Find the date for the first Monday after a given date
Question:
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?
Answers:
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)
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
.
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)
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)
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)
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)
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)
import datetime
d = datetime.date(2011, 7, 2)
while d.weekday() != 0:
d += datetime.timedelta(1)
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
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
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))
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)
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
.
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)
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)
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)
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)
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)
import datetime
d = datetime.date(2011, 7, 2)
while d.weekday() != 0:
d += datetime.timedelta(1)
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
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
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))