Get Date in a particular format in python

Question

I have to get the date of the day approximately one month before from today in python. The resulting string should be of the format yyyy/mm/dd

The date has to be approximate, so things like considering the leap year and all can be avoided (or if easy can be incorporated as well)

I have written the following code for the above problem

from datetime import date
tup=list(date.today().timetuple()[0:3])
if (tup[1]==1):
    tup[1]=12
    tup[0]-=1
else:
   tup[1]-=1
   if tup[1]==2:
      tup[2]=min(28,tup[2])
   else if tup[1] in [4,6,9,11]:
      tup[2]=min(30,tup[2])
print '/'.join([str(x) for x in tup])

Is there a more elegant way to solve the above problem??

khampson · Accepted Answer · 2014-07-29 04:58:59Z

3

You can use the datetime class in conjunction with the timedelta function.

i.e.

>>> from datetime import date, timedelta
>>> days=30
>>> d = date.today() - timedelta(days=days)
>>> print d
2014-06-29

Then, to format:

from datetime import datetime
dt = datetime.strftime(d, "%Y/%m/%d")
>>> print dt
2014/06/29

edited Jul 29, 2014 at 4:58

answered Jul 29, 2014 at 4:52

khampson

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alecxe · Accepted Answer · 2014-07-29 04:54:44Z

2

Just an alternative solution using a third-party package called Delorean:

>>> from delorean import Delorean
>>> d = Delorean()
>>> d.last_month().date.strftime('%Y/%m/%d')
'2014/06/29'

answered Jul 29, 2014 at 4:54

alecxe

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1 Comment

alphacentauri Over a year ago

Will using a third-party package on a server be an additional overhead?

Andy · Accepted Answer · 2014-07-29 05:39:52Z

0

You can use DateUtil's relativedelta function for this functionality as well. It handles days in a month and leap years appropriately, like alecxe's library does.

If you don't have the DateUtil library, first install that

pip install python-dateutil

Then you can utilize it like this:

>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> now = datetime.datetime.today()
>>> now + relativedelta(months=1)
datetime.datetime(2014, 8, 29, 0, 32, 51, 843000)
>>> now + relativedelta(months=-1)
datetime.datetime(2014, 6, 29, 0, 32, 51, 843000)

Finally, to format it, you need to use the strftime function.

>>> (now + relativedelta(months=-1)).strftime('%Y/%m/%d')
'2014/06/29'

answered Jul 29, 2014 at 5:39

Andy♦

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Comments

YSK · Accepted Answer · 2014-07-29 05:46:02Z

0

datetime is the correct module here..

import datetime
datetime.datetime.now().strftime("%H:%M:%S.%f")

Hope this helps.

answered Jul 29, 2014 at 5:46

YSK

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