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.
Tomerikoo
16.6k15 gold badges37 silver badges54 bronze badges
answered Mar 23, 2012 at 22:26
Simeon VisserSimeon Visser
115k18 gold badges175 silver badges178 bronze badges
6
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'
Uri
24k9 gold badges40 silver badges68 bronze badges
answered Apr 8, 2015 at 15:43
seddonymseddonym
15.5k6 gold badges63 silver badges71 bronze badges
3
Use date.weekday[]
when Monday is 0 and Sunday is 6
or
date.isoweekday[]
when Monday is 1 and Sunday is 7
answered Mar 23, 2012 at 22:24
orlporlp
109k33 gold badges203 silver badges305 bronze badges
1
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"]]
answered Mar 23, 2012 at 22:36
Ashwini ChaudharyAshwini Chaudhary
236k55 gold badges443 silver badges495 bronze badges
0
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']
answered Jun 15, 2013 at 5:18
3
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, 2019 at 9:48
Vlad BezdenVlad Bezden
75.6k23 gold badges235 silver badges174 bronze badges
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
answered Jul 21, 2020 at 16:46
F.E.AF.E.A
1511 silver badge4 bronze badges
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]
answered Oct 29, 2015 at 14:01
RodrigoRodrigo
1451 silver badge3 bronze badges
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.
answered Mar 12, 2017 at 0:01
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]
answered Feb 20, 2019 at 5:49
Shiv948Shiv948
4014 silver badges10 bronze badges
1
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]
answered Apr 22, 2014 at 22:38
mathwizurdmathwizurd
1,28713 silver badges14 bronze badges
1
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
//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
answered May 11, 2015 at 17:59
Barry AndersenBarry Andersen
4472 gold badges6 silver badges8 bronze badges
1
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
answered Jan 21, 2021 at 10:23
seunggabiseunggabi
1,50610 silver badges11 bronze badges
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]]
answered Dec 22, 2017 at 3:51
AnaCronIsmAnaCronIsm
511 silver badge2 bronze badges
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, 2019 at 9:10
1
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
answered May 9, 2014 at 1:47
ox.ox.
3,3291 gold badge19 silver badges18 bronze badges
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
answered Jan 23, 2019 at 7:39
1
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 3, 2019 at 18:04
nsky80nsky80
1055 silver badges8 bronze badges
2
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.
answered Jul 6, 2020 at 8:51
Gonçalo PeresGonçalo Peres
7,0073 gold badges38 silver badges66 bronze badges
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"]]
Neuron
4,5574 gold badges32 silver badges53 bronze badges
answered Jul 10, 2017 at 2:05
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, 2020 at 0:36
RomanRoman
16.3k11 gold badges78 silver badges83 bronze badges
If u are Chinese user, u can use this package: //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
answered Apr 8 at 2:54
Fan YangFan Yang
4906 silver badges8 bronze badges
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
answered Jun 16 at 15:24
LazyerLazyer
7241 gold badge5 silver badges15 bronze badges
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, 2018 at 18:34
Ravi BhushanRavi Bhushan
8222 gold badges10 silver badges18 bronze badges
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