of resampling a DataFrame with (semi)irregular periods?

I have a dataset at a daily interval and want it to resample to what sometimes (in s

Answer 1

Using HYRY's data and solution up to the computation of the d variable, we can also do the following in pandas 0.11-dev or later (regardless of numpy version):

In [18]: from datetime import timedelta

In [23]: pd.Series([ timedelta(int(i)) for i in d ])
Out[23]: 
0             00:00:00
1     1 days, 00:00:00
2     2 days, 00:00:00
3     3 days, 00:00:00
4     4 days, 00:00:00
5     5 days, 00:00:00
6     6 days, 00:00:00
7     7 days, 00:00:00
8     8 days, 00:00:00
9     9 days, 00:00:00
10            00:00:00

47    6 days, 00:00:00
48    7 days, 00:00:00
49    8 days, 00:00:00
50    9 days, 00:00:00
Length: 51, dtype: timedelta64[ns]

The date is constructed similary to above

date = pd.Series(df.index) - pd.Series([ timedelta(int(i)) for i in d ])
df.groupby(date.values).mean()

Answer 2

If you use numpy 1.7, you can use datetime64 & timedelta64 arrays to do the calculation:

create the sample data:

import pandas as pd
import numpy as np

begin = pd.datetime(2013,1,1)
end = pd.datetime(2013,2,20)

dtrange = pd.date_range(begin, end)

p1 = np.random.rand(len(dtrange)) + 5
p2 = np.random.rand(len(dtrange)) + 10

df = pd.DataFrame({'p1': p1, 'p2': p2}, index=dtrange)

calculate the dekad's date:

d = df.index.day - np.clip((df.index.day-1) // 10, 0, 2)*10 - 1
date = df.index.values - np.array(d, dtype="timedelta64[D]")
df.groupby(date).mean()

The output is:

                 p1         p2
2013-01-01  5.413795  10.445640
2013-01-11  5.516063  10.491339
2013-01-21  5.539676  10.528745
2013-02-01  5.783467  10.478001
2013-02-11  5.358787  10.579149