How do you add a day in python?

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

ekhumoro

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asked Jul 29, 2011 at 9:17

MuraliKrishnaMuraliKrishna

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4

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)

answered Jul 29, 2011 at 10:03

Botond BéresBotond Béres

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9

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)

pieguy

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answered Jul 29, 2011 at 9:20

4

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)

answered Aug 29, 2014 at 14:10

fantabolousfantabolous

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3

This might help:

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

answered Oct 1, 2020 at 12:07

2

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, 2018 at 19:51

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

answered Jun 25, 2015 at 12:56

Atul ArvindAtul Arvind

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1

I guess you are missing something like that:

from datetime import timedelta

answered Jul 29, 2011 at 9:20

vstmvstm

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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

answered Nov 14, 2014 at 8:01

Guray CelikGuray Celik

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1

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.

answered Sep 27, 2017 at 10:42

toto_ticototo_tico

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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)

answered Mar 22, 2021 at 19:24

Amar KumarAmar Kumar

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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

taras

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answered May 29, 2019 at 7:02

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

answered Aug 23, 2019 at 13:22

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])

Andrew Myers

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answered Oct 28, 2019 at 10:56

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, 2021 at 12:54

bvmcodebvmcode

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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

answered Jul 19 at 6:52

AlixaProDevAlixaProDev

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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)

answered Jun 15, 2021 at 13:20

1

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