Just avoid timedelta‘s F-ignorance filter knows about F, but timedelta does not. The trick is to keep the F out of the timedelta argument list: ThatModel.objects.filter( last_datetime__lte=now + datetime.timedelta(seconds=1)*F(“interval”)) This will work with PostgreSQL, but, alas, not with SQlite.
I think you can add to column TDate column Hour converted to_timedelta with unit=”h”: df = pd.DataFrame({‘TDate’:[‘2005-01-03′,’2005-01-04′,’2005-01-05’], ‘Hour’:[4,5,6]}) df[‘TDate’] = pd.to_datetime(df.TDate) print (df) Hour TDate 0 4 2005-01-03 1 5 2005-01-04 2 6 2005-01-05 df[‘TDate’] += pd.to_timedelta(df.Hour, unit=”h”) print (df) Hour TDate 0 4 2005-01-03 04:00:00 1 5 2005-01-04 05:00:00 2 6 2005-01-05 06:00:00
Here’s an example of how you might do this by iterating over each row in the dataframe. I created new data for the example for simplicity: df = pd.DataFrame({‘TimeDelta’: np.random.normal( 900, 60, size=100)}) print df.head() TimeDelta 0 971.021295 1 734.359861 2 867.000397 3 992.166539 4 853.281131 So let’s do an accumulator loop with your desired … Read more
If you are able to install the excellent dateutil package, you can do this: >>> from dateutil import relativedelta as rdelta >>> from datetime import date >>> d1 = date(2001,5,1) >>> d2 = date(2012,1,1) >>> rd = rdelta.relativedelta(d2,d1) >>> “{0.years} years and {0.months} months”.format(rd) ’10 years and 8 months’
It’s strange that Python returns the value as a datetime.timedelta. It probably should return a datetime.time. Anyway, it looks like it’s returning the elapsed time since midnight (assuming the column in the table is 6:00 PM). In order to convert to a datetime.time, you can do the following:: value = datetime.timedelta(0, 64800) (datetime.datetime.min + value).time() … Read more
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
from datetime import datetime, timedelta nine_hours_from_now = datetime.now() + timedelta(hours=9) #datetime.datetime(2012, 12, 3, 23, 24, 31, 774118) And then use string formatting to get the relevant pieces: >>> ‘{:%H:%M:%S}’.format(nine_hours_from_now) ’23:24:31′ If you’re only formatting the datetime then you can use: >>> format(nine_hours_from_now, ‘%H:%M:%S’) ’23:24:31′ Or, as @eumiro has pointed out in comments – strftime
You can use sort_values with groupby and aggregating diff: df[‘diff’] = df.sort_values([‘id’,’time’]).groupby(‘id’)[‘time’].diff() print (df) id time diff 0 A 2016-11-25 16:32:17 NaT 1 A 2016-11-25 16:36:04 00:00:35 2 A 2016-11-25 16:35:29 00:03:12 3 B 2016-11-25 16:35:24 NaT 4 B 2016-11-25 16:35:46 00:00:22 If need remove rows with NaT in column diff use dropna: df = … 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
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