How do I find "number of seconds since the beginning of the day UTC timezone" in Python? I looked at the docs and didn't understand how to get this using datetime.timedelta.
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Here's one way to do it.
from datetime import datetime, time
utcnow = datetime.utcnow()
midnight_utc = datetime.combine(utcnow.date(), time(0))
delta = utcnow - midnight_utc
print delta.seconds # <-- careful
EDIT As suggested, if you want microsecond precision, or potentially crossing a 24-hour period (i.e. delta.days > 0), use total_seconds() or the formula given by @unutbu.
print delta.total_seconds() # 2.7
print delta.days * 24 * 60 * 60 + delta.seconds + delta.microseconds / 1e6 # < 2.7
Joe Holloway
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use [`delta.total_seconds()`](http://stackoverflow.com/questions/8072740/pythonnumber-of-seconds-since-the-beginning-of-the-day-utc-timezone/8072776#8072776) – jfs Nov 09 '11 at 22:57
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If he/she knows it's < 24 hours since the beginning of the day, it's safe to use `timedelta.seconds` correct? Just wondering if there's another reason to avoid it for this answer. – Joe Holloway Nov 09 '11 at 23:20
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[Leap seconds](https://en.wikipedia.org/wiki/Leap_second) are not supported so it is safe to assume that `delta.days` is `0`. But `.total_seconds()` is better semantically and it is a less burden for a reader of the code. – jfs Nov 09 '11 at 23:44
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The number of seconds in a datetime.timedelta, x, is given by timedelta.total_seconds:
x.total_seconds()
This function was introduced in Python2.7. For older versions of python, you just have to compute it yourself: total_seconds = x.days*24*60*60 + x.seconds + x.microseconds/1e6.
0
import time
t = time.gmtime()
seconds_since_utc_midnight = t.tm_sec + (t.tm_min * 60) + (t.tm_hour * 3600)
for localtime, we can use time.localtime() instead of time.gmtime()
sonofusion82
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