How do I get the day of week given a date?
PythonDateDatetimeTimeWeekdayPython Problem Overview
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)
Python Solutions
Solution 1 - Python
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 - Python
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 - Python
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 - Python
Use date.weekday() when Monday is 0 and Sunday is 6
or
date.isoweekday() when Monday is 1 and Sunday is 7
Solution 5 - Python
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"))
Solution 6 - Python
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')
Solution 7 - Python
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
Solution 8 - Python
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
Solution 9 - Python
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]
Solution 10 - Python
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)
Solution 11 - Python
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.
Solution 12 - Python
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)
Solution 13 - Python
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
Solution 14 - Python
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)]
Solution 15 - Python
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
Solution 16 - Python
import datetime
int(datetime.datetime.today().strftime('%w'))+1
this should give you your real day number - 1 = sunday, 2 = monday, etc...
Solution 17 - Python
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
Solution 18 - Python
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"))
Solution 19 - Python
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
Solution 20 - Python
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
Solution 21 - Python
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.
Solution 22 - Python
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
Solution 23 - Python
Using Canlendar Module
import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
Solution 24 - Python
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)
Solution 25 - Python
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
Solution 26 - Python
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
Solution 27 - Python
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.")
Solution 28 - Python
use this code:
import pandas as pd
from datetime import datetime
print(pd.DatetimeIndex(df['give_date']).day)