from datetime import datetime, timedelta t = np.arange(datetime(1985,7,1), datetime(2015,7,1), timedelta(days=1)).astype(datetime) The key point here is to use astype(datetime), otherwise the result will be datetime64.
I found isodate library to do exactly what I want isodate.parse_duration(‘PT1H5M26S’) You can read the source code for the function here
If you are using Python 2.7 or higher you can use timedelta.total_seconds() to get a float representation of the timedelta as a positive or negative number of seconds. >>> datetime.timedelta(-1, 86100).total_seconds() -300.0 You should be able to use this to calculate a number of minutes fairly easily. If you are not using Python 2.7 you … Read more
The trick is to create a fixed interval and multiply it with the number of seconds in the column: SELECT start_time, expiration_time_seconds, start_time + expiration_time_seconds * interval ‘1 second’ FROM whatever ORDER BY start_time; start_time | expiration_time_seconds | end_time —————————-|————————-|—————————- 2014-08-05 08:23:32.428452 | 172800 | 2014-08-07 08:23:32.428452 2014-08-10 09:49:51.082456 | 3600 | 2014-08-10 10:49:51.082456 2014-08-13 … Read more
If you know your origin timezone and the new timezone that you want to convert it to, it turns out to be very straightforward: Make two pytz.timezone objects, one for the current timezone and one for the new timezone e.g. pytz.timezone(“US/Pacific”). You can find a list of all official timezones in pytz library: import pytz; … Read more
datetime.time objects do not support addition with datetime.timedeltas. There is one natural definition though, clock arithmetic. You could compute it like this: import datetime as dt now = dt.datetime.now() delta = dt.timedelta(hours = 12) t = now.time() print(t) # 12:39:11.039864 print((dt.datetime.combine(dt.date(1,1,1),t) + delta).time()) # 00:39:11.039864 dt.datetime.combine(…) lifts the datetime.time t to a datetime.datetime object, the … Read more
If it is just for the display, this idea works : avgString = str(avg).split(“.”)[0] The idea is to take only what is before the point. It will return 01:23:45 for 01:23:45.1235
You have mixed something up with your variables, you can subtract timedelta d from datetime.datetime.now() with no issue: import datetime tod = datetime.datetime.now() d = datetime.timedelta(days = 50) a = tod – d print(a) 2014-12-13 22:45:01.743172
You could use the total_seconds method: time_d_float = time_d.total_seconds()
You’ll have to create a new timedelta with the specified amount of time: d > timedelta(minutes=1) Or this slightly more complete script will help elaborate: import datetime from time import sleep start = datetime.datetime.now() sleep(3) stop = datetime.datetime.now() elapsed = stop – start if elapsed > datetime.timedelta(minutes=1): print “Slept for > 1 minute” if elapsed … Read more