Convert pandas timezone-aware DateTimeIndex to naive timestamp, but in certain timezone

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To answer my own question, this functionality has been added to pandas in the meantime. Starting from pandas 0.15.0, you can use tz_localize(None) to remove the timezone resulting in local time.
See the whatsnew entry: http://pandas.pydata.org/pandas-docs/stable/whatsnew.html#timezone-handling-improvements

So with my example from above:

In [4]: t = pd.date_range(start="2013-05-18 12:00:00", periods=2, freq='H',
                          tz= "Europe/Brussels")

In [5]: t
Out[5]: DatetimeIndex(['2013-05-18 12:00:00+02:00', '2013-05-18 13:00:00+02:00'],
                       dtype="datetime64[ns, Europe/Brussels]", freq='H')

using tz_localize(None) removes the timezone information resulting in naive local time:

In [6]: t.tz_localize(None)
Out[6]: DatetimeIndex(['2013-05-18 12:00:00', '2013-05-18 13:00:00'], 
                      dtype="datetime64[ns]", freq='H')

Further, you can also use tz_convert(None) to remove the timezone information but converting to UTC, so yielding naive UTC time:

In [7]: t.tz_convert(None)
Out[7]: DatetimeIndex(['2013-05-18 10:00:00', '2013-05-18 11:00:00'], 
                      dtype="datetime64[ns]", freq='H')

This is much more performant than the datetime.replace solution:

In [31]: t = pd.date_range(start="2013-05-18 12:00:00", periods=10000, freq='H',
                           tz="Europe/Brussels")

In [32]: %timeit t.tz_localize(None)
1000 loops, best of 3: 233 µs per loop

In [33]: %timeit pd.DatetimeIndex([i.replace(tzinfo=None) for i in t])
10 loops, best of 3: 99.7 ms per loop

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