This question already has answers here:
Python date string to date object
(9 answers)
Closed 5 years ago.
I'm a Python newbie and don't how to convert a Python 3.5x string
'2017-04-19 00:23'
into a date and time like
April 19, 2017 12:23 am
and even get individual units like
April
19
2017
12:23 am
or get day of week for 4/19/217
Wednesday
Use python datetime module, something like this :
from datetime import datetime
date_str = '2017-04-19 00:23'
date_obj = datetime.strptime(date_str, '%Y-%m-%d %H:%M')
# To get a particular part of the date in a particular format such as "Wednesday" for the "Datetime Object"
print(date_obj.strftime('%A'))
print(date_obj.strftime('%c'))
This will result in :
Wednesday
Wed Apr 19 00:23:00 2017
Check out the documentation.
This question already has answers here:
How do I calculate the date six months from the current date using the datetime Python module?
(47 answers)
Closed 7 years ago.
So I am trying to find a way to increment a datetime object by one month. However, it seems this is not so simple, according to this question.
I was hoping for something like:
import datetime as dt
now = dt.datetime.now()
later = now + dt.timedelta(months=1)
But that doesn't work. I was also hoping to be able to go to the same day (or the closest alternative) in the next month if possible. For example, a datetime object set at January 1st would increment to Feb 1st whereas a datetime object set at February 28th would increment to March 31st as opposed to March 28th or something.
To be clear, February 28th would (typically) map to March 31st because it is the last day of the month, and thus it should go to the last day of the month for the next month. Otherwise it would be a direct link: the increment should go to the day in the next month with the same numbered day.
Is there a simple way to do this in the current release of Python?
Check out from dateutil.relativedelta import *
for adding a specific amount of time to a date, you can continue to use timedelta for the simple stuff i.e.
import datetime
from dateutil.relativedelta import *
use_date = datetime.datetime.now()
use_date = use_date + datetime.timedelta(minutes=+10)
use_date = use_date + datetime.timedelta(hours=+1)
use_date = use_date + datetime.timedelta(days=+1)
use_date = use_date + datetime.timedelta(weeks=+1)
or you can start using relativedelta
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(years=+1)
for the last day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
Right now this will provide 29/02/2016
for the penultimate day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
use_date = use_date+relativedelta(days=-1)
last Friday of the next month:
use_date = use_date+relativedelta(months=+1, day=31, weekday=FR(-1))
2nd Tuesday of next month:
new_date = use_date+relativedelta(months=+1, day=1, weekday=TU(2))
As #mrroot5 points out dateutil's rrule functions can be applied, giving you an extra bang for your buck, if you require date occurences.
for example:
Calculating the last day of the month for 9 months from the last day of last month.
Then, calculate the 2nd Tuesday for each of those months.
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import datetime
use_date = datetime(2020,11,21)
#Calculate the last day of last month
use_date = use_date+relativedelta(months=-1)
use_date = use_date+relativedelta(day=31)
#Generate a list of the last day for 9 months from the calculated date
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, bymonthday=(-1,)))
print("Last day")
for ld in x:
print(ld)
#Generate a list of the 2nd Tuesday in each of the next 9 months from the calculated date
print("\n2nd Tuesday")
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, byweekday=TU(2)))
for tuesday in x:
print(tuesday)
Last day
2020-10-31 00:00:00
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00
2nd Tuesday
2020-11-10 00:00:00
2020-12-08 00:00:00
2021-01-12 00:00:00
2021-02-09 00:00:00
2021-03-09 00:00:00
2021-04-13 00:00:00
2021-05-11 00:00:00
2021-06-08 00:00:00
2021-07-13 00:00:00
rrule could be used to find the next date occurring on a particular day.
e.g. the next 1st of January occurring on a Monday (Given today is the 4th November 2021)
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import *
year = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0].year
year
2024
or the next 5 x 1st of January's occurring on a Monday
years = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0:5]
for i in years:print(i.year)
...
2024
2029
2035
2046
2052
The first Month next Year that starts on a Monday:
>>> month = rrule(YEARLY,dtstart=datetime.date(2023, 1, 1),bymonthday=1,byweekday=MO)[0]
>>> month.strftime('%Y-%m-%d : %B')
'2023-05-01 : May'
If you need the months that start on a Monday between 2 dates:
months = rrule(YEARLY,dtstart=datetime.date(2025, 1, 1),until=datetime.date(2030, 1, 1),bymonthday=1,byweekday=MO)
>>> for m in months:
... print(m.strftime('%Y-%m-%d : %B'))
...
2025-09-01 : September
2025-12-01 : December
2026-06-01 : June
2027-02-01 : February
2027-03-01 : March
2027-11-01 : November
2028-05-01 : May
2029-01-01 : January
2029-10-01 : October
This is by no means an exhaustive list of what is available.
Documentation is available here: https://dateutil.readthedocs.org/en/latest/
Note: This answer shows how to achieve this using only the datetime and calendar standard library (stdlib) modules - which is what was explicitly asked for. The accepted answer shows how to better achieve this with one of the many dedicated non-stdlib libraries. If you can use non-stdlib libraries, by all means do so for these kinds of date/time manipulations!
How about this?
def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12
new_day = orig_date.day
# while day is out of range for month, reduce by one
while True:
try:
new_date = datetime.date(new_year, new_month, new_day)
except ValueError as e:
new_day -= 1
else:
break
return new_date
EDIT:
Improved version which:
keeps the time information if given a datetime.datetime object
doesn't use try/catch, instead using calendar.monthrange from the calendar module in the stdlib:
import datetime
import calendar
def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12
last_day_of_month = calendar.monthrange(new_year, new_month)[1]
new_day = min(orig_date.day, last_day_of_month)
return orig_date.replace(year=new_year, month=new_month, day=new_day)
Question: Is there a simple way to do this in the current release of Python?
Answer: There is no simple (direct) way to do this in the current release of Python.
Reference: Please refer to docs.python.org/2/library/datetime.html, section 8.1.2. timedelta Objects. As we may understand from that, we cannot increment month directly since it is not a uniform time unit.
Plus: If you want first day -> first day and last day -> last day mapping you should handle that separately for different months.
>>> now
datetime.datetime(2016, 1, 28, 18, 26, 12, 980861)
>>> later = now.replace(month=now.month+1)
>>> later
datetime.datetime(2016, 2, 28, 18, 26, 12, 980861)
EDIT: Fails on
y = datetime.date(2016, 1, 31); y.replace(month=2) results in ValueError: day is out of range for month
Ther is no simple way to do it, but you can use your own function like answered below.
This question already has answers here:
Convert string date into date format in python?
(3 answers)
Closed 7 years ago.
How to convert 2015 June 1 into date format in python like date_object = datetime.date(2014, 12, 4)
You can use the format - '%Y %B %d' along with datetime.datetime.strptime() method to convert string to date. Where %Y is 4 digit year, %B is complete month name, and %d is date.
Example/Demo -
>>> datetime.datetime.strptime('2015 June 1','%Y %B %d')
datetime.datetime(2015, 6, 1, 0, 0)
>>> datetime.datetime.strptime('2015 June 1','%Y %B %d').date()
datetime.date(2015, 6, 1)
Use the first one, if you are content with datetime object, if you want the date() object itself, you can use the second one.
You can use the date constructor
>>> from datetime import date
>>> date_object = date(year=2015, month=6, day=1)
>>> print date_object
2015-06-01
This question already has answers here:
How to get the last day of the month?
(44 answers)
Closed 8 years ago.
As every month have different days in it, so i can't apply timedelta=30.
I want to get three variables
month_start,
month_end ,
month_days = month_end - month_start
Which will be correspond to start date of month and end date of month. and their interval will be number of days in the month.
for instance , for march : month_days = 31, april : month_days = 30
Use calendar module to get days from months
>>> import datetime
>>> import calendar
>>> now = datetime.datetime.now()
>>> print calendar.monthrange(now.year, now.month)[1]
31
For 2015 Feb month
>>> calendar.monthrange(2015, 2)
(6, 28)
https://docs.python.org/2/library/calendar.html
How can I get week start dates of each week in a year, considering start day of the week is Monday in python?
This assumes start day is Sunday:
>>>import datetime as datetime
>>>dt = datetime .date(2013,12,30)
>>>dt.isocalendar()[1]
1
However, result shouldn't be 1, because 30-12-2013 is still in 2013.
I don't think behaviour of isocalendar can be changed.
From : http://docs.python.org/2/library/datetime.html#datetime.date.isocalendar
The first week of an ISO year is the first (Gregorian) calendar week of a year containing a Thursday.
But strftime can display week number :
%U "All days in a new year preceding the first Sunday are considered to be in week 0."
%W "All days in a new year preceding the first Monday are considered to be in week 0."
>>> import datetime
>>> dt = datetime.date(2013,12,30)
>>> dt.isocalendar()
(2014, 1, 1)
>>> dt.strftime("%U")
'52'
But to respond to your first question, why don't you just use datetime.timedelta ?
>>> import datetime
>>> dt = datetime.date(2013,12,30)
>>> w = datetime.timedelta(weeks=1)
>>> dt - w
datetime.date(2013, 12, 23)
>>> dt + w
datetime.date(2014, 1, 6)
>>> dt + 10 * w
datetime.date(2014, 3, 10)