This function is supposed to add days to a from_date, adding n-number of days (add_days) without counting weekends and Holidays (from my list), but to stop at the end of the month if the from_date day will exceed the current month.
For example if I print:
print date_by_adding_business_days(datetime.date(2015,01,22), 3,Holidays)
will print: 2015-01-27, that is good, works for holidays also.
If I will try :
print date_by_adding_business_days(datetime.date(2015,01,22), 12,Holidays)
It will stop at the end of the month as I want : 2015-01-30.
But if my from_date will be :
print date_by_adding_business_days(datetime.date(2015,01,30), 12,Holidays)
It will return None, instead of 2015-01-30 that I need.
My problem is when my from_date is the end of the month, my function will return None. I want to return that exact from_date if my from_date is at the end of the month(won't depend of how much days I want to add.
This is my function:
def date_by_adding_business_days(from_date, add_days,holidays):
business_days_to_add = add_days
current_date = from_date
result = None
_, days_in_month = calendar.monthrange(current_date.year, current_date.month)
while business_days_to_add > 0 and current_date.day < days_in_month:
current_date += datetime.timedelta(days=1)
weekday = current_date.weekday()
if weekday >= 5:
continue
if current_date in holidays:
continue
business_days_to_add -= 1
result = current_date
return result
Holidays =[datetime.date(2015,10,14),datetime.date(2015,10,15)]
print date_by_adding_business_days(datetime.date(2015,01,30), 12,Holidays)
You codes does not check the start date.
>>> # The last day in the month is not a business day so it should returns `2015-01-30`
... print date_by_adding_business_days(datetime.date(2015, 01, 30), 12, Holidays)
None
>>> # The last day in the month is not a business day so it should returns `None`
... print date_by_adding_business_days(datetime.date(2015, 01, 31), 12, Holidays)
None
>>> # The last day in the month is a business day so it should returns `2016-10-31`
... print date_by_adding_business_days(datetime.date(2016, 10, 30), 12, Holidays)
2016-10-31
>>> # The last day in the month is a business day so it should returns `2016-10-31
... print date_by_adding_business_days(datetime.date(2016, 10, 31), 12, Holidays)
None
>>> # Last two days in the month is not business days so it should returns `None`
... print date_by_adding_business_days(datetime.date(2016, 07, 30), 12, Holidays)
None
The solution result = current_date does not work as it will always return the start date although there are no business day.
>>> # The last day in the month is not a business day so it should returns `2015-01-30`
... print date_by_adding_business_days(datetime.date(2015, 01, 30), 12, Holidays)
2015-01-30
>>> # The last day in the month is not a business day so it should returns `None`
... print date_by_adding_business_days(datetime.date(2015, 01, 31), 12, Holidays)
2015-01-31
>>> # The last day in the month is a business day so it should returns `2016-10-31`
... print date_by_adding_business_days(datetime.date(2016, 10, 30), 12, Holidays)
2016-10-31
>>> # The last day in the month is a business day so it should returns `2016-10-31
... print date_by_adding_business_days(datetime.date(2016, 10, 31), 12, Holidays)
2016-10-31
>>> # Last two days in the month is not business days so it should returns `None`
... print date_by_adding_business_days(datetime.date(2016, 07, 30), 12, Holidays)
2016-07-30
I have rewritten your code so it check the start date and will return the none in the absence of a result.
def date_by_adding_business_days(from_date, add_days, holidays):
business_days_to_add = add_days
current_date = from_date
result = None
_, days_in_month = calendar.monthrange(current_date.year, current_date.month)
while business_days_to_add > 0:
if current_date.weekday() < 5 and current_date not in holidays:
business_days_to_add -= 1
result = current_date
if current_date.day == days_in_month:
break
current_date += datetime.timedelta(days=1)
return result
-
>>> # The last day in the month is a not business day so it should returns `2015-01-30`
... print date_by_adding_business_days(datetime.date(2015, 01, 30), 12, Holidays)
2015-01-30
>>> # The last day in the month is not a business day so it should returns `None`
... print date_by_adding_business_days(datetime.date(2015, 01, 31), 12, Holidays)
None
>>> # The last day in the month is a business day so it should returns `2016-10-31`
... print date_by_adding_business_days(datetime.date(2016, 10, 30), 12, Holidays)
2016-10-31
>>> # The last day in the month is a business day so it should returns `2016-10-31
... print date_by_adding_business_days(datetime.date(2016, 10, 31), 12, Holidays)
2016-10-31
>>> # Last two days in the month is not business days so it should returns `None`
... print date_by_adding_business_days(datetime.date(2016, 07, 30), 12, Holidays)
None
Related
Here is the problem:
If the int values [0,7) (0, 1, 2, 3, 4, 5, 6) refer to Monday through
Sunday, and today is Monday, what day of the week will it be in 999
days?
Here is how I solved it:
import datetime
#Capture the First Date
day1 = datetime.date(2021, 1, 25)
print('day1:', day1.ctime())
# Capture the Second Date
day2 = datetime.date(2023, 10, 21)
print('day2:', day2.ctime())
# Find the difference between the dates
print('Number of Days:', day1-day2)
Returns:
day1: Mon Jan 25 00:00:00 2021
day2: Sat Oct 21 00:00:00 2023
Number of Days: -999 days, 0:00:00
Use timedelta to add n days from your "start date":
from datetime import date, timedelta
current = date.today()
future = current + timedelta(days=999)
print(f"{current=}", current.weekday(), current.strftime("%A"))
print(f"{future=}", future.weekday(), future.strftime("%A"))
Output:
current=datetime.date(2021, 1, 24) 6 Sunday
future=datetime.date(2023, 10, 20) 4 Friday
.weekday() returns the day of the week as an integer.
.strftime("%A") will format a date object as a Weekday name.
I'm currently unsure on the logic to be used for the below problem and new to programming as well.(Currently learning python)
Trying to iterate thru every date for a given month - say 05/01 -- 05/31 and print it out in the below format.
Monday thru Friday dates are to be printed separately.
Saturday & Sunday dates are to be printed separately.
If the month starts on say Friday - 05/01/2020, ouput should be like
as, its the last weekday of that week.
For the month of April 2020, output would be like below, as April month's 1st week started on Wednesday.
I managed to comeup with the below try, but not sure how to proceed further.
import sys
from datetime import date, datetime, timedelta
year = int(sys.argv[1])
month = int(sys.argv[2])
st_dt = int(sys.argv[3])
en_dt = int(sys.argv[4])
first_date = datetime(year, month, st_dt).date()
get_first_day = datetime(year, month, st_dt).isoweekday()
def daterange(startDate, endDate, delta=timedelta(days=1)):
currentDate = startDate
while currentDate <= endDate:
yield currentDate
currentDate += delta
for date in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)):
print(date)
date.py 2020 5 1 31 # script
Came up with a standalone 'if loop' and as i said before, not sure how to construct the bigger picture :(
if get_first_day == 1:
#print("Monday")
sec_d = first_date + timedelta(days=4)
elif get_first_day == 2:
sec_d = first_date + timedelta(days=3)
elif get_first_day == 3:
sec_d = first_date + timedelta(days=2)
elif get_first_day == 4:
sec_d = first_date + timedelta(days=2)
elif get_first_day == 5:
sec_d = first_date
#print("Friday")
else:
pass
print(f"Second date:{sec_d} ") -- which gave -- > Second date:2020-05-01
You could keep the dates in a dictionary, dictionary key is tuple of calendar week and type of day (weekend, day of the week).
Each day is saved by in the allDays dictionary, grouped by the combination of weeknum and type of day as key:
('18', 'weekend'): [datetime.date(2020, 5, 2), datetime.date(2020, 5, 3)],
('18', 'working'): [datetime.date(2020, 5, 1)],
('19', 'weekend'): [datetime.date(2020, 5, 9), datetime.date(2020, 5, 10)],
('19', 'working'): [datetime.date(2020, 5, 4), ...
So you just need to take out the fist and last item of each dict item:
import sys
from datetime import date, datetime, timedelta
year, month, st_dt, en_dt = 2020, 5, 1, 31
first_date = datetime(year, month, st_dt).date()
get_first_day = datetime(year, month, st_dt).isoweekday()
def daterange(startDate, endDate, delta=timedelta(days=1)):
currentDate = startDate
while currentDate <= endDate:
yield currentDate
currentDate += delta
allDays = {}
_lastDayType = None
for dte in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)):
if 0 <= dte.weekday() < 5:
_dayType = 'working'
else:
_dayType = 'weekend'
_weeknum = dte.strftime("%V") # number of calendar week
_key = (_weeknum, _dayType)
if _key not in allDays: # create an empty list if unique key doesnt exist
allDays[_key] = []
allDays[_key].append(dte) # add the dates ...
for k,v in allDays.items():
if len(v) == 1:
first, last = v[0], v[0]
else:
first, last = v[0], v[-1]
print("%s >> %s" % (first, last))
Output:
2020-05-01 >> 2020-05-01
2020-05-02 >> 2020-05-03
2020-05-04 >> 2020-05-08
2020-05-09 >> 2020-05-10
2020-05-11 >> 2020-05-15
2020-05-16 >> 2020-05-17
2020-05-18 >> 2020-05-22
2020-05-23 >> 2020-05-24
2020-05-25 >> 2020-05-29
2020-05-30 >> 2020-05-31
I am writing code to take data from the last year. I want to round up the earlier date like so: If it is July 14 2015, I want data from August 1st 2014-July 14,2015
df = pd.read_csv('MyData.csv')
df['recvd_dttm'] = pd.to_datetime(df['recvd_dttm'])
range_max = datetime.datetime.now()
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)+ pd.tseries.offsets.MonthEnd(1) + pd.tseries.offsets.DateOffset(days=1)
if datetime.datetime.now() == is_month_end:
# take slice with final week of data
df = df[(df['recvd_dttm'] >= range_min) &
(df['recvd_dttm'] <= range_max)]
My problem is that when it is July 31, 2015, my code goes to the end of the next month, essentially cutting out an entire month.
I am trying to make a for loop to fix this problem.
If it is the end of the month:
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)
else:
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)+ pd.tseries.offsets.MonthEnd(1) + pd.tseries.offsets.DateOffset(days=1)
How do I tell python to check for the end of the month? MonthEnd is only an offset function.
We can avoid importing the calendar module with a short function that only leverages datetime.
If tomorrow's month is not the same as today's month, then that means today is the last day of the current month. We can check this programmatically with a short function such as
import datetime
def end_of_month(dt):
todays_month = dt.month
tomorrows_month = (dt + datetime.timedelta(days=1)).month
return tomorrows_month != todays_month
Now, for your specific use case:
now = datetime.datetime.now()
if end_of_month(now):
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)
else:
range_min = range_max - pd.tseries.offsets.DateOffset(years=1) +pd.tseries.offsets.MonthEnd(1) + pd.tseries.offsets.DateOffset(days=1)
I simply would use the monthrange method of calendar module to find last day number of the month:
def check_if_last_day_of_week(date):
import datetime
import calendar
# calendar.monthrange return a tuple (weekday of first day of the
# month, number
# of days in month)
last_day_of_month = calendar.monthrange(date.year, date.month)[1]
# here i check if date is last day of month
if date == datetime.date(date.year, date.month, last_day_of_month):
return True
return False
>>> date = datetime.date(2018, 12, 31)
>>> check_if_last_day_of_week(date)
True
If the next day is a different month, it means it is the last day of a month.
def check_if_last_day_of_month(to_date):
delta = datetime.timedelta(days=1)
next_day = to_date + delta
if to_date.month != next_day.month:
return True
return False
I was using Pandas and I did not want to include another library, so I used this to check whether is the last day of the month and last day of the year:
import pandas as pd
my_date = '31-12-2021'
current_data = pd.to_datetime(my_date, format='%d-%m-%Y')
current_month = current_data.month
current_year = current_data.year
following_day = current_data + pd.DateOffset(1)
tomorrows_month = following_day.month
tomorrows_year = following_day.year
is_last_day_of_month = True if tomorrows_month != current_month else False
is_last_day_of_year = True if tomorrows_year != current_year else False
Here's a pure python approach that also takes into account leap years for february:
# total days in every month during non leap years
M_DAYS = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
def isleap(year):
"""Return True for leap years, False for non-leap years."""
return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
def days_in_month(year, month):
"""Returns total number of days in a month accounting for leap years."""
return M_DAYS[month] + (month == 2 and isleap(year))
def is_monthend(ref_date):
"""Checks whether a date is also a monthend"""
return ref_date.day == days_in_month(ref_date.year, ref_date.month)
Alright, here's what I did. Found the calendar module that BryanOakley suggested and made this loop. It checks the current day and checks if it is the same as the last day of the month, and chooses the range_min accordingly.
if datetime.datetime.now().day == calendar.monthrange(date.year, date.month)[1]:
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)+ pd.tseries.offsets.DateOffset(days=1)
else:
range_min = range_max - pd.tseries.offsets.DateOffset(years=1)+ pd.tseries.offsets.MonthEnd(1) + pd.tseries.offsets.DateOffset(days=1)
import datetime
def find_curr_month_end_date(curr_date):
if(curr_date.month != 12):
next_month_first_date= curr_date.replace(day=1).replace(month=curr_date.month+1)
else:
next_month_first_date= curr_date.replace(day=1).replace(month=1).replace(year=curr_date.year+1)
curr_month_end_date = next_month_first_date - datetime.timedelta(days=1)
return curr_month_end_date
curr_date = datetime.datetime.today()
# or curr_date = datetime.datetime.strptime("2020-12-16","%Y-%m-%d")
curr_month_end_date =
find_curr_month_end_date(curr_date)
Here is a short function to accomplish this. It requires the dateutil module so that you can do relative date math.
import datetime
from dateutil.relativedelta import relativedelta
def lastyear_period_start(current_date):
last_year = current_date - relativedelta(months=11)
return datetime.date(last_year.year, last_year.month, 1)
It can be utilized like so:
dates = [
datetime.datetime(2010, 2, 27),
datetime.datetime(2011, 2, 27),
datetime.datetime(2012, 2, 27),
datetime.datetime(2013, 2, 27),
datetime.datetime(2014, 2, 27),
datetime.datetime(2010, 7, 27),
datetime.datetime(2011, 7, 27),
datetime.datetime(2012, 7, 27),
datetime.datetime(2013, 7, 27),
datetime.datetime(2014, 7, 27),
datetime.datetime(2015, 7, 14),
datetime.datetime(2015, 7, 31),
datetime.datetime(2011, 2, 28),
datetime.datetime(2012, 2, 29),
datetime.datetime(2013, 2, 28),
]
for d in dates:
print d, lastyear_period_start(d)
This prints that following
2010-02-27 00:00:00 2009-03-01
2011-02-27 00:00:00 2010-03-01
2012-02-27 00:00:00 2011-03-01
2013-02-27 00:00:00 2012-03-01
2014-02-27 00:00:00 2013-03-01
2010-07-27 00:00:00 2009-08-01
2011-07-27 00:00:00 2010-08-01
2012-07-27 00:00:00 2011-08-01
2013-07-27 00:00:00 2012-08-01
2014-07-27 00:00:00 2013-08-01
2015-07-14 00:00:00 2014-08-01
2015-07-31 00:00:00 2014-08-01
2011-02-28 00:00:00 2010-03-01
2012-02-29 00:00:00 2011-03-01
2013-02-28 00:00:00 2012-03-01
In the function we're doing two simple steps
last_year = current_date - relativedelta(months=11)
First we find out what the date was 11 months ago, based on the date passed to the function
return datetime.date(last_year.year, last_year.month, 1)
Then we return the first day of that month.
In the output above you can see this accounts for leap years as well.
I think everyone kwows when it's pi-Day (If you don't know it's on 14 March each year). When you have a result in python like:
(2016, 4, 4)
(This stands for the April 4, 2016). How can I find in a fast way when it's the next pi-Day. In this example the answer would be:
(2017, 3, 14)
Is there any formula I can use? Many thanks!
Using the datetime module you'd:
Get todays date
Get this years Pi-day (take the year from today's date)
If this date is in the past, add one year
In code:
from datetime import date
def next_pi_date():
today = date.today()
pi_date = date(today.year, 3, 14)
if pi_date < today:
pi_date = pi_date.replace(year=pi_date.year + 1)
return pi_date
Demo:
>>> from datetime import date
>>> def next_pi_date():
... today = date.today()
... pi_date = date(today.year, 3, 14)
... if pi_date < today:
... pi_date = pi_date.replace(year=pi_date.year + 1)
... return pi_date
...
>>> next_pi_date()
datetime.date(2015, 3, 14)
import datetime
dates = [
datetime.datetime(2014,2,4),
datetime.datetime(2014,3,4),
datetime.datetime(2014,3,27),
datetime.datetime(2014,4,4),
]
for d in dates:
if d.month < 3:
print datetime.datetime(d.year, 3, 14)
elif d.month == 3 and d.day <= 14:
print datetime.datetime(d.year, 3, 14)
else:
print datetime.datetime(d.year+1, 3, 14)
Outputs:
2014-03-14 00:00:00
2014-03-14 00:00:00
2015-03-14 00:00:00
2015-03-14 00:00:00
This is checking that the month is less than 3 (March) and if it is, prints this year's Pi Day. If it is March, it makes sure it's less than the 14th. If it's after March, it uses next March 14th.
Given a particular date, say 2011-07-02, how can I find the date of the next Monday (or any weekday day for that matter) after that date?
import datetime
def next_weekday(d, weekday):
days_ahead = weekday - d.weekday()
if days_ahead <= 0: # Target day already happened this week
days_ahead += 7
return d + datetime.timedelta(days_ahead)
d = datetime.date(2011, 7, 2)
next_monday = next_weekday(d, 0) # 0 = Monday, 1=Tuesday, 2=Wednesday...
print(next_monday)
Here's a succinct and generic alternative to the slightly weighty answers above.
def onDay(date, day):
"""
Returns the date of the next given weekday after
the given date. For example, the date of next Monday.
NB: if it IS the day we're looking for, this returns 0.
consider then doing onDay(foo, day + 1).
"""
days = (day - date.weekday() + 7) % 7
return date + datetime.timedelta(days=days)
Try
>>> dt = datetime(2011, 7, 2)
>>> dt + timedelta(days=(7 - dt.weekday()))
datetime.datetime(2011, 7, 4, 0, 0)
using, that the next monday is 7 days after the a monday, 6 days after a tuesday, and so on, and also using, that Python's datetime type reports monday as 0, ..., sunday as 6.
This is example of calculations within ring mod 7.
import datetime
def next_day(given_date, weekday):
day_shift = (weekday - given_date.weekday()) % 7
return given_date + datetime.timedelta(days=day_shift)
now = datetime.date(2018, 4, 15) # sunday
names = ['monday', 'tuesday', 'wednesday', 'thursday', 'friday',
'saturday', 'sunday']
for weekday in range(7):
print(names[weekday], next_day(now, weekday))
will print:
monday 2018-04-16
tuesday 2018-04-17
wednesday 2018-04-18
thursday 2018-04-19
friday 2018-04-20
saturday 2018-04-21
sunday 2018-04-15
As you see it's correctly give you next monday, tuesday, wednesday, thursday friday and saturday. And it also understood that 2018-04-15 is a sunday and returned current sunday instead of next one.
I'm sure you'll find this answer extremely helpful after 7 years ;-)
Another alternative uses rrule
from dateutil.rrule import rrule, WEEKLY, MO
from datetime import date
next_monday = rrule(freq=WEEKLY, dtstart=date.today(), byweekday=MO, count=1)[0]
rrule docs: https://dateutil.readthedocs.io/en/stable/rrule.html
Another simple elegant solution is to use pandas offsets.
I find it very helpful and robust when playing with dates.
If you want the first Sunday just modify the frequency to freq='W-SUN'.
If you want a couple of next Sundays, change the offsets.Day(days).
Using pandas offsets allow you to ignore holidays, work only with Business Days and more.
You can also apply this method easily on a whole DataFrame using the apply method.
import pandas as pd
import datetime
# 1. Getting the closest monday from a given date
date = datetime.date(2011, 7, 2)
closest_monday = pd.date_range(start=date, end=date + pd.offsets.Day(6), freq="W-MON")[
0
]
# 2. Adding a 'ClosestMonday' column with the closest monday for each row in
# a pandas df using apply. Requires you to have a 'Date' column in your df
def get_closest_monday(row):
return pd.date_range(
start=row.Date, end=row.Date + pd.offsets.Day(6), freq="W-MON"
)[0]
df = pd.DataFrame([datetime.date(2011, 7, 2)], columns=["Date"])
df["ClosestMonday"] = df.apply(lambda row: get_closest_monday(row), axis=1)
print(df)
You can start adding one day to date object and stop when it's monday.
>>> d = datetime.date(2011, 7, 2)
>>> while d.weekday() != 0: #0 for monday
... d += datetime.timedelta(days=1)
...
>>> d
datetime.date(2011, 7, 4)
import datetime
d = datetime.date(2011, 7, 2)
while d.weekday() != 0:
d += datetime.timedelta(1)
dateutil has a special feature for this kind of operation and it's the most elegant way I have ever seen yet.
from datetime import datetime
from dateutil.relativedelta import relativedelta, MO
first_monday_date = (datetime(2011,7,2) + relativedelta(weekday=MO(0))).date()
if you want datetime just
first_monday_date = datetime(2011,7,2) + relativedelta(weekday=MO(0))
weekday = 0 ## Monday
dt = datetime.datetime.now().replace(hour=0, minute=0, second=0) ## or any specific date
days_remaining = (weekday - dt.weekday() - 1) % 7 + 1
next_dt = dt + datetime.timedelta(days_remaining)
Generally to find any date from day of week from today:
def getDateFromDayOfWeek(dayOfWeek):
week_days = ["monday", "tuesday", "wednesday",
"thursday", "friday", "saturday", "sunday"]
today = datetime.datetime.today().weekday()
requiredDay = week_days.index(dayOfWeek)
if today>requiredDay:
noOfDays=7-(today-requiredDay)
print("noDays",noOfDays)
else:
noOfDays = requiredDay-today
print("noDays",noOfDays)
requiredDate = datetime.datetime.today()+datetime.timedelta(days=noOfDays)
return requiredDate
print(getDateFromDayOfWeek('sunday').strftime("%d/%m/%y"))
Gives output in format of Day/Month/Year
This will give the first next Monday after given date:
import datetime
def get_next_monday(year, month, day):
date0 = datetime.date(year, month, day)
next_monday = date0 + datetime.timedelta(7 - date0.weekday() or 7)
return next_monday
print get_next_monday(2011, 7, 2)
print get_next_monday(2015, 8, 31)
print get_next_monday(2015, 9, 1)
2011-07-04
2015-09-07
2015-09-07
via list comprehension?
from datetime import *
[datetime.today()+timedelta(days=x) for x in range(0,7) if (datetime.today()+timedelta(days=x)).weekday() % 7 == 0]
(0 at the end is for next monday, returns current date when run on monday)