How do I get the day of week given a date?

Question

I want to find out the following: given a date (datetime object), what is the corresponding day of the week?

For instance, Sunday is the first day, Monday: second day.. and so on

And then if the input is something like today's date.

Example

>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday()  # what I look for

The output is maybe 6 (since it's Friday)

Solution 1

Use weekday():

>>> import datetime
>>> datetime.datetime.today()
datetime.datetime(2012, 3, 23, 23, 24, 55, 173504)
>>> datetime.datetime.today().weekday()
4

From the documentation:

Return the day of the week as an integer, where Monday is 0 and Sunday is 6.

Solution 2

If you'd like to have the date in English:

from datetime import date
import calendar
my_date = date.today()
calendar.day_name[my_date.weekday()]  #'Wednesday'

Solution 3

If you'd like to have the date in English:

from datetime import datetime
datetime.today().strftime('%A')
'Wednesday'

Read more: https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior

Solution 4

Use date.weekday() when Monday is 0 and Sunday is 6

or

date.isoweekday() when Monday is 1 and Sunday is 7

Solution 5

I solved this for a CodeChef question.

import datetime
dt = '21/03/2012'
day, month, year = (int(x) for x in dt.split('/'))    
ans = datetime.date(year, month, day)
print (ans.strftime("%A"))

View more solutions

Comments

[radtek]

One important thing to note is that in JavaScript 0 = Sunday, Python starts with 0 = Monday. Something that I ran into, front-end vs back-end..

[Johnny Utahh]

This seems to be the best answer to generate an English, day-of-week date. I'm guessing it's not upvoted more simply because the answer is ~1 month old, while the question is ~3 years old.

[ox.]

The question asks for Sunday == 1, Monday == 2, and Friday == 6.

[mrooney]

If you'd like Sunday to be day 0: int(datetime.datetime.today().strftime('%w'))

[Aman Kumar]

To start from 1, we can use isoweekday in place of weekday; 1 = Monday

[Jens]

Because the op asked with regards to a datetime object (not a date object) I’d like to mention that the datetime class sports the same weekday() and isoweekday() methods.

[himanshu219]

why do we need to do aux+100 / 400 instead of aux/400 can you please explain

[Tom Russell]

'A' for effort! You might move statements, like those assigning to fg and fj, inside the conditional to prevent unnecessary computations.

[Lekr0]

There is no need of using DayL array as you can directly get day name by using strftime("%A") instead of weekday()

[Nathan Tew]

I find it much more effective to just do my_date.strftime('%A')

[Nebulosar]

Why would you use +1? It is common sence that the weeknumbering in python starts at sundat as 0 and monday as 1.

[chabir]

python3: just change all '/' with '//' in the function above and it will work like a charm.

[Liu Yu]

In Pandas 1.0 and after, weekday_name() has been changed to day_name()

[Cornelis]

@NathanTew 1) is "my_date" a variable? 2) can this be used in a query/aggregation? I want to count the average number of meals ordered on mondays/tuesdays/wednesdays. Each order is on a new line, and just like this question, the weekday has to be retrieved from the data-time column of the csv file that we've just uploaded to elasticsearch.

[Cornelis]

Will this edit the my elasticsearch data/columns? For me this would enable the use of bucket aggregations on weekdays for example.

[DikShU]

but there was Wednesday on 05/08/2015

[rborodinov]

What about .strftime('%A') link to get a weekday name.

[not2savvy]

@himanshu219 Because there's an exception from the leap year rule every 400 years, and aux was derived from year -1700, so we need to add 100 to make it a multiple of 400. For example: 2000 - 1700 = 300, so + 100 gives us 400. Not sure though why aux is used to determine the leap year in that lime and not just year itself.

[Lennart Rolland]

Who in their right mind would think the week starts at Sunday. Of course it starts at Saturday xD

[xjlin0]

instead of .weekday() try isoweekday()

[nsky80]

Input given is DD MM YYYY format, perhaps you're interpreting it as MMDDYYYY format.