Part of the problem here is that the strings usually used to represent timezones are not actually unique. “EST” only means “America/New_York” to people in North America. This is a limitation in the C time API, and the Python solution is… to add full tz features in some future version any day now, if anyone is … Read more
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
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
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
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.
Pandas timestamp differences returns a datetime.timedelta object. This can easily be converted into hours by using the *as_type* method, like so import pandas df = pandas.DataFrame(columns=[‘to’,’fr’,’ans’]) df.to = [pandas.Timestamp(‘2014-01-24 13:03:12.050000’), pandas.Timestamp(‘2014-01-27 11:57:18.240000’), pandas.Timestamp(‘2014-01-23 10:07:47.660000’)] df.fr = [pandas.Timestamp(‘2014-01-26 23:41:21.870000’), pandas.Timestamp(‘2014-01-27 15:38:22.540000’), pandas.Timestamp(‘2014-01-23 18:50:41.420000’)] (df.fr-df.to).astype(‘timedelta64[h]’) to yield, 0 58 1 3 2 8 dtype: float64
Yes, definitely datetime is what you need here. Specifically, the datetime.strptime() method, which parses a string into a datetime object. from datetime import datetime s1 = ’10:33:26′ s2 = ’11:15:49′ # for example FMT = ‘%H:%M:%S’ tdelta = datetime.strptime(s2, FMT) – datetime.strptime(s1, FMT) That gets you a timedelta object that contains the difference between the … Read more
In Python 3.3+: from datetime import datetime, timezone def utc_to_local(utc_dt): return utc_dt.replace(tzinfo=timezone.utc).astimezone(tz=None) In Python 2/3: import calendar from datetime import datetime, timedelta def utc_to_local(utc_dt): # get integer timestamp to avoid precision lost timestamp = calendar.timegm(utc_dt.timetuple()) local_dt = datetime.fromtimestamp(timestamp) assert utc_dt.resolution >= timedelta(microseconds=1) return local_dt.replace(microsecond=utc_dt.microsecond) Using pytz (both Python 2/3): import pytz local_tz = pytz.timezone(‘Europe/Moscow’) # … Read more
The easiest way is to use to_datetime: df[‘col’] = pd.to_datetime(df[‘col’]) It also offers a dayfirst argument for European times (but beware this isn’t strict). Here it is in action: In [11]: pd.to_datetime(pd.Series([’05/23/2005′])) Out[11]: 0 2005-05-23 00:00:00 dtype: datetime64[ns] You can pass a specific format: In [12]: pd.to_datetime(pd.Series([’05/23/2005′]), format=”%m/%d/%Y”) Out[12]: 0 2005-05-23 dtype: datetime64[ns]