Making a time adding function in python

Question

I'm trying to build a function that recieves a date and adds days, updating everything in case it changes, so far i've come up with this:

def addnewDate(date, numberOfDays):

    date = date.split(":")   
    day = int(date[0])
    month = int(date[1])
    year = int(date[2])
    new_days = 0
    l = 0
    l1 = 28
    l2 = 30
    l3 = 31
    #l's are the accordingly days of the month

    while numberOfDays > l:
        numberOfDays  = numberOfDays - l 
        if month != 12:
            month += 1
        else:
            month = 1
            year += 1

        if month in [1, 3, 5, 7, 8, 10, 12]:
            l = l3
        elif month in [4, 6, 9, 11]:
            l = l2
        else:
            l = l1

    return  str(day) + ':' + str(month) + ':' + str(year) #i'll deal 
    #with fact that it doesn't put the 0's in the < 10 digits later

Desired output:

addnewDate('29:12:2016', 5):

'03:01:2017'

I think the problem is with either the variables, or the position i'm using them in, kinda lost though..

Thanks in advance!

p.s I can't use python build in functions :)

Oh ye, forgot bout that, but for the sake of simplicity, let's assume it only goes to 28 — stasisOo
– stasisOo, Commented Dec 15, 2016 at 21:16

Jean-François Fabre · Accepted Answer · 2016-12-17 08:42:13Z

1

Since you cannot use standard library, here's my attempt. I hope I did not forget anything.

define a table for month lengths
tweak it if leap year detected (every 4 year, but special cases)
work on zero-indexed days & months, much easier
add the number of days. If lesser that current month number of days, end, else, substract current month number of days and retry (while loop)
when last month reached, increase year
add 1 to day and month in the end

code:

def addnewDate(date, numberOfDays):
    month_days = [31,28,31,30,31,30,31,31,30,31,30,31]

    date = date.split(":")
    day = int(date[0])-1
    month = int(date[1])-1
    year = int(date[2])
    if year%4==0 and year%400!=0:
        month_days[1]+=1

    new_days = 0
    #l's are the accordingly days of the month

    day += numberOfDays

    nb_days_month = month_days[month]

    done = False   # since you don't want to use break, let's create a flag
    while not done:
        nb_days_month = month_days[month]
        if day < nb_days_month:
            done = True
        else:
            day -= nb_days_month
            month += 1
            if month==12:
                year += 1
                month = 0


return  "{:02}:{:02}:{:04}".format(day+1,month+1,year)

test (may be not exhaustive):

for i in ("28:02:2000","28:02:2004","28:02:2005","31:12:2012","03:02:2015"):
    print(addnewDate(i,2))
    print(addnewDate(i,31))

result:

of course, this is just for fun. Else use time or datetime modules!

edited Dec 17, 2016 at 8:42

answered Dec 15, 2016 at 21:30

Jean-François Fabre♦

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5 Comments

stasisOo Over a year ago

Thanks for the quick reply, i'll look into it but i don't think i'll be able to use it fully at least, i can't use built in modules from python because i'ts a school work, therefor can't use break either..

Jean-François Fabre Over a year ago

okay, that's stupid, but whatever, check my edit. No trace of break anymore. You teacher should know better...

stasisOo Over a year ago

I'll check it out :) this is for a task assignemnt to deliveries, half of the code i posted was advised by him, so i'm kinda trying to make that work since it should have able to work (also for the future if i need it) to add stupid stuff like 400 days, thanks for your help

stasisOo Over a year ago

sorry to bother you again, i'm probably gonna use your code (!) but i'm not understanding very well why you subtracted one then added it at the end. Also correct me if i'm wrong, the 13rd date at the list is because of the 0 month when the year changes right?

Jean-François Fabre Over a year ago

I found it easier to work 0-indexed, then restore the 1 indexed in the end. specially when you compare with numbers of days. and the list is too long. Editing.

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