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
You’ll need to use the pytz module (available from PyPI): import pytz from datetime import datetime est = pytz.timezone(‘US/Eastern’) utc = pytz.utc fmt=”%Y-%m-%d %H:%M:%S %Z%z” winter = datetime(2016, 1, 24, 18, 0, 0, tzinfo=utc) summer = datetime(2016, 7, 24, 18, 0, 0, tzinfo=utc) print(winter.strftime(fmt)) print(summer.strftime(fmt)) print(winter.astimezone(est).strftime(fmt)) print(summer.astimezone(est).strftime(fmt)) which will print: 2016-01-24 18:00:00 UTC+0000 2016-07-24 18:00:00 … Read more
You have to use strptime to convert a string into a date. The comparaison operator only applies between datetime. date = datetime.strptime(‘2018-11-10 10:55:31’, ‘%Y-%m-%d %H:%M:%S’) then can you do if past > date : print(“This is older than 90 days”)
I would actually follow a different approach and rely on timedelta and date addition to determine the non-overlapping ranges Implementation def date_range(start, end, intv): from datetime import datetime start = datetime.strptime(start,”%Y%m%d”) end = datetime.strptime(end,”%Y%m%d”) diff = (end – start ) / intv for i in range(intv): yield (start + diff * i).strftime(“%Y%m%d”) yield end.strftime(“%Y%m%d”) Execution … Read more
Here is a 1-line solution using nothing more than the standard datetime library: (dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1) Examples: >>> dt = datetime.datetime(2016, 2, 29) >>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)) 2016-03-01 00:00:00 >>> dt = datetime.datetime(2019, 12, 31) >>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)) 2020-01-01 00:00:00 >>> dt = datetime.datetime(2019, 12, 1) >>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)) 2020-01-01 00:00:00
import numpy as np import pandas as pd import matplotlib.pyplot as plt import matplotlib.dates as mdates # sample data N = 30 drange = pd.date_range(“2014-01″, periods=N, freq=”MS”) np.random.seed(365) # for a reproducible example of values values = {‘values’:np.random.randint(1,20,size=N)} df = pd.DataFrame(values, index=drange) fig, ax = plt.subplots() ax.plot(df.index, df.values) ax.set_xticks(df.index) # use formatters to specify major … Read more
parsedatetime – Python module that is able to parse ‘human readable’ date/time expressions. #!/usr/bin/env python from datetime import datetime import parsedatetime as pdt # $ pip install parsedatetime cal = pdt.Calendar() now = datetime.now() print(“now: %s” % now) for time_string in [“tomorrow at 6am”, “next moday at noon”, “2 min ago”, “3 weeks ago”, “1 … Read more
This question is already answered, but for the benefit of others looking at this question I’d like to add the following suggestion: Instead of doing the slicing yourself as suggested in the accepted answer, you might also use strptime() which is (IMHO) easier to read and perhaps the preferred way to do this conversion. import … Read more
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 found isodate library to do exactly what I want isodate.parse_duration(‘PT1H5M26S’) You can read the source code for the function here