Understanding timedelta

末鹿安然 提交于 2019-11-28 03:31:42
utdemir

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

why do I have to pass seconds = uptime to timedelta

Because timedelta objects can be passed seconds, milliseconds, days, etc... so you need to specify what are you passing in (this is why you use the explicit key). Typecasting to int is superfluous as they could also accept floats.

and why does the string casting works so nicely that I get HH:MM:SS ?

It's not the typecasting that formats, is the internal __str__ method of the object. In fact you will achieve the same result if you write:

print datetime.timedelta(seconds=int(uptime))
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