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
Related
I want last friday of each month for upcoming three months.
Friday_date = datetime.date.today()
while Friday_date.weekday() != 4:
Friday_date += datetime.timedelta(1)
This gives me the nearest friday. I want to make sure this is the last friday of this month so that i can add 28 days to get next friday.
The easiest way to do this is to use the module dateutil:
>>> from dateutil.relativedelta import FR, relativedelta
>>> datetime.date.today()+relativedelta(day=31, weekday=FR(-1))
datetime.date(2021, 6, 25)
Don't assume you can get the last Friday of subsequent months just by adding 28 days. It won't always work. Adding 28 days to the last Friday of February 2024 gives you this:
>>> datetime.date(2024,2,1)+relativedelta(day=31, weekday=FR(-1), days=28)
datetime.date(2024, 3, 22)
but the last Friday of that month is 29 March. Let dateutil do that correctly for you:
>>> datetime.date(2024,2,1)+relativedelta(day=31, weekday=FR(-1), months=1)
datetime.date(2024, 3, 29)
If needed with standard library only, here is with calendar and datetime:
import calendar
from datetime import date
today = date.today()
year, month = today.year, today.month
n_months = 4
friday = calendar.FRIDAY
for _ in range(n_months):
# get last friday
c = calendar.monthcalendar(year, month)
day_number = c[-1][friday] or c[-2][friday]
# display the found date
print(date(year, month, day_number))
# refine year and month
if month < 12:
month += 1
else:
month = 1
year += 1
where the line c[-1][friday] or c[-2][friday] first checks the last week of the month: is Friday nonzero there? if so take it, else look at the week before where there must be a Friday.
This prints
2021-06-25
2021-07-30
2021-08-27
2021-09-24
This formula gets you the day of the last Friday of any given month:
import calendar
year = 2021
month = 6
last_day = calendar.monthrange(year, month)[1]
last_weekday = calendar.weekday(year, month, last_day)
last_friday = last_day - ((7 - (4 - last_weekday)) % 7)
# ^ ^
# | Friday
# days in a week
This is my first coffee, so this can probably be condensed a bit, but it illustrates the logic. last_day is the last calendar day (30 for June), last_weekday is what weekday it is (2 for Wednesday), and based on that we simply calculate how many days to subtract to land on the last Friday (25).
If you want to know the last friday you can do this :
from datetime import date
from datetime import timedelta
today = date.today()
offset = (today.weekday() - 4) % 7
last_wednesday = today - timedelta(days=offset)
This question already has answers here:
Previous weekday in Python
(4 answers)
Closed 1 year ago.
I need to find the last weekday (without considering holidays). Is there any more elegant/robust way of implementing this than the one I'm currently using?
from datetime import date
b = date.today() - timedelta(days=1)
while b.weekday()>=5:
b = b - timedelta(days=1)
Something like this might work (untested):
from datetime import date, timedelta
b = date.today() - timedelta(days=1)
current_day = b.weekday() # 0 for Monday, 6 for Sunday
# go back 1 day for Sat, 2 days for Sun, 0 for other days
days_to_go_back = max(current_day - 4, 0)
b -= timedelta(days=days_to_go_back)
This question already has answers here:
calculate the difference between two datetime.date() dates in years and months
(3 answers)
Closed 4 years ago.
I have 2 person's birth information, I want to do some analysis on them.
Like, the difference between their age, seconds, years+months+days.
I tried this:
from datetime import date
a = date(1991, 07, 20)
b = date(1999, 06, 06)
print((a-b).days)
-2878
this gives me 2878 days, but i want to calculate years + months + days
i tried to divide 2878/365, but i want the exact calculations
How can i approach this?
Expected Output:
7 years x months x days
Use datetime and dateutil:
from datetime import datetime
from dateutil import relativedelta
date1 = datetime(1991, 7, 20)
date2 = datetime(1999, 6, 6)
diff = relativedelta.relativedelta(date2, date1)
years = diff.years
months = diff.months
days = diff.days
print('{} years {} months {} days'.format(years, months, days))
# 7 years 10 months 17 days
For strict differences, i.e. differences between years, months and days, you can use the attributes of timedelta objects.
from datetime import date
a = date(1991, 7, 20)
b = date(1999, 6, 6)
months = a.month - b.month
years = a.year - b.year
days = a.day - b.day
print('{0} years, {1} months, {2} days'.format(years, months, days))
-8 years, 1 months, 14 days
For time-aware differences, you can use 3rd party dateutil as per #Austin's solution.
This question already has answers here:
How to convert integer timestamp into a datetime
(3 answers)
Closed 5 years ago.
How do I convert an int like 1485714600 such that my result ends up being Monday, January 30, 2017 12:00:00 AM?
I've tried using datetime.datetime but it gives me results like '5 days, 13:23:07'
Like this?
>>> from datetime import datetime
>>> datetime.fromtimestamp(1485714600).strftime("%A, %B %d, %Y %I:%M:%S")
'Sunday, January 29, 2017 08:30:00'
What you describe here is a (Unix) timestamp (the number of seconds since January 1st, 1970). You can use:
datetime.datetime.fromtimestamp(1485714600)
This will generate:
>>> import datetime
>>> datetime.datetime.fromtimestamp(1485714600)
datetime.datetime(2017, 1, 29, 19, 30)
You can get the name of the day by using .strftime('%A'):
>>> datetime.datetime.fromtimestamp(1485714600).strftime('%A')
'Sunday'
Or you can call weekday() to obtain an integers between 0 and 6 (both inclusive) that maps thus from monday to sunday:
>>> datetime.datetime.fromtimestamp(1485714600).weekday()
6
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.