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
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
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')
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.
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]
Setting the tz attribute of the index explicitly seems to work:
ts_utc = ts.tz_convert("UTC")
ts_utc.index.tz = None
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.)
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)
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())
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.
Related
I want to change Datetime (2014-12-23 00:00:00) into unixtime. I tried it with the Datetime function but it didn´t work. I got the Datetime stamps in an array.
Zeit =np.array(Jahresgang1.ix[ :,'Zeitstempel'])
t = pd.to_datetime(Zeit, unit='s')
unixtime = pd.DataFrame(t)
print unixtime
Thanks a lot
I think you can subtract the date 1970-1-1 to create a timedelta and then access the attribute total_seconds:
In [130]:
s = pd.Series(pd.datetime(2012,1,1))
s
Out[130]:
0 2012-01-01
dtype: datetime64[ns]
In [158]:
(s - dt.datetime(1970,1,1)).dt.total_seconds()
Out[158]:
0 1325376000
dtype: float64
to emphasize EdChum's first comment, you can directly get Unix time like
import pandas as pd
s = pd.to_datetime(["2014-12-23 00:00:00"])
unix = s.astype("int64")
print(unix)
# Int64Index([1419292800000000000], dtype='int64')
or for a pd.Timestamp:
print(pd.to_datetime("2014-12-23 00:00:00").value)
# 1419292800000000000
Notes
the output precision is nanoseconds - if you want another, divide appropriately, e.g. by 10⁹ to get seconds, 10⁶ to get milliseconds etc.
this assumes the input date/time to be UTC, unless a time zone / UTC offset is specified
I need help converting into python/pandas date time format. For example, my times are saved like the following line:
2017-01-01 05:30:24.468911+00:00
.....
2017-05-05 01:51:31.351718+00:00
and I want to know the simplest way to convert this into date time format for essentially performing operations with time (like what is the range in days of my dataset to split up my dataset into chunks by time, what's the time difference from one time to another)? I don't mind losing some of the significance for the times if that makes things easier. Thank you so much!
Timestamp will convert it for you.
>>> pd.Timestamp('2017-01-01 05:30:24.468911+00:00')
Timestamp('2017-01-01 05:30:24.468911+0000', tz='UTC')
Let's say you have a dataframe that includes your timestamp column (let's call it stamp). You can use apply on that column together with Timestamp:
df = pd.DataFrame(
{'stamp': ['2017-01-01 05:30:24.468911+00:00',
'2017-05-05 01:51:31.351718+00:00']})
>>> df
stamp
0 2017-01-01 05:30:24.468911+00:00
1 2017-05-05 01:51:31.351718+00:00
>>> df['stamp'].apply(pd.Timestamp)
0 2017-01-01 05:30:24.468911+00:00
1 2017-05-05 01:51:31.351718+00:00
Name: stamp, dtype: datetime64[ns, UTC]
You could also use Timeseries:
>>> pd.TimeSeries(df.stamp)
0 2017-01-01 05:30:24.468911+00:00
1 2017-05-05 01:51:31.351718+00:00
Name: stamp, dtype: object
Once you have a Timestamp object, it is pretty efficient to manipulate. You can just difference their values, for example.
You may also want to have a look at this SO answer which discusses timezone unaware values to aware.
Let's say I have two strings 2017-06-06 and 1944-06-06 and I wanted to get the difference (what Python calls a timedelta) between the two.
First, I'll need to import datetime. Then I'll need to get both of those strings into datetime objects:
>>> a = datetime.datetime.strptime('2017-06-06', '%Y-%m-%d')
>>> b = datetime.datetime.strptime('1944-06-06', '%Y-%m-%d')
That will give us two datetime objects that can be used in arithmetic functions that will return a timedelta object:
>>> c = abs((a-b).days)
This will give us 26663, and days is the largest resolution that timedelta supports: documentation
Since the Pandas tag is there:
df = pd.DataFrame(['2017-01-01 05:30:24.468911+00:00'])
df.columns = ['Datetime']
df['Datetime'] = pd.to_datetime(df['Datetime'], format='%Y-%m-%d %H:%M:%S.%f', utc=True)
print(df.dtypes)
Is it possible to convert a pd.DatetimeIndex consisting of timestamps in a single timezone to one where each timestamp has its own, in some cases distinct timezone?
Here is an example of what I would like to have:
type(df.index)
pandas.tseries.index.DatetimeIndex
df.index[0]
Timestamp('2015-06-07 23:00:00+0100', tz='Europe/London')
df.index[1]
Timestamp('2015-06-08 00:01:00+0200', tz='Europe/Brussels')
You can have an index contain Timestamps with different timezones. But you would have to explicity construct it as an Index.
In [33]: pd.Index([pd.Timestamp('2015-06-07 23:00:00+0100', tz='Europe/London'),pd.Timestamp('2015-06-08 00:01:00+0200', tz='Europe/Brussels')],dtype='object')
Out[33]: Index([2015-06-07 23:00:00+01:00, 2015-06-08 00:01:00+02:00], dtype='object')
In [34]: list(pd.Index([pd.Timestamp('2015-06-07 23:00:00+0100', tz='Europe/London'),pd.Timestamp('2015-06-08 00:01:00+0200', tz='Europe/Brussels')],dtype='object'))
Out[34]:
[Timestamp('2015-06-07 23:00:00+0100', tz='Europe/London'),
Timestamp('2015-06-08 00:01:00+0200', tz='Europe/Brussels')]
This is a very odd thing to do, and completely non-performant. You generally want to have a single timezone represented (UTC or other). In 0.17.0, you can represent efficiently a single column with a timezone, so one way of accomplishing what I think your goal is to segregate the different timezones into different columns. See the docs
If you're happy for it to not be an Index, but just a regular Series this should be OK:
pd.Series([pd.Timestamp('2015-06-07 23:00:00+0100', tz='Europe/London'),
pd.Timestamp('2015-06-08 00:01:00+0200', tz='Europe/Brussels')])
Adding timestamps with different timezones into the same DatetimeIndex automatically yields a DatetimeIndex with UTC as the default timezone. For example:
In [269] index = pandas.DatetimeIndex([Timestamp('2015-06-07 23:00:00+0100')])
In [270] index
Out[270] DatetimeIndex(['2015-06-07 23:00:00+01:00'], dtype='datetime64[ns, pytz.FixedOffset(60)]', freq=None)
In [271] index2 = DatetimeIndex([Timestamp('2015-06-08 00:01:00+0200')])
In [272] index2
Out[272] DatetimeIndex(['2015-06-08 00:01:00+02:00'], dtype='datetime64[ns, pytz.FixedOffset(120)]', freq=None)
In [273] index.append(index2) # returns single index containing both data
Out[273] DatetimeIndex(['2015-06-07 22:00:00+00:00', '2015-06-07 22:01:00+00:00'], dtype='datetime64[ns, UTC]', freq=None)
Notice how the result is a UTC DatetimeIndex with the correct UTC timestamps preserved.
Similarly:
In [279] pandas.to_datetime([Timestamp('2015-06-07 23:00:00+0100'), Timestamp('2015-06-08 00:01:00+0200')], utc=True) # utc=True is needed
Out[279] DatetimeIndex(['2015-06-07 22:00:00+00:00', '2015-06-07 22:01:00+00:00'], dtype='datetime64[ns, UTC]', freq=None)
This is not a bad thing as you get to preserve the correct time while having the ability to use the indexing prowess of DatetimeIndex (e.g. slice by date range) and at the same time you can easily convert the timestamps to any other timezone (unless you really need to know the original timezone of each timestamp, then this won't be ideal).
I am trying to get the time index of a dataframe as a series, but it appears to be dropping the timezone when I call the method to_series. Below is an example. Is this a bug or am I doing something incorrectly?
rows = 50
df = pd.DataFrame(np.random.randn(rows,2), columns=list('AB'), index=pd.date_range('1/1/2000', periods=rows, freq='1H', tz=pytz.UTC))
print df.index[-1]
# 2000-01-03 01:00:00+00:00
print df.index.to_series()[-1]
# 2000-01-03 01:00:00
print df.index[-1].tzinfo
# UTC
print df.index.to_series()[-1].tzinfo
#None
No, you are not doing something wrong, and neither it is a bug.
It is a currently known limitation of pandas/numpy: timezones aware datetime data are only supported in the index. In a series, the data are stored as numpy datetime64 types, which does not support timezones. There is an open issue for adding this timezone feature also to series: https://github.com/pydata/pandas/issues/8260
A workaround is to store your data as object dtype instead of datetime64 (then they will be stored as Timestamp objects, a subclass of datetime.datetime). This will enable you to keep the timezone info.
to_series has a keep_tz keyword argument to obtain this (see docstring):
In [34]: df = df.tz_convert('US/Eastern')
In [35]: df.index.to_series()[-1]
Out[35]: Timestamp('2000-01-03 01:00:00')
In [36]: df.index.to_series(keep_tz=True)[-1]
Out[36]: Timestamp('2000-01-02 20:00:00-0500', tz='US/Eastern', offset='H')
What is going on here?
I need to generate a dataframe with beginning of month dates,,(1-1-2014 To 12-1-2014) fwiw I use the fcast_year variable elsewhere where I need the end of month, hence doing date math
from pandas.tseries.offsets import *
fcast_yr=pd.to_datetime('2014-12-31')
x=(fcast_yr + pd.DateOffset(days= -30)) # to set x to 2014-12-01
d=pd.date_range((x +pd.DateOffset(months=-10)), periods=12, freq='MS') #"MS" means start of month!!
print d.values
Gives these end of month values....yech!!
['2014-01-31T18:00:00.000000000-0600' '2014-02-28T18:00:00.000000000-0600'
'2014-03-31T19:00:00.000000000-0500' '2014-04-30T19:00:00.000000000-0500'
'2014-05-31T19:00:00.000000000-0500' '2014-06-30T19:00:00.000000000-0500'
'2014-07-31T19:00:00.000000000-0500' '2014-08-31T19:00:00.000000000-0500'
'2014-09-30T19:00:00.000000000-0500' '2014-10-31T19:00:00.000000000-0500'
'2014-11-30T18:00:00.000000000-0600' '2014-12-31T18:00:00.000000000-0600']
Using 13.0 pf Pandas
You don't need to coerce the timestamp to the begin month; the frequency will do it (but your answer is correct).
The 'values' are just the way numpy represents dates (they are UTC).
In [8]: pd.date_range((Timestamp('20141231') +pd.DateOffset(months=-11)), periods=12, freq='MS')
Out[8]:
<class 'pandas.tseries.index.DatetimeIndex'>
[2014-02-01, ..., 2015-01-01]
Length: 12, Freq: MS, Timezone: None