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

user_14373754 · Answer

This might help:

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

user_1060349 · Answer

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.

user_11183478 · Answer

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)

user_9724968 · Answer

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

user_9962315 · Answer

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

user_1820640 · Answer

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])

user_3775431 · Answer

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()

user_9939873 · Answer

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

user_15106666 · Answer

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)

Adding days to a date in Python

16 Answers

Adding 5 days to current date.

subtracting 5 days from current date.