Understanding timedelta

Paul picture Paul · Jul 19, 2011 · Viewed 213.2k times · Source

Given the python code below, please help me understand what is happening there.

start_time = time.time()
time.sleep(42)
end_time = time.time()

uptime = end_time - start_time

human_uptime = str(datetime.timedelta(seconds=int(uptime)))

So I get the difference between start time and end time, on line 5 I round up the duration by casting and what now, what's the further explanation?

I know what delta means(average or difference), but why do I have to pass seconds = uptime to timedelta and why does the string casting works so nicely that I get HH:MM:SS ?

Answer

utdemir picture utdemir · Jul 19, 2011

Because timedelta is defined like:

class datetime.timedelta([days,] [seconds,] [microseconds,] [milliseconds,] [minutes,] [hours,] [weeks])

All arguments are optional and default to 0.

You can easily say "Three days and four milliseconds" with optional arguments that way.

>>> datetime.timedelta(days=3, milliseconds=4)
datetime.timedelta(3, 0, 4000)
>>> datetime.timedelta(3, 0, 0, 4) #no need for that.
datetime.timedelta(3, 0, 4000)

And for str casting, it returns a nice formatted value instead of __repr__ to improve readability. From docs:

str(t) Returns a string in the form [D day[s], ][H]H:MM:SS[.UUUUUU], where D is negative for negative t. (5)

>>> datetime.timedelta(seconds = 42).__repr__()
'datetime.timedelta(0, 42)'
>>> datetime.timedelta(seconds = 42).__str__()
'0:00:42'

Checkout documentation:

http://docs.python.org/library/datetime.html#timedelta-objects