Get First Date and Last Date of Current Quarter in Python? - python-2.7

Question

How can i get the Current Quarter year and then First Date and last Date of Current Quarter Year in Python?
i want by importing datetime
import datetime
People look into Stack overflow need straight forward answer and which should be very simple. Which ever link you provided it having lot of Comments. SO, users has to go through all the comments to find correct answer. I am writing simple and straight forward answer.

Answer 1

I believe that none of the current answers are still valid in Python 3, so since this is the top hit in google for first and last day of quarter, I will provide a solution that works in Python 3 (mostly Ahmet's with // instead of /):
from datetime import date as date_class
from datetime import timedelta, datetime
def get_quarter(p_date: date_class) -> int:
return (p_date.month - 1) // 3 + 1
def get_first_day_of_the_quarter(p_date: date_class):
return datetime(p_date.year, 3 * ((p_date.month - 1) // 3) + 1, 1)
def get_last_day_of_the_quarter(p_date: date_class):
quarter = get_quarter(p_date)
return datetime(p_date.year + 3 * quarter // 12, 3 * quarter % 12 + 1, 1) + timedelta(days=-1)
assert get_quarter(datetime(year=2021, month=10, day=5).date()) == 4
assert get_quarter(datetime(year=2020, month=9, day=25).date()) == 3
assert get_quarter(datetime(year=2020, month=12, day=11).date()) == 4
assert get_quarter(datetime(year=2020, month=1, day=2).date()) == 1
assert get_first_day_of_the_quarter(datetime(2020, 10, 5).date()) == datetime(2020, 10, 1)
assert get_first_day_of_the_quarter(datetime(2020, 9, 25).date()) == datetime(2020, 7, 1)
assert get_first_day_of_the_quarter(datetime(2020, 12, 11).date()) == datetime(2020, 10, 1)
assert get_first_day_of_the_quarter(datetime(2020, 1, 2).date()) == datetime(2020, 1, 1)
assert get_last_day_of_the_quarter(datetime(2020, 10, 5).date()) == datetime(2020, 12, 31)
assert get_last_day_of_the_quarter(datetime(2020, 9, 25).date()) == datetime(2020, 9, 30)
assert get_last_day_of_the_quarter(datetime(2020, 12, 11).date()) == datetime(2020, 12, 31)
assert get_last_day_of_the_quarter(datetime(2020, 1, 2).date()) == datetime(2020, 3, 31)
assert get_last_day_of_the_quarter(datetime(2020, 5, 6).date()) == datetime(2020, 6, 30)

Answer 2

Having the first day is the same with #Karishh's solution. But, for the last date, Python2.7 causes a problem for fourth quarter. Because 12+1=13 and datetime does not accept 13 as a month. So you need to do some tricks to handle it.
import datetime
def get_quarter(date):
return (date.month - 1) / 3 + 1
def get_first_day_of_the_quarter(date):
quarter = get_quarter(date)
return datetime.datetime(date.year, 3 * quarter - 2, 1)
def get_last_day_of_the_quarter(date):
quarter = get_quarter(date)
month = 3 * quarter
remaining = month / 12
return datetime.datetime(date.year + remaining, month % 12 + 1, 1) + datetime.timedelta(days=-1)

Answer 3

Any how i found some simple solution in c# and converted it into python,
from datetime import datetime,timedelta
current_date=datetime.now()
currQuarter = (current_date.month - 1) / 3 + 1
dtFirstDay = datetime(current_date.year, 3 * currQuarter - 2, 1)
dtLastDay = datetime(current_date.year, 3 * currQuarter + 1, 1) + timedelta(days=-1)

Answer 4

Here's a one/two liner to get the start/end date of the current quarter.
from datetime import datetime
import math
from dateutil.relativedelta import relativedelta # requires python-dateutil
start_of_quarter = datetime(year=datetime.now().year, month=((math.floor(((datetime.now().month - 1) / 3) + 1) - 1) * 3) + 1, day=1)
end_of_quarter = start_of_quarter + relativedelta(months=3, seconds=-1)

Answer 5

pendulum has a much more intuitive implementation.
import pendulum
dt = pendulum.datetime(2021, 3, 23)
print(dt.first_of('quarter'))
print(dt.last_of('quarter'))
2021-01-01T00:00:00+00:00
2021-03-31T00:00:00+00:00

Answer 6

Building off of the answer from Krishh, but addressing several issues found:
calculating last day in Quarter 4
Wasae Shoaib's comment about raising a ValueError
TypeError with a float being passed instead of an integer
Using relativedeta instead and shifting the correctly calculated start date by three months, we end up with a much more reliable way to get at the quarter end date.
from datetime import datetime
from dateutil.relativedelta import relativedelta
current_date = datetime.now()
currQuarter = int((current_date.month - 1) / 3 + 1)
dtFirstDay = datetime(current_date.year, 3 * currQuarter - 2, 1)
dtLastDay = dtFirstDay + relativedelta(months=3, days=-1)

Answer 7

I did of a lot of tests to find the solution that fit my need and I will be happy if it helps someone else :
datval = fields.date.today()
if datval.month < 4 :
self.start_date = fields.date.today().replace(month=10, day=1)
self.end_date = fields.date.today().replace(month=12, day=31)
elif datval.month < 7 :
self.start_date = fields.date.today().replace(month=1, day=1)
self.end_date = fields.date.today().replace(month=3, day=31)
elif datval.month < 10 :
self.start_date = fields.date.today().replace(month=4, day=1)
self.end_date = fields.date.today().replace(month=6, day=30)
else :
self.start_date = fields.date.today().replace(month=7, day=1)
self.end_date = fields.date.today().replace(month=9, day=30)

Answer 8

Get start and end points for: week, month, quarter and year
https://gist.github.com/dejurin/236b398dc4b8064685702a27a3df612b
from datetime import date
from dateutil.relativedelta import relativedelta
def start_end_day(sunmon: bool = True):
today = date.today()
curr_quarter = int((today.month - 1) / 3 + 1)
dayofweek = [today.weekday(),today.isoweekday()][sunmon]
week_start = today - relativedelta(days=dayofweek)
week_end = week_start + relativedelta(days=6)
month_start = date(today.year,today.month, 1)
month_end = month_start + relativedelta(months=1, days=-1)
quarter_start = date(today.year, 3 * curr_quarter - 2, 1)
quarter_end = quarter_start + relativedelta(months=3, days=-1)
year_start = date(today.year, 1, 1)
year_end = year_start + relativedelta(years=1, days=-1)
return ((week_start,week_end),(month_start,month_end),(quarter_start,quarter_end),(year_start,year_end))
"""
Current date: 18/02/2022
"""
"""
((datetime.date(2022, 2, 13), datetime.date(2022, 2, 19)),
(datetime.date(2022, 2, 1), datetime.date(2022, 2, 28)),
(datetime.datetime(2022, 1, 1, 0, 0), datetime.datetime(2022, 3, 31, 0, 0))
(datetime.date(2022, 1, 1), datetime.date(2022, 12, 31)))
"""