In Python, How to Convert Seconds Since Epoch to a 'Datetime' Object

In Python, how do you convert seconds since epoch to a `datetime` object?

datetime.datetime.fromtimestamp will do, if you know the time zone, you could produce the same output as with time.gmtime

>>> datetime.datetime.fromtimestamp(1284286794)
datetime.datetime(2010, 9, 12, 11, 19, 54)

or

>>> datetime.datetime.utcfromtimestamp(1284286794)
datetime.datetime(2010, 9, 12, 10, 19, 54)

Converting Epoch time into the datetime

To convert your time value (float or int) to a formatted string, use:

import time

time.strftime('%Y-%m-%d %H:%M:%S', time.localtime(1347517370))

For example:

import time

my_time = time.strftime('%Y-%m-%d %H:%M:%S', time.localtime(1347517370))

print(my_time)

Python convert a series of seconds since epoch time to datetime series

you can use to_datetime(..., unit='s'):

df['datetime'] = pd.to_datetime(df['epochtime'], unit='s')

Time comparison:

In [158]: df = pd.DataFrame({'epochtime': pd.date_range('2001-01-01', freq='1S', periods=10**5)}).astype(np.int64)//10**9

In [159]:

In [159]: df.head()
Out[159]:
   epochtime
0  978307200
1  978307201
2  978307202
3  978307203
4  978307204

In [160]:

In [160]: len(df)
Out[160]: 100000

In [161]:

In [161]: %timeit df['datetime'] = pd.to_datetime(df['epochtime'], unit='s')
100 loops, best of 3: 16.9 ms per loop

In [162]:

In [162]: %%timeit
   .....: for i in range(len(df)):
   .....:     df['datetime'].iloc[i] = datetime.fromtimestamp(df2['epochtime'].iloc[i])
   .....:
c:\envs\py35\lib\site-packages\pandas\core\indexing.py:128: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)
1 loop, best of 3: 54.5 s per loop

Conclusion:
@Natecat,
as you can see 16.9 ms vs. 54.5 s

In Python, how do you convert a `datetime` object to seconds?

For the special date of January 1, 1970 there are multiple options.

For any other starting date you need to get the difference between the two dates in seconds. Subtracting two dates gives a timedelta object, which as of Python 2.7 has a total_seconds() function.

>>> (t-datetime.datetime(1970,1,1)).total_seconds()
1256083200.0

The starting date is usually specified in UTC, so for proper results the datetime you feed into this formula should be in UTC as well. If your datetime isn't in UTC already, you'll need to convert it before you use it, or attach a tzinfo class that has the proper offset.

As noted in the comments, if you have a tzinfo attached to your datetime then you'll need one on the starting date as well or the subtraction will fail; for the example above I would add tzinfo=pytz.utc if using Python 2 or tzinfo=timezone.utc if using Python 3.

Python - Convert seconds from epoch time into human readable time

a = last_epoch #last epoch recorded
b = time.time() #current epoch time
c = b - a #returns seconds
days = c // 86400
hours = c // 3600 % 24
minutes = c // 60 % 60
seconds = c % 60

Convert python datetime to epoch with strftime

If you want to convert a python datetime to seconds since epoch you could do it explicitly:

>>> (datetime.datetime(2012,4,1,0,0) - datetime.datetime(1970,1,1)).total_seconds()
1333238400.0

In Python 3.3+ you can use timestamp() instead:

>>> datetime.datetime(2012,4,1,0,0).timestamp()
1333234800.0

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Why you should not use datetime.strftime('%s')

Python doesn't actually support %s as an argument to strftime (if you check at http://docs.python.org/library/datetime.html#strftime-and-strptime-behavior it's not in the list), the only reason it's working is because Python is passing the information to your system's strftime, which uses your local timezone.

>>> datetime.datetime(2012,04,01,0,0).strftime('%s')
'1333234800'

Seconds since date (not epoch!) to date

You could try this:

from datetime import datetime, timedelta

t0 = datetime(1981, 1, 1)
seconds = 1000000
dt = t0 + timedelta(seconds=seconds)
print dt
# 1981-01-12 13:46:40

Here t0 is set to a datetime object representing the "epoch" of 1981-01-01. This is your reference datetime to which you can add a timedelta object initialised with an arbitrary number of seconds. The result is a datetime object representing the required date time.

N.B. this assumes UTC time

How can I convert a datetime object to milliseconds since epoch (unix time) in Python?

It appears to me that the simplest way to do this is

import datetime

epoch = datetime.datetime.utcfromtimestamp(0)

def unix_time_millis(dt):
    return (dt - epoch).total_seconds() * 1000.0

Convert timestamp since epoch to datetime.datetime

I would use the time module

>>> import time
>>> time.gmtime(1346114717972/1000.)
time.struct_time(tm_year=2012, tm_mon=8, tm_mday=28, tm_hour=0, tm_min=45, tm_sec=17, tm_wday=1, tm_yday=241, tm_isdst=0)  

This shows the timestamp in UTC/GMT time.

The timestamp is divided by 1000 as the stamps you have provided are in milliseconds since the epoch, not seconds.

Then use strftime to format like so:

>>> time.strftime('%m/%d/%Y %H:%M:%S',  time.gmtime(1346114717972/1000.))
'08/28/2012 00:45:17'

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