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python get the quarterly dates from a date range

python get the quarterly dates from a date range
example :
start date = 01/04/2020
end date = 01/04/2021
Here I need to get the Quaternary dates from this date range.

Try:
start_date = "01/04/2020"
end_date = "01/04/2021"
pd.date_range(start_date, end_date, freq='Q')
DatetimeIndex(['2020-03-31', '2020-06-30', '2020-09-30', '2020-12-31'], dtype='datetime64[ns]', freq='Q-DEC')

pd.date_range(start date, end date, freq='3m').to_period('Q')

With pure Python:
import datetime
start_date_str = "01/04/2020"
end_date_str = "01/04/2021"
start_date = datetime.datetime.strptime(start_date_str, "%d/%m/%Y").date()
end_date = datetime.datetime.strptime(end_date_str, "%d/%m/%Y").date()
print(f"Quarters within {start_date_str} and {end_date_str}:")
start_of_quarter = start_date
while True:
far_future = start_of_quarter + datetime.timedelta(days=93)
start_of_next_quarter = far_future.replace(day=1)
end_of_quarter = start_of_next_quarter - datetime.timedelta(days=1)
if end_of_quarter > end_date:
break
print(f"\t{start_of_quarter:%d/%m/%Y} - {end_of_quarter:%d/%m/%Y}")
start_of_quarter = start_of_next_quarter

How to replace the day in a date with another date?

I'm trying to replace the day in my if statement for my date but I keep getting this output for my year.
05/15/5 besides 05/15/2020 . Code is below:
today_date = datetime.datetime.now()
date = today_date.date()
formatted_date = datetime.date.strftime(date, "%m/%d/%Y")
mmonth = date.month
myear = date.year
mdate = date.day
if mdate < 7:
m0weekend = formatted_date.replace(str(myear),str(mmonth),1)
else:
m0weekend = formatted_date.replace(str(myear),str(mmonth),15)

it's easier to replace the day before converting to a string:
date = date.replace(day=1)
or, in your case:
if mdate < 7:
m0weekend = date.replace(day=1)
else:
m0weekend = date.replace(day=15)

formatted_date is actually a string.
You are using the str.replace() method not the datetime.date.replace() method.
import datetime
today_date = datetime.datetime.now()
pre_formatted_date = today_date.date()
mmonth = pre_formatted_date.month
myear = pre_formatted_date.year
mdate = pre_formatted_date.day
if mdate < 7:
pre_formatted_date = pre_formatted_date.replace(day=1)
else:
pre_formatted_date = pre_formatted_date.replace(day=15)
print(pre_formatted_date)
formatted_date = pre_formatted_date.strftime("%m/%d/%Y")
print(formatted_date)
Which has the following output:
2020-05-15
05/15/2020

You might get today datetime.date directly from datetime rather than creating datetime.datetime and converting to date. After you have today you might create needed datetime.date and turn it into str, i.e.:
import datetime
today = datetime.date.today()
date = datetime.date(today.year, today.month, 1 if today.day < 7 else 15)
formatted_date = datetime.date.strftime(date, "%m/%d/%Y")
print(formatted_date) # 05/15/2020

Getting dates in python between a past datestamp and the present

Language Python
I am wondering if anyone can help me print out some dates.
i cam trying to create a loop in which i pass in a date say 01/1/2017 and the loop will then output the first and last day in every month between then and the present day.
Example
01/01/2017
31/01/2017
01/02/2017
28/02/2017
etc
Any help will be appreciated

Hope this will help,
Code:
from datetime import date
from dateutil.relativedelta import relativedelta
from calendar import monthrange
d1 = date(2018, 2, 26)
d2 = date.today()
def print_month_day_range(date):
first_day = date.replace(day = 1)
last_day = date.replace(day = monthrange(date.year, date.month)[1])
print (first_day.strftime("%d/%m/%Y"))
print (last_day.strftime("%d/%m/%Y"))
print_month_day_range(d1)
while (d1 < d2):
d1 = d1 + relativedelta(months=1)
print_month_day_range(d1)
Output:
01/02/2018
28/02/2018
01/03/2018
31/03/2018
...
01/07/2018
31/07/2018

you can use calendar module. Below is the code:
import datetime
from calendar import monthrange
strt_date = '01/1/2017'
iter_year = datetime.datetime.strptime(strt_date, '%m/%d/%Y').year
iter_month = datetime.datetime.strptime(strt_date, '%m/%d/%Y').month
cur_year = datetime.datetime.today().year
cur_month = datetime.datetime.today().month
while cur_year > iter_year or (cur_year == iter_year and iter_month <= cur_month):
number_of_days_in_month = monthrange(iter_year, iter_month)[1]
print '/'.join([str(iter_month) , '01', str(iter_year)])
print '/'.join([str(iter_month), str(number_of_days_in_month), str(iter_year)])
if iter_month == 12:
iter_year += 1
iter_month = 1
else:
iter_month += 1

this could work, as long as the first date you give is always the first of the month
from datetime import datetime
from datetime import timedelta
date_string = '01/01/2017'
date = datetime.strptime(date_string, '%d/%m/%Y').date()
today = datetime.now().date()
months = range(1,13)
years = range(date.year, today.year + 1)
for y in years:
for m in months:
new_date = date.replace(month=m, year=y)
last_day = new_date - timedelta(days=1)
if (date < new_date) & (new_date <= today):
print last_day.strftime('%d/%m/%Y')
print new_date.strftime('%d/%m/%Y')
elif (date <= new_date) & (new_date <= today):
print new_date.strftime('%d/%m/%Y')

This code would print the first and last days of all months in a year.
Maybe add some logic to iterate through the years
import datetime
def first_day_of_month(any_day):
first_month = any_day.replace(day=1)
return first_month
def last_day_of_month(any_day):
next_month = any_day.replace(day=28) + datetime.timedelta(days=4)
return next_month - datetime.timedelta(days=next_month.day)
for month in range(1, 13):
print first_day_of_month(datetime.date(2017, month, 1))
print last_day_of_month(datetime.date(2017, month, 1))

how to convert string to datetime.timedelta()?

how can i convert my string of date to a datetime.timedelta() in Python?
I have this code :
import datetime
date_select = '2011-12-1'
delta = datetime.timedelta(days=1)
target_date = date_select + delta
print target_date
thanks in advance ...

You wouldn't convert date_select to a timedelta, instead, you need a datetime object, which can be added to a timedelta to produce an updated datetime object:
from datetime import datetime, timedelta
date_select = datetime.strptime('2011-12-1', '%Y-%m-%d')
delta = timedelta(days=1)
target_date = date_select + delta
print target_date
Or, if you prefer, without the fancy from ... import ... import line:
import datetime # <- LOOK HERE, same as in your example
date_select = datetime.datetime.strptime('2011-12-1', '%Y-%m-%d')
delta = datetime.timedelta(days=1)
target_date = date_select + delta
print target_date

You use strptime to do this.
from datetime import datetime
target_date = datetime.strptime(date_select, '%Y-%m-%d')

from datetime import datetime, timedelta
date_select = '2011-12-1'
new_data = datetime.strptime(date_select, '%Y-%m-%d')
delta = timedelta(days=1)
target_date = date_select + delta
print target_date
You will get 2011-12-02 00:00:00; to strip off the '00:00:00' and get only the date, just add .date() to target_date
print target_date.date()
This should give you the only the date = 2011-12-02

How to get sat and sun dates from given date ranges in python

How to get saturday and sunday dates from given date range.
for example :
start_date = 2011-09-01
end_date = 2011-09-15
Now it should give the out put
2011-09-03
2011-09-04
2011-09-10
2011-09-11
Any help really appreciate.

>>> import datetime
>>> start = datetime.datetime.strptime("2011-09-01", "%Y-%m-%d")
>>> end = datetime.datetime.strptime("2011-09-15", "%Y-%m-%d")
>>> while start <= end:
... if start.weekday() in (5, 6):
... print start.strftime("%Y-%m-%d")
... start += datetime.timedelta(days=1)
...
2011-09-03
2011-09-04
2011-09-10
2011-09-11
>>>

Well, I think this should work:
from datetime import datetime, timedelta
def daterange(start_date, end_date):
for n in range((end_date - start_date).days):
yield start_date + timedelta(n)
start_date = '2011-09-01'
end_date = '2011-09-15'
format = '%Y-%m-%d'
start_date_object = datetime.strptime(start_date, format)
end_date_object = datetime.strptime(end_date, format)
for day in daterange(start_date_object, end_date_object):
if day.weekday() in [5, 6]:
print day.strftime(format)