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)

score 1363 · Accepted Answer · edited Sep 01 '20 at 13:04

1363

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.

edited Sep 01 '20 at 13:04

Tomerikoo

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answered Mar 23 '12 at 22:26

Simeon Visser

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502

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.. – radtek Jun 09 '14 at 15:39
24

If you'd like Sunday to be day 0: `int(datetime.datetime.today().strftime('%w'))` – mrooney Dec 13 '16 at 08:22
47

To start from 1, we can use isoweekday in place of weekday; 1 = Monday – Aman Kumar Jul 07 '17 at 06:25
29

What about `.strftime('%A')` [link](https://docs.python.org/3/library/datetime.html#strftime-and-strptime-format-codes) to get a weekday name. – rborodinov Dec 17 '20 at 15:25
6

Who in their right mind would think the week starts at Sunday. Of course it starts at Saturday xD – Mr. Developerdude Jul 23 '21 at 05:02
10

instead of .weekday() try isoweekday() – xjlin0 Aug 18 '21 at 23:34
date.isoweekday() does start on sunday – alex Mar 31 '23 at 14:59

score 389 · Answer 2 · edited Nov 23 '18 at 15:03

389

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'

edited Nov 23 '18 at 15:03

Uri

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answered Apr 08 '15 at 15:43

seddonym

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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. – Johnny Utahh May 13 '15 at 19:31
77

I find it much more effective to just do `my_date.strftime('%A')` – Nathan Tew Jan 23 '19 at 01:45
@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 Oct 24 '20 at 21:32

score 233 · Answer 3 · edited Apr 12 '22 at 21:49

233

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

edited Apr 12 '22 at 21:49

answered Jul 25 '18 at 10:07

Kannan Ramaswamy

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score 114 · Answer 4 · edited May 18 '21 at 16:43

114

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

or

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

edited May 18 '21 at 16:43

Divyanshu Shekhar

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answered Mar 23 '12 at 22:24

orlp

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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()`](https://docs.python.org/3/library/datetime.html#datetime.datetime.weekday) and [`isoweekday()`](https://docs.python.org/3/library/datetime.html#datetime.datetime.isoweekday) methods. – Jens Aug 16 '17 at 05:16

score 55 · Answer 5 · edited Jun 06 '20 at 11:59

55

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"))

edited Jun 06 '20 at 11:59

Chris Tang

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answered Mar 23 '12 at 22:36

Ashwini Chaudhary

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Arnaldo P. Figueira Figueira · Answer 6 · 2014-02-25T04:02:01.233

36

A solution whithout imports for dates after 1700/1/1

def weekDay(year, month, day):
    offset = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334]
    week   = ['Sunday', 
              'Monday', 
              'Tuesday', 
              'Wednesday', 
              'Thursday',  
              'Friday', 
              'Saturday']
    afterFeb = 1
    if month > 2: afterFeb = 0
    aux = year - 1700 - afterFeb
    # dayOfWeek for 1700/1/1 = 5, Friday
    dayOfWeek  = 5
    # partial sum of days betweem current date and 1700/1/1
    dayOfWeek += (aux + afterFeb) * 365                  
    # leap year correction    
    dayOfWeek += aux / 4 - aux / 100 + (aux + 100) / 400     
    # sum monthly and day offsets
    dayOfWeek += offset[month - 1] + (day - 1)               
    dayOfWeek %= 7
    return dayOfWeek, week[dayOfWeek]

print weekDay(2013, 6, 15) == (6, 'Saturday')
print weekDay(1969, 7, 20) == (0, 'Sunday')
print weekDay(1945, 4, 30) == (1, 'Monday')
print weekDay(1900, 1, 1)  == (1, 'Monday')
print weekDay(1789, 7, 14) == (2, 'Tuesday')

edited Feb 25 '14 at 04:02

answered Jun 15 '13 at 05:18

Arnaldo P. Figueira Figueira

1,100
14
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why do we need to do aux+100 / 400 instead of aux/400 can you please explain – himanshu219 Apr 07 '18 at 05:25
3

python3: just change all '/' with '//' in the function above and it will work like a charm. – chabir Apr 22 '20 at 15:53
1

@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. – not2savvy Dec 21 '20 at 20:54
What is `offset` here? – Shyam3089 Jan 22 '23 at 11:17

score 18 · Answer 7 · answered Apr 12 '19 at 09:48

18

If you have dates as a string, it might be easier to do it using pandas' Timestamp

import pandas as pd
df = pd.Timestamp("2019-04-12")
print(df.dayofweek, df.weekday_name)

Output:

4 Friday

answered Apr 12 '19 at 09:48

Vlad Bezden

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score 15 · Answer 8 · answered Jul 21 '20 at 16:46

Here's a simple code snippet to solve this problem

import datetime

intDay = datetime.date(year=2000, month=12, day=1).weekday()
days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
print(days[intDay])

The output should be:

Friday

Rodrigo · Answer 9 · 2015-10-29T14:06:34.490

12

This is a solution if the date is a datetime object.

import datetime
def dow(date):
    days=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
    dayNumber=date.weekday()
    print days[dayNumber]

edited Oct 29 '15 at 14:06

answered Oct 29 '15 at 14:01

Rodrigo

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score 10 · Answer 10 · answered Mar 12 '17 at 00:01

datetime library sometimes gives errors with strptime() so I switched to dateutil library. Here's an example of how you can use it :

from dateutil import parser
parser.parse('January 11, 2010').strftime("%a")

The output that you get from this is 'Mon'. If you want the output as 'Monday', use the following :

parser.parse('January 11, 2010').strftime("%A")

This worked for me pretty quickly. I was having problems while using the datetime library because I wanted to store the weekday name instead of weekday number and the format from using the datetime library was causing problems. If you're not having problems with this, great! If you are, you cand efinitely go for this as it has a simpler syntax as well. Hope this helps.

score 10 · Answer 11 · edited Feb 20 '19 at 06:18

10

Say you have timeStamp: String variable, YYYY-MM-DD HH:MM:SS

step 1: convert it to dateTime function with blow code...

df['timeStamp'] = pd.to_datetime(df['timeStamp'])

Step 2 : Now you can extract all the required feature as below which will create new Column for each of the fild- hour,month,day of week,year, date

df['Hour'] = df['timeStamp'].apply(lambda time: time.hour)
df['Month'] = df['timeStamp'].apply(lambda time: time.month)
df['Day of Week'] = df['timeStamp'].apply(lambda time: time.dayofweek)
df['Year'] = df['timeStamp'].apply(lambda t: t.year)
df['Date'] = df['timeStamp'].apply(lambda t: t.day)

edited Feb 20 '19 at 06:18

Siong Thye Goh

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answered Feb 20 '19 at 05:49

Shiv948

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Will this edit the my elasticsearch data/columns? For me this would enable the use of bucket aggregations on weekdays for example. – Cornelis Oct 24 '20 at 21:37

seunggabi · Answer 12 · 2022-01-16T10:27:24.310

This don't need to day of week comments.
I recommend this code~!

import datetime


DAY_OF_WEEK = {
    "MONDAY": 0,
    "TUESDAY": 1,
    "WEDNESDAY": 2,
    "THURSDAY": 3,
    "FRIDAY": 4,
    "SATURDAY": 5,
    "SUNDAY": 6
}

def string_to_date(dt, format='%Y%m%d'):
    return datetime.datetime.strptime(dt, format)

def date_to_string(date, format='%Y%m%d'):
    return datetime.datetime.strftime(date, format)

def day_of_week(dt):
    return string_to_date(dt).weekday()


dt = '20210101'
if day_of_week(dt) == DAY_OF_WEEK['SUNDAY']:
    None

mathwizurd · Answer 13 · 2014-04-24T00:58:32.610

7

Assuming you are given the day, month, and year, you could do:

import datetime
DayL = ['Mon','Tues','Wednes','Thurs','Fri','Satur','Sun']
date = DayL[datetime.date(year,month,day).weekday()] + 'day'
#Set day, month, year to your value
#Now, date is set as an actual day, not a number from 0 to 6.

print(date)

edited Apr 24 '14 at 00:58

answered Apr 22 '14 at 22:38

mathwizurd

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1

There is no need of using DayL array as you can directly get day name by using `strftime("%A")` instead of `weekday()` – Lekr0 Jan 02 '19 at 16:30

score 6 · Answer 14 · answered May 11 '15 at 17:59

If you have reason to avoid the use of the datetime module, then this function will work.

Note: The change from the Julian to the Gregorian calendar is assumed to have occurred in 1582. If this is not true for your calendar of interest then change the line if year > 1582: accordingly.

def dow(year,month,day):
    """ day of week, Sunday = 1, Saturday = 7
     http://en.wikipedia.org/wiki/Zeller%27s_congruence """
    m, q = month, day
    if m == 1:
        m = 13
        year -= 1
    elif m == 2:
        m = 14
        year -= 1
    K = year % 100    
    J = year // 100
    f = (q + int(13*(m + 1)/5.0) + K + int(K/4.0))
    fg = f + int(J/4.0) - 2 * J
    fj = f + 5 - J
    if year > 1582:
        h = fg % 7
    else:
        h = fj % 7
    if h == 0:
        h = 7
    return h

'A' for effort! You might move statements, like those assigning to `fg` and `fj`, inside the conditional to prevent unnecessary computations. — Tom Russell, Dec 12 '18 at 19:38

score 5 · Answer 15 · answered Dec 22 '17 at 03:51

If you're not solely reliant on the datetime module, calendar might be a better alternative. This, for example, will provide you with the day codes:

calendar.weekday(2017,12,22);

And this will give you the day itself:

days = ["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
days[calendar.weekday(2017,12,22)]

Or in the style of python, as a one liner:

["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"][calendar.weekday(2017,12,22)]

score 4 · Answer 16 · answered May 20 '19 at 09:10

4

import datetime
int(datetime.datetime.today().strftime('%w'))+1

this should give you your real day number - 1 = sunday, 2 = monday, etc...

answered May 20 '19 at 09:10

neoghost

41
5

Why would you use `+1`? It is common sence that the weeknumbering in python starts at sundat as 0 and monday as 1. – Nebulosar Dec 10 '19 at 08:45

score 3 · Answer 17 · answered May 09 '14 at 01:47

To get Sunday as 1 through Saturday as 7, this is the simplest solution to your question:

datetime.date.today().toordinal()%7 + 1

All of them:

import datetime

today = datetime.date.today()
sunday = today - datetime.timedelta(today.weekday()+1)

for i in range(7):
    tmp_date = sunday + datetime.timedelta(i)
    print tmp_date.toordinal()%7 + 1, '==', tmp_date.strftime('%A')

Output:

1 == Sunday
2 == Monday
3 == Tuesday
4 == Wednesday
5 == Thursday
6 == Friday
7 == Saturday

The question asks for Sunday == 1, Monday == 2, and Friday == 6. — ox., Jul 26 '16 at 02:31

score 3 · Answer 18 · edited Aug 16 '21 at 08:53

3

Here is how to convert a list of little endian string dates to datetime:

import datetime, time
ls = ['31/1/2007', '14/2/2017']
for d in ls:    
    dt = datetime.datetime.strptime(d, "%d/%m/%Y")
    print(dt)
    print(dt.strftime("%A"))

edited Aug 16 '21 at 08:53

Neuron

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answered Jul 10 '17 at 02:05

Brij Bhushan Nanda

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score 3 · Answer 19 · answered Jan 23 '19 at 07:39

We can take help of Pandas:

import pandas as pd

As mentioned above in the problem We have:

datetime(2017, 10, 20)

If execute this line in the jupyter notebook we have an output like this:

datetime.datetime(2017, 10, 20, 0, 0)

Using weekday() and weekday_name:

If you want weekdays in integer number format then use:

pd.to_datetime(datetime(2017, 10, 20)).weekday()

The output will be:

And if you want it as name of the day like Sunday, Monday, Friday, etc you can use:

pd.to_datetime(datetime(2017, 10, 20)).weekday_name

The output will be:

'Friday'

If having a dates column in Pandas dataframe then:

Now suppose if you have a pandas dataframe having a date column like this: pdExampleDataFrame['Dates'].head(5)

0   2010-04-01
1   2010-04-02
2   2010-04-03
3   2010-04-04
4   2010-04-05
Name: Dates, dtype: datetime64[ns]

Now If we want to know the name of the weekday like Monday, Tuesday, ..etc we can use .weekday_name as follows:

pdExampleDataFrame.head(5)['Dates'].dt.weekday_name

the output will be:

0    Thursday
1      Friday
2    Saturday
3      Sunday
4      Monday
Name: Dates, dtype: object

And if we want the integer number of weekday from this Dates column then we can use:

pdExampleDataFrame.head(5)['Dates'].apply(lambda x: x.weekday())

The output will look like this:

0    3
1    4
2    5
3    6
4    0
Name: Dates, dtype: int64

In Pandas 1.0 and after, weekday_name() has been changed to day_name() — Liu Yu, Sep 14 '20 at 06:27

score 3 · Answer 20 · answered Feb 03 '19 at 18:04

3

import datetime
import calendar

day, month, year = map(int, input().split())
my_date = datetime.date(year, month, day)
print(calendar.day_name[my_date.weekday()])

Output Sample

08 05 2015
Friday

answered Feb 03 '19 at 18:04

nsky80

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but there was Wednesday on 05/08/2015 – DikShU Nov 12 '20 at 03:17
Input given is DD MM YYYY format, perhaps you're interpreting it as MMDDYYYY format. – nsky80 Jan 18 '22 at 18:04

Gonçalo Peres · Answer 21 · 2020-07-06T09:22:32.657

If you want to generate a column with a range of dates (Date) and generate a column that goes to the first one and assigns the Week Day (Week Day), do the following (I will used the dates ranging from 2008-01-01 to 2020-02-01):

import pandas as pd
dr = pd.date_range(start='2008-01-01', end='2020-02-1')
df = pd.DataFrame()
df['Date'] = dr
df['Week Day'] = pd.to_datetime(dr).weekday

The output is the following:

The Week Day varies from 0 to 6, where 0 corresponds to Monday and 6 to Sunday.

Lazyer · Answer 22 · 2022-07-13T00:37:17.640

3

Here's a fresh way. Sunday is 0.

from datetime import datetime
today = datetime(year=2022, month=6, day=17)
print(today.toordinal()%7)  # 5
yesterday = datetime(year=1, month=1, day=1)
print(today.toordinal()%7)  # 1

edited Jul 13 '22 at 00:37

answered Jun 16 '22 at 15:24

Lazyer

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score 2 · Answer 23 · answered Aug 11 '20 at 00:36

2

A simple, straightforward and still not mentioned option:

import datetime
...
givenDateObj = datetime.date(2017, 10, 20)
weekday      = givenDateObj.isocalendar()[2] # 5
weeknumber   = givenDateObj.isocalendar()[1] # 42

answered Aug 11 '20 at 00:36

Roman

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score 2 · Answer 24 · answered Apr 08 '22 at 02:54

If u are Chinese user, u can use this package: https://github.com/LKI/chinese-calendar

import datetime

# 判断 2018年4月30号 是不是节假日
from chinese_calendar import is_holiday, is_workday
april_last = datetime.date(2018, 4, 30)
assert is_workday(april_last) is False
assert is_holiday(april_last) is True

# 或者在判断的同时，获取节日名
import chinese_calendar as calendar  # 也可以这样 import
on_holiday, holiday_name = calendar.get_holiday_detail(april_last)
assert on_holiday is True
assert holiday_name == calendar.Holiday.labour_day.value

# 还能判断法定节假日是不是调休
import chinese_calendar
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 1)) is False
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 2)) is True

score 1 · Answer 25 · answered Feb 21 '18 at 18:34

1

Using Canlendar Module

import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])

answered Feb 21 '18 at 18:34

Ravi Bhushan

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score 1 · Answer 26 · answered Mar 04 '18 at 08:18

Here is my python3 implementation.

months = {'jan' : 1, 'feb' : 4, 'mar' : 4, 'apr':0, 'may':2, 'jun':5, 'jul':6, 'aug':3, 'sep':6, 'oct':1, 'nov':4, 'dec':6}
dates = {'Sunday':1, 'Monday':2, 'Tuesday':3, 'Wednesday':4, 'Thursday':5, 'Friday':6, 'Saterday':0}
ranges = {'1800-1899':2, '1900-1999':0, '2000-2099':6, '2100-2199':4, '2200-2299':2}

def getValue(val, dic):
    if(len(val)==4):
        for k,v in dic.items():
            x,y=int(k.split('-')[0]),int(k.split('-')[1])
            val = int(val)
            if(val>=x and val<=y):
                return v
    else:
        return dic[val]

def getDate(val):
    return (list(dates.keys())[list(dates.values()).index(val)]) 



def main(myDate):
    dateArray = myDate.split('-')
    # print(dateArray)
    date,month,year = dateArray[2],dateArray[1],dateArray[0]
    # print(date,month,year)

    date = int(date)
    month_v = getValue(month, months)
    year_2 = int(year[2:])
    div = year_2//4
    year_v = getValue(year, ranges)
    sumAll = date+month_v+year_2+div+year_v
    val = (sumAll)%7
    str_date = getDate(val)

    print('{} is a {}.'.format(myDate, str_date))

if __name__ == "__main__":
    testDate = '2018-mar-4'
    main(testDate)

score 1 · Answer 27 · edited Nov 21 '21 at 15:31

import numpy as np

def date(df):
    df['weekday'] = df['date'].dt.day_name()

    conditions = [(df['weekday'] == 'Sunday'),
              (df['weekday'] == 'Monday'),
              (df['weekday'] == 'Tuesday'),
              (df['weekday'] == 'Wednesday'),
              (df['weekday'] == 'Thursday'),
              (df['weekday'] == 'Friday'),
              (df['weekday'] == 'Saturday')]

    choices = [0, 1, 2, 3, 4, 5, 6]

    df['week'] = np.select(conditions, choices)

    return df

score 0 · Answer 28 · answered Jun 10 '20 at 05:30

Below is the code to enter date in the format of DD-MM-YYYY you can change the input format by changing the order of '%d-%m-%Y' and also by changing the delimiter.

import datetime
try:
    date = input()
    date_time_obj = datetime.datetime.strptime(date, '%d-%m-%Y')
    print(date_time_obj.strftime('%A'))
except ValueError:
    print("Invalid date.")

score 0 · Answer 29 · answered Aug 02 '23 at 05:22

It will print the present week's Monday and Friday, and by adjusting days value, we can also print the remaining days of the week

from datetime import timedelta
import datetime as dt
today = dt.date.today()
monday = today - timedelta(days=today.weekday())
friday = monday + timedelta(days=4)

score -1 · Answer 30 · edited Sep 09 '19 at 18:35

-1

use this code:

import pandas as pd
from datetime import datetime
print(pd.DatetimeIndex(df['give_date']).day)

edited Sep 09 '19 at 18:35

Nikaido

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answered Sep 09 '19 at 16:21

Jeevan kumar

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score -1 · Answer 31 · answered Nov 22 '22 at 11:34

In MATLAB, Gauss' method

day_name={'Sun','Mon','Tue','Wed','Thu','Fri','Sat'}
month_offset=[0 3 3 6 1 4 6 2 5 0 3 5];  % common year

% input date
y1=2022
m1=11
d1=22

% is y1 leap
if mod(y1,4)==0 && mod(y1,100)==0 && mod(y1,400)==0
    month_offset=[0 3 4 0 2 5 0 3 6 1 4 6];  % offset for leap year
end

% Gregorian calendar
weekday_gregor=rem( d1+month_offset(m1) + 5*rem(y1-1,4) +  4*rem(y1-1,100) + 6*rem(y1-1,400),7)

day_name{weekday_gregor+1}

0: Sunday 1: Monday .. 6: Saturday

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

Example

31 Answers31

Output Sample

Linked

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