Find Monday's date with Python
Question:
How do I find the previous Monday’s date, based off of the current date using Python? I thought maybe I could use: datetime.weekday() to do it, but I am getting stuck.
I basically want to find today’s date and Mondays date to construct a date range query in django using: created__range=(start_date, end_date).
Answers:
>>> import datetime
>>> today = datetime.date.today()
>>> today + datetime.timedelta(days=-today.weekday(), weeks=1)
datetime.date(2009, 10, 26)
Some words of explanation:
Take todays date. Subtract the number of days which already passed this week (this gets you ‘last’ monday). Add one week.
Edit: The above is for ‘next monday’, but since you were looking for ‘last monday’ you could use
today - datetime.timedelta(days=today.weekday())
Using timedeltas and datetime module:
import datetime
datetime.date.today()+datetime.timedelta(days=-datetime.date.today().weekday())
d = datetime.datetime.today().weekday()
gives you today’s day of the week, counting 0 (monday) to 6 (sunday)
datetime.datetime.today() + datetime.timedelta(days=(7-d)%7)
(7-d)%7 gives you days until Monday, or leaves you where you are if today is Monday
Note: The OP says in the comments, “I was looking for the past Monday“. I take this to mean we are looking for the last Monday that occurred strictly before today.
The calculation is a little difficult to get right using only the datetime module (especially given the above interpretation of “past Monday” and if you wish to avoid clunky if-statements). For example, if today is a Monday such as 2013-12-23,
today - DT.timedelta(days=today.weekday())
returns 2013-12-23, which is the same day as today (not the past Monday).
The advantage of using the dateutil module is that you don’t have to do tricky mental calculations nor force the reader to do the same to get the right date. dateutil does it all for you:
import dateutil.relativedelta as rdelta
import datetime as DT
today = DT.date(2013, 12, 23) # Monday
past_monday = today + rdelta.relativedelta(days=-1, weekday=rdelta.MO(-1))
print(past_monday)
# 2013-12-16
next_monday = today + rdelta.relativedelta(days=1, weekday=rdelta.MO(+1))
print(next_monday)
# 2013-12-30
Note that days=-1 is needed to guarantee that past_monday is a different day than today.
ChristopheD’s post is close to what you want. I don’t have enough rep to make a comment 🙁
Instead of (which actually gives you the next upcoming monday):
>>> today + datetime.timedelta(days=-today.weekday(), weeks=1)
datetime.date(2009, 10, 26)
I would say:
>>> last_monday = today - datetime.timedelta(days=today.weekday())
If you want the previous week, add the ‘weeks=1’ parameter.
This makes the code more readable since you are subtracting a timedelta. This clears up any confusion caused by adding a timedelta that has negative and positive offsets.
I think the easiest way is using python-dateutil like this:
from datetime import date
from dateutil.relativedelta import relativedelta, MO
today = date.today()
last_monday = today + relativedelta(weekday=MO(-1))
print last_monday
You can use Natty. I tried parsedatetime and dateparser. Comparing these three, I think Natty is the best one.
To get your result, use like this:
>>> from natty import DateParser
>>> dp = DateParser('last monday')
>>> dp.result()
[datetime.datetime(2016, 8, 1, 17, 35, 5, tzinfo=tzlocal())] #Today is 9th of August 2016 5.35 PM
Github Link : https://github.com/eadmundo/python-natty
Try it, It can do more!
For future googlers who show up on this page looking for a way to get “the most recent Sunday”, rather than “the most recent Monday”, you need to do some additional math because datetime.weekday() treats Monday as the first day of the week:
today = datetime.date.today()
weekday = today.weekday() + 1
start_day = today - datetime.timedelta(days=weekday % 7)
If today is Tuesday, this sets start_day to last Sunday. If today is Sunday, this sets start_day to today. Take away the % 7 if you want “last Sunday” to be a week ago if it’s currently Sunday.
How do I find the previous Monday’s date, based off of the current date using Python? I thought maybe I could use: datetime.weekday() to do it, but I am getting stuck.
I basically want to find today’s date and Mondays date to construct a date range query in django using: created__range=(start_date, end_date).
>>> import datetime
>>> today = datetime.date.today()
>>> today + datetime.timedelta(days=-today.weekday(), weeks=1)
datetime.date(2009, 10, 26)
Some words of explanation:
Take todays date. Subtract the number of days which already passed this week (this gets you ‘last’ monday). Add one week.
Edit: The above is for ‘next monday’, but since you were looking for ‘last monday’ you could use
today - datetime.timedelta(days=today.weekday())
Using timedeltas and datetime module:
import datetime
datetime.date.today()+datetime.timedelta(days=-datetime.date.today().weekday())
d = datetime.datetime.today().weekday()
gives you today’s day of the week, counting 0 (monday) to 6 (sunday)
datetime.datetime.today() + datetime.timedelta(days=(7-d)%7)
(7-d)%7 gives you days until Monday, or leaves you where you are if today is Monday
Note: The OP says in the comments, “I was looking for the past Monday“. I take this to mean we are looking for the last Monday that occurred strictly before today.
The calculation is a little difficult to get right using only the datetime module (especially given the above interpretation of “past Monday” and if you wish to avoid clunky if-statements). For example, if today is a Monday such as 2013-12-23,
today - DT.timedelta(days=today.weekday())
returns 2013-12-23, which is the same day as today (not the past Monday).
The advantage of using the dateutil module is that you don’t have to do tricky mental calculations nor force the reader to do the same to get the right date. dateutil does it all for you:
import dateutil.relativedelta as rdelta
import datetime as DT
today = DT.date(2013, 12, 23) # Monday
past_monday = today + rdelta.relativedelta(days=-1, weekday=rdelta.MO(-1))
print(past_monday)
# 2013-12-16
next_monday = today + rdelta.relativedelta(days=1, weekday=rdelta.MO(+1))
print(next_monday)
# 2013-12-30
Note that days=-1 is needed to guarantee that past_monday is a different day than today.
ChristopheD’s post is close to what you want. I don’t have enough rep to make a comment 🙁
Instead of (which actually gives you the next upcoming monday):
>>> today + datetime.timedelta(days=-today.weekday(), weeks=1) datetime.date(2009, 10, 26)
I would say:
>>> last_monday = today - datetime.timedelta(days=today.weekday())
If you want the previous week, add the ‘weeks=1’ parameter.
This makes the code more readable since you are subtracting a timedelta. This clears up any confusion caused by adding a timedelta that has negative and positive offsets.
I think the easiest way is using python-dateutil like this:
from datetime import date
from dateutil.relativedelta import relativedelta, MO
today = date.today()
last_monday = today + relativedelta(weekday=MO(-1))
print last_monday
You can use Natty. I tried parsedatetime and dateparser. Comparing these three, I think Natty is the best one.
To get your result, use like this:
>>> from natty import DateParser
>>> dp = DateParser('last monday')
>>> dp.result()
[datetime.datetime(2016, 8, 1, 17, 35, 5, tzinfo=tzlocal())] #Today is 9th of August 2016 5.35 PM
Github Link : https://github.com/eadmundo/python-natty
Try it, It can do more!
For future googlers who show up on this page looking for a way to get “the most recent Sunday”, rather than “the most recent Monday”, you need to do some additional math because datetime.weekday() treats Monday as the first day of the week:
today = datetime.date.today()
weekday = today.weekday() + 1
start_day = today - datetime.timedelta(days=weekday % 7)
If today is Tuesday, this sets start_day to last Sunday. If today is Sunday, this sets start_day to today. Take away the % 7 if you want “last Sunday” to be a week ago if it’s currently Sunday.