Adding days to a date in Python

Question

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

General clue: if you get the error `name 'timedelta' is not defined`, that means that you haven't defined `timedelta` anywhere. Python is usually pretty informative about its error messages. — Katriel, Jul 29 '11 at 10:06
Search didn't work? All of these code examples would have helped: http://stackoverflow.com/search?q=python+timedelta. There appear to be over 200 questions just like this one. — S.Lott, Jul 29 '11 at 10:08
possible duplicate of [add days to a date in Python using loops, ranges, and slicing](http://stackoverflow.com/questions/5206759/add-days-to-a-date-in-python-using-loops-ranges-and-slicing) — S.Lott, Jul 29 '11 at 10:11
You want to add five days, but then you have timedelta(days=10)…I'm confused about where the 10 came from and why it isn't 5 — FeifanZ, Jan 07 '13 at 00:53

score 912 · Accepted Answer · edited Sep 25 '15 at 09:43

912

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)

edited Sep 25 '15 at 09:43

Mathieu Rodic

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answered Jul 29 '11 at 10:03

Botond Béres

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9

datetime.datetime - why twice? – paulmorriss Aug 11 '14 at 09:11
112

importing like "from datetime import datetime, timedelta" would add readibility to the code – Manel Clos Nov 12 '14 at 13:31
12

@paulmorriss: You are calling the `strptime` method on the `datetime` class in the `datetime` module, so you need to specify `datetime.datetime`. – Graeme Perrow Jan 06 '15 at 22:43
If explicit is better, u should avoid to short'n it. But, agree with `from datetime import datetime, timedelta`, it's a bit clear and did remove duplicated words at all. – m3nda Feb 29 '16 at 12:59
2

Shouldn't it be `%m/%d/%Y` and not `%m/%d/%y`? – Kevin Vasko Feb 27 '17 at 22:33
21

Can we all agree that naming a commonly-used class the same name as the module containing it is a dumb idea? What is `datetime`? You can't rely on convention to know, but always have to look at the imports. – Xiong Chiamiov Jun 05 '17 at 17:47
32

Long tail legacy problem there. it "should" be `from datetime import DateTime` since classes are CamelCased, but datetime precedes PEP8. – Aaron McMillin Jun 26 '17 at 20:28
5

You can also do `from datetime import datetime as DateTime, timedelta as TimeDelta`, so you have came case and yet import the module `import datetime` avoiding name conflict. – toto_tico Sep 27 '17 at 10:34
2

In python3, I have to do `from datetime import datetime, timedelta` and `end_date = date_1 + timedelta(days=2)` – LaTechneuse Jan 28 '18 at 21:14

score 214 · Answer 2 · edited Jan 29 '23 at 02:35

214

Import timedelta and date first.

from datetime import timedelta, date

And date.today() will return today's datetime, which you can then add a timedelta to it:

end_date = date.today() + timedelta(days=10)

edited Jan 29 '23 at 02:35

Swift

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answered Jul 29 '11 at 09:20

DrTyrsa

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datetime.date.today() instead of Date.today() – elsadek Aug 06 '14 at 15:19
2

@dan-klasson It doesn't work for me, `date` object don't have `timedelta` method. What Python version are you using? – DrTyrsa Jul 17 '17 at 06:32
@DrTyrsa My bad. Should be: `from datetime import timedelta, date; date.today() + timedelta(days=10)` – dan-klasson Jul 17 '17 at 07:07
I am using python 3.7. Worked for me. Thanks a lot. – Geshan Ravindu Jun 26 '21 at 11:23

fantabolous · Answer 3 · 2019-11-25T11:38:33.737

44

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)

edited Nov 25 '19 at 11:38

answered Aug 29 '14 at 14:10

fantabolous

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Worked fine for me. Thanks – renny Apr 30 '19 at 09:18
7

just want to note that installing pandas just for this is seriously overkill. – Christopher Hunter Mar 29 '21 at 22:04
This was the best answer for me because I was already using pandas and didn't want to import yet another library. – autonopy Jul 30 '21 at 03:07

Pradeep Subash · Answer 4 · 2021-04-19T09:22:40.970

39

This might help:

from datetime import date, timedelta
date1 = date(2011, 10, 10)
date2 = date1 + timedelta(days=5)
print (date2)

edited Apr 19 '21 at 09:22

answered Oct 01 '20 at 12:07

Pradeep Subash

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3

there is a small typo, in the third line it should be date2 = date1 + timedelta(days=5) – Vyshak Puthusseri Apr 07 '21 at 14:44
1

you can also do `date1 += timedelta(days=5)`. – not2qubit Apr 13 '22 at 18:09

score 19 · Answer 5 · answered Feb 20 '18 at 19:51

19

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')

answered Feb 20 '18 at 19:51

Jorge Omar MH

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score 15 · Answer 6 · edited Jun 20 '20 at 09:12

15

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

edited Jun 20 '20 at 09:12

Community

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answered Jun 25 '15 at 12:56

Atul Arvind

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relativedelta is especially useful when doing operations on month year etc. – forkadam Aug 20 '20 at 04:11

score 14 · Answer 7 · answered Jul 29 '11 at 09:20

14

I guess you are missing something like that:

from datetime import timedelta

answered Jul 29 '11 at 09:20

vstm

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Guray Celik · Answer 8 · 2014-11-14T08:17:45.350

9

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

edited Nov 14 '14 at 08:17

answered Nov 14 '14 at 08:01

Guray Celik

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Good code. But your IF to test the addDays in get_date is not necessary – Eduardo Nov 24 '14 at 15:09

toto_tico · Answer 9 · 2019-01-16T21:14:19.583

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.

score 4 · Answer 10 · answered Mar 22 '21 at 19:24

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)

score 1 · Answer 11 · answered Jul 19 '22 at 06:52

I just came across this old thread:

I have checked but most of the answers are the same. I liked two answers from all these so I thought to check the efficiency of these two approaches.

First Approach: using the DateTime module Second Approach: using the panda's library

So I run the test about 10k times and The pandas library method was much slower. So I suggest using the built-in DateTime module.

from datetime import date, timedelta
import pandas as pd
import timeit

def using_datetime():
    pre_date = date(2013, 10, 10)
    day_date = pre_date + timedelta(days=5)
    return day_date

def using_pd():
    start_date = "10/10/2022"
    pd_date = pd.to_datetime(start_date)
    end_date = pd_date + pd.DateOffset(days=5)
    return end_date
    

for func in [using_datetime, using_pd]:
    print(f"{func.__name__} Time Took: ",  timeit.timeit(stmt=func, number=10000))
    
# Output 
# using_datetime Time Took:  0.009390000021085143
# using_pd Time Took:  2.1051381999859586

score 0 · Answer 12 · edited May 29 '19 at 07:21

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

score 0 · Answer 13 · answered Aug 23 '19 at 13:22

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 project
GetDate("2019-08-08").current() - this will convert your string date to time aware object with timezone you defined in pt 1. Default string format is format="%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 day
GetDate("2019-08-08").get_month_last_day() returns given date month last day
GetDate("2019-08-08").minimum_time() returns given date day start
GetDate("2019-08-08").maximum_time() returns given date day end
GetDate("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 hours
GetDate("2019-08-08").time_delta_seconds({number_of_seconds}) similar to pt 7 but working on seconds

score 0 · Answer 14 · edited Aug 11 '20 at 02:52

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])

score 0 · Answer 15 · answered Mar 19 '21 at 12:54

0

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()

answered Mar 19 '21 at 12:54

yoyoyoyo123

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2
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score 0 · Answer 16 · answered Jun 07 '23 at 07:33

You can use it this way, It will work:

If you need only date, then:

import datetime

StartDate = "10/10/11"
Date = datetime.datetime.strptime(StartDate, "%m/%d/%y").date()
print("Date: ", Date)

EndDate = Date + datetime.timedelta(days=10)
print("EndDate: ", EndDate)

If you need date time both, then:

import datetime

StartDate = "10/10/11"
Date = datetime.datetime.strptime(StartDate, "%m/%d/%y")
print("Date: ", Date)

EndDate = Date + datetime.timedelta(days=10)
print("EndDate: ", EndDate)

score 0 · Answer 17 · answered Jul 25 '23 at 04:17

You forgot to import timedelta function from datetime

This is how you can fix it:

import re
from datetime import datetime, timedelta

StartDate = "10/10/11"

Date = datetime.strptime(StartDate, "%m/%d/%y")

EndDate = Date.today()+timedelta(days=10)

print(EndDate)

This is the output from the code above: 2012-08-03 23:13:53.485750

score -19 · Answer 18 · answered Jun 15 '21 at 13:20

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)

There is a built-in timedelta object that takes care of everything you're trying to do here, and then some. — Thumper, May 16 '22 at 01:26

Adding days to a date in Python

18 Answers18

Adding 5 days to current date.

subtracting 5 days from current date.

Linked

Related