Adding days to a date in Python
PythonDateDatetimeTimedeltaPython Problem Overview
I have a date "10/10/11(m-d-y)" and I want to add 5 days to it using a Python script. Please consider a general solution that works on the month ends also.
I am using following code:
import re
from datetime import datetime
StartDate = "10/10/11"
Date = datetime.strptime(StartDate, "%m/%d/%y")
print Date -> is printing '2011-10-10 00:00:00'
Now I want to add 5 days to this date. I used the following code:
EndDate = Date.today()+timedelta(days=10)
Which returned this error:
name 'timedelta' is not defined
Python Solutions
Solution 1 - Python
The previous answers are correct but it's generally a better practice to do:
import datetime
Then you'll have, using datetime.timedelta:
date_1 = datetime.datetime.strptime(start_date, "%m/%d/%y")
end_date = date_1 + datetime.timedelta(days=10)
Solution 2 - Python
Import timedelta and date first.
from datetime import timedelta, date
And date.today() will return today's datetime, may be you want
EndDate = date.today() + timedelta(days=10)
Solution 3 - Python
If you happen to already be using pandas, you can save a little space by not specifying the format:
import pandas as pd
startdate = "10/10/2011"
enddate = pd.to_datetime(startdate) + pd.DateOffset(days=5)
Solution 4 - Python
This might help:
from datetime import date, timedelta
date1 = date(2011, 10, 10)
date2 = date1 + timedelta(days=5)
print (date2)
Solution 5 - Python
If you want add days to date now, you can use this code
from datetime import datetime
from datetime import timedelta
date_now_more_5_days = (datetime.now() + timedelta(days=5) ).strftime('%Y-%m-%d')
Solution 6 - Python
Here is another method to add days on date using dateutil's relativedelta.
from datetime import datetime
from dateutil.relativedelta import relativedelta
print 'Today: ',datetime.now().strftime('%d/%m/%Y %H:%M:%S')
date_after_month = datetime.now()+ relativedelta(days=5)
print 'After 5 Days:', date_after_month.strftime('%d/%m/%Y %H:%M:%S')
Output:
> Today: 25/06/2015 15:56:09 > > After 5 Days: 30/06/2015 15:56:09
Solution 7 - Python
I guess you are missing something like that:
from datetime import timedelta
Solution 8 - Python
Here is a function of getting from now + specified days
import datetime
def get_date(dateFormat="%d-%m-%Y", addDays=0):
timeNow = datetime.datetime.now()
if (addDays!=0):
anotherTime = timeNow + datetime.timedelta(days=addDays)
else:
anotherTime = timeNow
return anotherTime.strftime(dateFormat)
Usage:
addDays = 3 #days
output_format = '%d-%m-%Y'
output = get_date(output_format, addDays)
print output
Solution 9 - Python
In order to have have a less verbose code, and avoid name conflicts between datetime and datetime.datetime, you should rename the classes with CamelCase names.
from datetime import datetime as DateTime, timedelta as TimeDelta
So you can do the following, which I think it is clearer.
date_1 = DateTime.today()
end_date = date_1 + TimeDelta(days=10)
Also, there would be no name conflict if you want to import datetime later on.
Solution 10 - Python
Try this:
Adding 5 days to current date.
from datetime import datetime, timedelta
current_date = datetime.now()
end_date = current_date + timedelta(days=5) # Adding 5 days.
end_date_formatted = end_date.strftime('%Y-%m-%d')
print(end_date_formatted)
subtracting 5 days from current date.
from datetime import datetime, timedelta
current_date = datetime.now()
end_date = current_date + timedelta(days=-5) # Subtracting 5 days.
end_date_formatted = end_date.strftime('%Y-%m-%d')
print(end_date_formatted)
Solution 11 - Python
using timedeltas you can do:
import datetime
today=datetime.date.today()
time=datetime.time()
print("today :",today)
# One day different .
five_day=datetime.timedelta(days=5)
print("one day :",five_day)
#output - 1 day , 00:00:00
# five day extend .
fitfthday=today+five_day
print("fitfthday",fitfthday)
# five day extend .
fitfthday=today+five_day
print("fitfthday",fitfthday)
#output -
today : 2019-05-29
one day : 5 days, 0:00:00
fitfthday 2019-06-03
Solution 12 - Python
Generally you have'got an answer now but maybe my class I created will be also helpfull. For me it solves all my requirements I have ever had in my Pyhon projects.
class GetDate:
def __init__(self, date, format="%Y-%m-%d"):
self.tz = pytz.timezone("Europe/Warsaw")
if isinstance(date, str):
date = datetime.strptime(date, format)
self.date = date.astimezone(self.tz)
def time_delta_days(self, days):
return self.date + timedelta(days=days)
def time_delta_hours(self, hours):
return self.date + timedelta(hours=hours)
def time_delta_seconds(self, seconds):
return self.date + timedelta(seconds=seconds)
def get_minimum_time(self):
return datetime.combine(self.date, time.min).astimezone(self.tz)
def get_maximum_time(self):
return datetime.combine(self.date, time.max).astimezone(self.tz)
def get_month_first_day(self):
return datetime(self.date.year, self.date.month, 1).astimezone(self.tz)
def current(self):
return self.date
def get_month_last_day(self):
lastDay = calendar.monthrange(self.date.year, self.date.month)[1]
date = datetime(self.date.year, self.date.month, lastDay)
return datetime.combine(date, time.max).astimezone(self.tz)
How to use it
self.tz = pytz.timezone("Europe/Warsaw")- here you define Time Zone you want to use in projectGetDate("2019-08-08").current()- this will convert your string date to time aware object with timezone you defined in pt 1. Default string format isformat="%Y-%m-%d"but feel free to change it. (eg.GetDate("2019-08-08 08:45", format="%Y-%m-%d %H:%M").current())GetDate("2019-08-08").get_month_first_day()returns given date (string or object) month first dayGetDate("2019-08-08").get_month_last_day()returns given date month last dayGetDate("2019-08-08").minimum_time()returns given date day startGetDate("2019-08-08").maximum_time()returns given date day endGetDate("2019-08-08").time_delta_days({number_of_days})returns given date + add {number of days} (you can also call:GetDate(timezone.now()).time_delta_days(-1)for yesterday)GetDate("2019-08-08").time_delta_haours({number_of_hours})similar to pt 7 but working on hoursGetDate("2019-08-08").time_delta_seconds({number_of_seconds})similar to pt 7 but working on seconds
Solution 13 - Python
Sometimes we need to use searching by from date & to date. If we use date__range then we need to add 1 day to to_date otherwise queryset will be empty.
Example:
from datetime import timedelta
from_date = parse_date(request.POST['from_date'])
to_date = parse_date(request.POST['to_date']) + timedelta(days=1)
attendance_list = models.DailyAttendance.objects.filter(attdate__range = [from_date, to_date])
Solution 14 - Python
I already see a pandas example, but here a twist to it where you can directly import the Day class
from pandas.tseries.offsets import Day
date1 = datetime(2011, 10, 10)
date2 = date1 + 5 * Day()
Solution 15 - Python
class myDate:
def __init__(self):
self.day = 0
self.month = 0
self.year = 0
## for checking valid days month and year
while (True):
d = int(input("Enter The day :- "))
if (d > 31):
print("Plz 1 To 30 value Enter ........")
else:
self.day = d
break
while (True):
m = int(input("Enter The Month :- "))
if (m > 13):
print("Plz 1 To 12 value Enter ........")
else:
self.month = m
break
while (True):
y = int(input("Enter The Year :- "))
if (y > 9999 and y < 0000):
print("Plz 0000 To 9999 value Enter ........")
else:
self.year = y
break
## method for aday ands cnttract days
def adayDays(self, n):
## aday days to date day
nd = self.day + n
print(nd)
## check days subtract from date
if nd == 0: ## check if days are 7 subtracted from 7 then,........
if(self.year % 4 == 0):
if(self.month == 3):
self.day = 29
self.month -= 1
self.year = self. year
else:
if(self.month == 3):
self.day = 28
self.month -= 1
self.year = self. year
if (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
self.day = 30
self.month -= 1
self.year = self. year
elif (self.month == 2) or (self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 31
self.month -= 1
self.year = self. year
elif(self.month == 1):
self.month = 12
self.year -= 1
## nd == 0 if condition over
## after subtract days to day io goes into negative then
elif nd < 0 :
n = abs(n)## return positive if no is negative
for i in range (n,0,-1): ##
if self.day == 0:
if self.month == 1:
self.day = 30
self.month = 12
self.year -= 1
else:
self.month -= 1
if(self.month == 1) or (self.month == 3)or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month ==12):
self.day = 30
elif(self.month == 4)or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 29
elif(self.month == 2):
if(self.year % 4 == 0):
self.day == 28
else:
self.day == 27
else:
self.day -= 1
## enf of elif negative days
## adaying days to DATE
else:
cnt = 0
while (True):
if self.month == 2: # check leap year
if(self.year % 4 == 0):
if(nd > 29):
cnt = nd - 29
nd = cnt
self.month += 1
else:
self.day = nd
break
## if not leap year then
else:
if(nd > 28):
cnt = nd - 28
nd = cnt
self.month += 1
else:
self.day = nd
break
## checking month other than february month
elif(self.month == 1) or (self.month == 3) or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
if(nd > 31):
cnt = nd - 31
nd = cnt
if(self.month == 12):
self.month = 1
self.year += 1
else:
self.month += 1
else:
self.day = nd
break
elif(self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
if(nd > 30):
cnt = nd - 30
nd = cnt
self.month += 1
else:
self.day = nd
break
## end of month condition
## end of while loop
## end of else condition for adaying days
def formatDate(self,frmt):
if(frmt == 1):
ff=str(self.day)+"-"+str(self.month)+"-"+str(self.year)
elif(frmt == 2):
ff=str(self.month)+"-"+str(self.day)+"-"+str(self.year)
elif(frmt == 3):
ff =str(self.year),"-",str(self.month),"-",str(self.day)
elif(frmt == 0):
print("Thanky You.....................")
else:
print("Enter Correct Choice.......")
print(ff)
dt = myDate()
nday = int(input("Enter No. For Aday or SUBTRACT Days :: "))
dt.adayDays(nday)
print("1 : day-month-year")
print("2 : month-day-year")
print("3 : year-month-day")
print("0 : EXIT")
frmt = int (input("Enter Your Choice :: "))
dt.formatDate(frmt)