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

Lurlenelurline

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. – Kyl

Search didn't work? All of these code examples would have helped: stackoverflow.com/search?q=python+timedelta. There appear to be over 200 questions just like this one. – Astrometry

possible duplicate of add days to a date in Python using loops, ranges, and slicing – Astrometry

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 – Katusha

with same question, exclude the Weekend means? – Biblicist

Answer 1

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)

Cannular

datetime.datetime - why twice? – Koblas

importing like "from datetime import datetime, timedelta" would add readibility to the code – Pater

@paulmorriss: You are calling the strptime method on the datetime class in the datetime module, so you need to specify datetime.datetime. – Scrivener

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. – Avivah

Shouldn't it be %m/%d/%Y and not %m/%d/%y? – Molest

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. – Callan

Long tail legacy problem there. it "should" be from datetime import DateTime since classes are CamelCased, but datetime precedes PEP8. – Conjunctiva

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. – Drusilla

In python3, I have to do from datetime import datetime, timedelta and end_date = date_1 + timedelta(days=2) – Stoughton

That looks like PascalCase over camelCase – Originative

Answer 2

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)

Noshow

datetime.date.today() instead of Date.today() – Canikin

@dan-klasson It doesn't work for me, date object don't have timedelta method. What Python version are you using? – Noshow

@Noshow My bad. Should be: from datetime import timedelta, date; date.today() + timedelta(days=10) – Spinescent

I am using python 3.7. Worked for me. Thanks a lot. – Wycoff

Answer 3

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)

Valdez

Worked fine for me. Thanks – Synchrotron

just want to note that installing pandas just for this is seriously overkill. – Ideography

This was the best answer for me because I was already using pandas and didn't want to import yet another library. – Kristelkristen

Answer 4

This might help:

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

Strafford

you can also do date1 += timedelta(days=5). – Swaddle

Answer 5

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')

Tingaling

Answer 6

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

Phosgene

relativedelta is especially useful when doing operations on month year etc. – Mauricio

Answer 7

I guess you are missing something like that:

from datetime import timedelta

Tupler

Answer 8

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

Neoplasticism

Good code. But your IF to test the addDays in get_date is not necessary – Slaby

Answer 9

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.

Drusilla

Answer 10

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)

Kono

Answer 11

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

Interfertile

Answer 12

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

Bergen

Answer 13

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

Campagna

Answer 14

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

1. self.tz = pytz.timezone("Europe/Warsaw") - here you define Time Zone you want to use in project
2. 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())
3. GetDate("2019-08-08").get_month_first_day() returns given date (string or object) month first day
4. GetDate("2019-08-08").get_month_last_day() returns given date month last day
5. GetDate("2019-08-08").minimum_time() returns given date day start
6. GetDate("2019-08-08").maximum_time() returns given date day end
7. 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)
8. GetDate("2019-08-08").time_delta_haours({number_of_hours}) similar to pt 7 but working on hours
9. GetDate("2019-08-08").time_delta_seconds({number_of_seconds}) similar to pt 7 but working on seconds

Opalopalesce

Answer 15

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])

Ifill

Answer 16

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()

Ingesta

Answer 17

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)

Mallina

Answer 18

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)

Uremia

There is a built-in timedelta object that takes care of everything you're trying to do here, and then some. – Vesicate