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.
>>> today = datetime.datetime(2017, 10, 20) >>> today.get_weekday() # what I look for The output is maybe 6 (since it's Friday)
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.
int(datetime.datetime.today().strftime('%w'))
.strftime('%A') link to get a weekday name.
If you'd like to have the date in the language of your current locale:
from datetime import date import calendar my_date = date.today() calendar.day_name[my_date.weekday()] #'Wednesday' Documentation: https://docs.python.org/3/library/calendar.html#calendar.day_name
my_date.strftime('%A')
If you'd like to have the date in the language of your current locale:
>>> from datetime import datetime >>> datetime.now().strftime('%A') 'Wednesday' >>> datetime.now().strftime('%a') 'Wed' Read more: https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior
Use date.weekday() when Monday is 0 and Sunday is 6
or
date.isoweekday() when Monday is 1 and Sunday is 7
datetime object (not a date object) I’d like to mention that the datetime class sports the same weekday() and isoweekday() methods.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")) 
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') 
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.
offset here?
offset is a table of the number of days between the first of the month and the first of the year, assuming it's not a leap year.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 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 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] 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.
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) 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 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) 
strftime("%A") instead of weekday()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 
fg and fj, inside the conditional to prevent unnecessary computations.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)] 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 
import datetime int(datetime.datetime.today().strftime('%w'))+1 this should give you your real day number - 1 = sunday, 2 = monday, etc...
+1? It is common sence that the weeknumbering in python starts at sundat as 0 and monday as 1.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:
4 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 
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()]) 08 05 2015 Friday 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.
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")) 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 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 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 Using Canlendar Module
import calendar a=calendar.weekday(year,month,day) days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"] print(days[a]) 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) 
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 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.") use this code:
import pandas as pd from datetime import datetime print(pd.DatetimeIndex(df['give_date']).day) 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) 