How to get the seconds since epoch from the time + date output of gmtime()?


How do you do reverse gmtime(), where you put the time + date and get the number of seconds?

I have strings like 'Jul 9, 2009 @ 20:02:58 UTC', and I want to get back the number of seconds between the epoch and July 9, 2009.

I have tried time.strftime but I don't know how to use it properly, or if it is the correct command to use.

1/14/2018 3:41:42 AM

Accepted Answer

If you got here because a search engine told you this is how to get the Unix timestamp, stop reading this answer. Scroll down one.

If you want to reverse time.gmtime(), you want calendar.timegm().

>>> calendar.timegm(time.gmtime())

You can turn your string into a time tuple with time.strptime(), which returns a time tuple that you can pass to calendar.timegm():

>>> import calendar
>>> import time
>>> calendar.timegm(time.strptime('Jul 9, 2009 @ 20:02:58 UTC', '%b %d, %Y @ %H:%M:%S UTC'))

More information about calendar module here

3/31/2020 2:21:15 PM

Note that time.gmtime maps timestamp 0 to 1970-1-1 00:00:00.

In [61]: import time       
In [63]: time.gmtime(0)
Out[63]: time.struct_time(tm_year=1970, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=3, tm_yday=1, tm_isdst=0)

time.mktime(time.gmtime(0)) gives you a timestamp shifted by an amount that depends on your locale, which in general may not be 0.

In [64]: time.mktime(time.gmtime(0))
Out[64]: 18000.0

The inverse of time.gmtime is calendar.timegm:

In [62]: import calendar    
In [65]: calendar.timegm(time.gmtime(0))
Out[65]: 0

ep = datetime.datetime(1970,1,1,0,0,0)
x = (datetime.datetime.utcnow()- ep).total_seconds()

This should be different from int(time.time()), but it is safe to use something like x % (60*60*24)

datetime — Basic date and time types:

Unlike the time module, the datetime module does not support leap seconds.


t = datetime.strptime('Jul 9, 2009 @ 20:02:58 UTC',"%b %d, %Y @ %H:%M:%S %Z")

There are two ways, depending on your original timestamp:

mktime() and timegm()


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