import time from datetime import datetime, date, time, timedelta dayDates = [] today = datetime.now() dayDates.append(today.strftime(“%m%d%Y”)) for i in range(0,14): day = today + datetime.timedelta(days=i) print day The error that you are getting says, that datetime has no attribute timedelta. It happens, because you have imported from datetime specific things. In order to access timedelta … 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 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
You can use pd.to_timedelta or np.timedelta64 to define a duration and divide by this: # set up as per @EdChum df[‘total_days_td’] = df[‘time_delta’] / pd.to_timedelta(1, unit=”D”) df[‘total_days_td’] = df[‘time_delta’] / np.timedelta64(1, ‘D’)
You can try by creating a new column with years in this way: df[‘diff_year’] = df[‘diff’] / np.timedelta64(1, ‘Y’)
To add timedeltas you can use the builtin operator +: result = timedelta1 + timedelta2 To add a lot of timedeltas you can use sum: result = sum(timedeltas, datetime.timedelta()) Or reduce: import operator result = reduce(operator.add, timedeltas)
Full code solution: import pandas as pd from datetime import timedelta df = pd.DataFrame([[‘2016-01-10’,28],[‘2016-05-11’,28],[‘2016-02-23’,15],[‘2015-12-08’,30]], columns = [‘ship_string’,’days_supply’]) df[‘ship_date’] = pd.to_datetime(df[‘ship_string’]) df[‘time_added’] = pd.to_timedelta(df[‘days_supply’],’d’) df[‘supply_ended’] = df[‘ship_date’] + df[‘time_added’] print df ship_string days_supply ship_date time_added supply_ended 0 2016-01-10 28 2016-01-10 28 days 2016-02-07 1 2016-05-11 28 2016-05-11 28 days 2016-06-08 2 2016-02-23 15 2016-02-23 15 days … Read more
It doesn’t make sense to convert a timedelta into a datetime, but it does make sense to pick an initial or starting datetime and add or subtract a timedelta from that. >>> import datetime >>> today = datetime.datetime.today() >>> today datetime.datetime(2010, 3, 9, 18, 25, 19, 474362) >>> today + datetime.timedelta(days=1) datetime.datetime(2010, 3, 10, 18, … Read more
Maybe these examples will help you get an idea: from dateutil.relativedelta import relativedelta import datetime date1 = datetime.datetime.strptime(“2015-01-30”, “%Y-%m-%d”).strftime(“%d-%m-%Y”) print(date1) today = datetime.date.today() print(today) addMonths = relativedelta(months=3) future = today + addMonths print(future) If you import datetime it will give you more options in managing date and time variables. In my example above I have … Read more
You could use python-dateutil. In [4]: from datetime import datetime In [5]: date1 = datetime.strptime(str(‘2011-08-15 12:00:00’), ‘%Y-%m-%d %H:%M:%S’) In [6]: date2 = datetime.strptime(str(‘2012-02-15’), ‘%Y-%m-%d’) In [7]: from dateutil import relativedelta In [8]: r = relativedelta.relativedelta(date1, date2) In [9]: r Out[9]: relativedelta(months=-5, days=-30, hours=-12)