In general, to make a naive datetime timezone-aware, use the localize method: import datetime import pytz unaware = datetime.datetime(2011, 8, 15, 8, 15, 12, 0) aware = datetime.datetime(2011, 8, 15, 8, 15, 12, 0, pytz.UTC) now_aware = pytz.utc.localize(unaware) assert aware == now_aware For the UTC timezone, it is not really necessary to use localize since … Read more
You can take the mean of Timedelta. So find the minimum value and subtract it from the series to get a series of Timedelta. Then take the mean and add it back to the minimum. dob = df_test.DOB m = dob.min() (m + (dob – m).mean()).to_pydatetime() datetime.datetime(2014, 7, 12, 0, 0) One-line df_test.DOB.pipe(lambda d: (lambda … Read more
You can use pd.offsets.MonthBegin() In [261]: d = pd.to_datetime([‘2011-09-30’, ‘2012-02-28’]) In [262]: d Out[262]: DatetimeIndex([‘2011-09-30’, ‘2012-02-28’], dtype=”datetime64[ns]”, freq=None) In [263]: d + pd.offsets.MonthBegin(1) Out[263]: DatetimeIndex([‘2011-10-01’, ‘2012-03-01′], dtype=”datetime64[ns]”, freq=None) You’ll find a lot of examples in the official Pandas docs
In general, to make a naive datetime timezone-aware, use the localize method: import datetime import pytz unaware = datetime.datetime(2011, 8, 15, 8, 15, 12, 0) aware = datetime.datetime(2011, 8, 15, 8, 15, 12, 0, pytz.UTC) now_aware = pytz.utc.localize(unaware) assert aware == now_aware For the UTC timezone, it is not really necessary to use localize since … Read more
I would suggest the following simple approach for conversion: from datetime import datetime, timedelta s = “20120213” # you could also import date instead of datetime and use that. date = datetime(year=int(s[0:4]), month=int(s[4:6]), day=int(s[6:8])) For adding/subtracting an arbitary amount of days (seconds work too btw.), you could do the following: date += timedelta(days=10) date -= … Read more
Use datetime.datetime.strptime() and call .time() on the result: >>> datetime.datetime.strptime(’03:55′, ‘%H:%M’).time() datetime.time(3, 55) The first argument to .strptime() is the string to parse, the second is the expected format.
This is because pandas falls back to dateutil.parser.parse for parsing the strings when it has a non-default format or when no format string is supplied (this is much more flexible, but also slower). As you have shown above, you can improve the performance by supplying a format string to to_datetime. Or another option is to … Read more
My original answer focused very specifically on the question as posed and didn’t accommodate time ranges that span midnight. As this is still the accepted answer 6 years later, I’ve incorporated @rouble’s answer below that expanded on mine to support midnight. from datetime import datetime, time def is_time_between(begin_time, end_time, check_time=None): # If check time is … Read more
>>> import datetime >>> import time >>> x = time.strptime(’00:01:00,000′.split(‘,’)[0],’%H:%M:%S’) >>> datetime.timedelta(hours=x.tm_hour,minutes=x.tm_min,seconds=x.tm_sec).total_seconds() 60.0
You can use round(freq). There is also a shortcut column.dt for datetime functions access (as @laurens-koppenol suggests). Here’s one-liner: df[‘old column’].dt.round(’15min’) String aliases for valid frequencies can be found here. Full working example: In [1]: import pandas as pd In [2]: df = pd.DataFrame([pd.Timestamp(‘2015-07-18 13:53:33.280’), pd.Timestamp(‘2015-07-18 13:33:33.330’)], columns=[‘old column’]) In [3]: df[‘new column’]=df[‘old column’].dt.round(’15min’) In … Read more