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

Question:

You can use the function tz_localize to make a Timestamp or DateTimeIndex timezone aware, but how can you do the opposite: how can you convert a timezone aware Timestamp to a naive one, while preserving its timezone?

An example:

In [82]: t = pd.date_range(start="2013-05-18 12:00:00", periods=10, freq='s', tz="Europe/Brussels")

In [83]: t
Out[83]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-18 12:00:00, ..., 2013-05-18 12:00:09]
Length: 10, Freq: S, Timezone: Europe/Brussels

I could remove the timezone by setting it to None, but then the result is converted to UTC (12 o’clock became 10):

In [86]: t.tz = None

In [87]: t
Out[87]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-18 10:00:00, ..., 2013-05-18 10:00:09]
Length: 10, Freq: S, Timezone: None

Is there another way I can convert a DateTimeIndex to timezone naive, but while preserving the timezone it was set in?


Some context on the reason I am asking this: I want to work with timezone naive timeseries (to avoid the extra hassle with timezones, and I do not need them for the case I am working on).
But for some reason, I have to deal with a timezone-aware timeseries in my local timezone (Europe/Brussels). As all my other data are timezone naive (but represented in my local timezone), I want to convert this timeseries to naive to further work with it, but it also has to be represented in my local timezone (so just remove the timezone info, without converting the user-visible time to UTC).

I know the time is actually internal stored as UTC and only converted to another timezone when you represent it, so there has to be some kind of conversion when I want to “delocalize” it. For example, with the python datetime module you can “remove” the timezone like this:

In [119]: d = pd.Timestamp("2013-05-18 12:00:00", tz="Europe/Brussels")

In [120]: d
Out[120]: <Timestamp: 2013-05-18 12:00:00+0200 CEST, tz=Europe/Brussels>

In [121]: d.replace(tzinfo=None)
Out[121]: <Timestamp: 2013-05-18 12:00:00> 

So, based on this, I could do the following, but I suppose this will not be very efficient when working with a larger timeseries:

In [124]: t
Out[124]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-18 12:00:00, ..., 2013-05-18 12:00:09]
Length: 10, Freq: S, Timezone: Europe/Brussels

In [125]: pd.DatetimeIndex([i.replace(tzinfo=None) for i in t])
Out[125]: 
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-18 12:00:00, ..., 2013-05-18 12:00:09]
Length: 10, Freq: None, Timezone: None
Asked By: joris

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Answers:

I think you can’t achieve what you want in a more efficient manner than you proposed.

The underlying problem is that the timestamps (as you seem aware) are made up of two parts. The data that represents the UTC time, and the timezone, tz_info. The timezone information is used only for display purposes when printing the timezone to the screen. At display time, the data is offset appropriately and +01:00 (or similar) is added to the string. Stripping off the tz_info value (using tz_convert(tz=None)) doesn’t doesn’t actually change the data that represents the naive part of the timestamp.

So, the only way to do what you want is to modify the underlying data (pandas doesn’t allow this… DatetimeIndex are immutable — see the help on DatetimeIndex), or to create a new set of timestamp objects and wrap them in a new DatetimeIndex. Your solution does the latter:

pd.DatetimeIndex([i.replace(tzinfo=None) for i in t])

For reference, here is the replace method of Timestamp (see tslib.pyx):

def replace(self, **kwds):
    return Timestamp(datetime.replace(self, **kwds),
                     offset=self.offset)

You can refer to the docs on datetime.datetime to see that datetime.datetime.replace also creates a new object.

If you can, your best bet for efficiency is to modify the source of the data so that it (incorrectly) reports the timestamps without their timezone. You mentioned:

I want to work with timezone naive timeseries (to avoid the extra hassle with timezones, and I do not need them for the case I am working on)

I’d be curious what extra hassle you are referring to. I recommend as a general rule for all software development, keep your timestamp ‘naive values’ in UTC. There is little worse than looking at two different int64 values wondering which timezone they belong to. If you always, always, always use UTC for the internal storage, then you will avoid countless headaches. My mantra is Timezones are for human I/O only.

Answered By: D. A.

Building on D.A.’s suggestion that “the only way to do what you want is to modify the underlying data” and using numpy to modify the underlying data…

This works for me, and is pretty fast:

def tz_to_naive(datetime_index):
    """Converts a tz-aware DatetimeIndex into a tz-naive DatetimeIndex,
    effectively baking the timezone into the internal representation.

    Parameters
    ----------
    datetime_index : pandas.DatetimeIndex, tz-aware

    Returns
    -------
    pandas.DatetimeIndex, tz-naive
    """
    # Calculate timezone offset relative to UTC
    timestamp = datetime_index[0]
    tz_offset = (timestamp.replace(tzinfo=None) - 
                 timestamp.tz_convert('UTC').replace(tzinfo=None))
    tz_offset_td64 = np.timedelta64(tz_offset)

    # Now convert to naive DatetimeIndex
    return pd.DatetimeIndex(datetime_index.values + tz_offset_td64)
Answered By: Jack Kelly

Setting the tz attribute of the index explicitly seems to work:

ts_utc = ts.tz_convert("UTC")
ts_utc.index.tz = None
Answered By: filmor

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
Answered By: joris

The most important thing is add tzinfo when you define a datetime object.

from datetime import datetime, timezone
from tzinfo_examples import HOUR, Eastern
u0 = datetime(2016, 3, 13, 5, tzinfo=timezone.utc)
for i in range(4):
     u = u0 + i*HOUR
     t = u.astimezone(Eastern)
     print(u.time(), 'UTC =', t.time(), t.tzname())
Answered By: Yuchao Jiang

Because I always struggle to remember, a quick summary of what each of these do:

>>> pd.Timestamp.now()  # naive local time
Timestamp('2019-10-07 10:30:19.428748')

>>> pd.Timestamp.utcnow()  # tz aware UTC
Timestamp('2019-10-07 08:30:19.428748+0000', tz='UTC')

>>> pd.Timestamp.now(tz='Europe/Brussels')  # tz aware local time
Timestamp('2019-10-07 10:30:19.428748+0200', tz='Europe/Brussels')

>>> pd.Timestamp.now(tz='Europe/Brussels').tz_localize(None)  # naive local time
Timestamp('2019-10-07 10:30:19.428748')

>>> pd.Timestamp.now(tz='Europe/Brussels').tz_convert(None)  # naive UTC
Timestamp('2019-10-07 08:30:19.428748')

>>> pd.Timestamp.utcnow().tz_localize(None)  # naive UTC
Timestamp('2019-10-07 08:30:19.428748')

>>> pd.Timestamp.utcnow().tz_convert(None)  # naive UTC
Timestamp('2019-10-07 08:30:19.428748')
Answered By: Juan A. Navarro

The accepted solution does not work when there are multiple different timezones in a Series. It throws ValueError: Tz-aware datetime.datetime cannot be converted to datetime64 unless utc=True

The solution is to use the apply method.

Please see the examples below:

# Let's have a series `a` with different multiple timezones. 
> a
0    2019-10-04 16:30:00+02:00
1    2019-10-07 16:00:00-04:00
2    2019-09-24 08:30:00-07:00
Name: localized, dtype: object

> a.iloc[0]
Timestamp('2019-10-04 16:30:00+0200', tz='Europe/Amsterdam')

# trying the accepted solution
> a.dt.tz_localize(None)
ValueError: Tz-aware datetime.datetime cannot be converted to datetime64 unless utc=True

# Make it tz-naive. This is the solution:
> a.apply(lambda x:x.tz_localize(None))
0   2019-10-04 16:30:00
1   2019-10-07 16:00:00
2   2019-09-24 08:30:00
Name: localized, dtype: datetime64[ns]

# a.tz_convert() also does not work with multiple timezones, but this works:
> a.apply(lambda x:x.tz_convert('America/Los_Angeles'))
0   2019-10-04 07:30:00-07:00
1   2019-10-07 13:00:00-07:00
2   2019-09-24 08:30:00-07:00
Name: localized, dtype: datetime64[ns, America/Los_Angeles]
Answered By: tozCSS

Late contribution but just came across something similar in Python datetime and pandas give different timestamps for the same date.

If you have timezone-aware datetime in pandas, technically, tz_localize(None) changes the POSIX timestamp (that is used internally) as if the local time from the timestamp was UTC. Local in this context means local in the specified timezone. Ex:

import pandas as pd

t = pd.date_range(start="2013-05-18 12:00:00", periods=2, freq='H', tz="US/Central")
# DatetimeIndex(['2013-05-18 12:00:00-05:00', '2013-05-18 13:00:00-05:00'], dtype='datetime64[ns, US/Central]', freq='H')

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

# offset in seconds according to timezone:
(t_loc.values-t.values)//1e9
# array([-18000, -18000], dtype='timedelta64[ns]')

Note that this will leave you with strange things during DST transitions, e.g.

t = pd.date_range(start="2020-03-08 01:00:00", periods=2, freq='H', tz="US/Central")
(t.values[1]-t.values[0])//1e9
# numpy.timedelta64(3600,'ns')

t_loc = t.tz_localize(None)
(t_loc.values[1]-t_loc.values[0])//1e9
# numpy.timedelta64(7200,'ns')

In contrast, tz_convert(None) does not modify the internal timestamp, it just removes the tzinfo.

t_utc = t.tz_convert(None)
(t_utc.values-t.values)//1e9
# array([0, 0], dtype='timedelta64[ns]')

My bottom line would be: stick with timezone-aware datetime if you can or only use t.tz_convert(None) which doesn’t modify the underlying POSIX timestamp. Just keep in mind that you’re practically working with UTC then.

(Python 3.8.2 x64 on Windows 10, pandas v1.0.5.)

Answered By: FObersteiner

How I handled this problem with a 15-min frequency datetimeindex in europe.

If you are in the situation where you have a timezone aware (Europe/Amsterdam in my case) index and want to convert it into a timezone naive index by transforming everything into local time, you will have dst problems, namely

  • there will be 1 hour missing on the last sunday of march (when europe switches to summer time)
  • there will be 1 hour duplicate on the last sunday of october (when europe switches to summer time)

Here is how you can handle it:

# make index tz naive
df.index = df.index.tz_localize(None)

# handle dst
if df.index[0].month == 3:
    # last sunday of march, one hour is lost
    df = df.resample("15min").pad()

if df.index[0].month == 10:
    # in october, one hour is added
    df = df[~df.index.duplicated(keep='last')]

Note: in my case, I run the above code on a df that contains only a single month, hence I do df.index[0].month to find out the month. If yours contains more months, you should probably index it differently to know when to do DST.

It consists of resampling from the last valid value in march, to avoid losing the 1 hour (in my case, all my data is in 15 min intervals, hence i resample like that. Resample for whatever your interval is). And for october, I drop duplicates.

Answered By: charelf
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