Find the date for the first Monday after a given date

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Method 1

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)

Method 2

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)

Method 3

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.

Method 4

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 😉

Method 5

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)

Method 6

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

Method 7

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)

Method 8

import datetime

d = datetime.date(2011, 7, 2)
while d.weekday() != 0:
    d += datetime.timedelta(1)

Method 9

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)

Method 10

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

Method 11

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))

Method 12

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

Method 13

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)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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