Python gets last month and year

Question

I am trying to get the last month and current year in the format: July 2016.

I tried (but this did not work) and it does not print July, but the number:

import datetime
now = datetime.datetime.now()
print now.year, now.month(-1)

Answer 1

now = datetime.datetime.now()
last_month = now.month-1 if now.month > 1 else 12
last_year = now.year - 1

to get the name of the month that you can use

"Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec".split()[last_month-1]

Answer 2

If you are manipulating dates, then the dateutil library is always great to have the convenience that Python stdlib does not cover easily.

from datetime import datetime
from dateutil.relativedelta import relativedelta

# Returns the same day of last month if possible otherwise end of month
# (eg: March 31st->29th Feb an July 31st->June 30th)
last_month = datetime.now() - relativedelta(months=1)

# Create string of month name and year...
text = format(last_month, '%B %Y')

Gives you:

'July 2016'

Answer 3

def subOneMonth(dt):
   day = dt.day
   res = dt.replace(day=1) - datetime.timedelta(days =1) 
   try:
     res.replace(day= day)
   except ValueError:
     pass
   return res

print subOneMonth(datetime.datetime(2016,07,11)).strftime('%d, %b %Y')
11, Jun 2016
print subOneMonth(datetime.datetime(2016,01,11)).strftime('%d, %b %Y')
11, Dec 2015
print subOneMonth(datetime.datetime(2016,3,31)).strftime('%d, %b %Y')
29, Feb 2016

Answer 4

, Pandas, , (). strftime.

import datetime as dt
import pandas as pd

>>> (pd.Period(dt.datetime.now(), 'M') - 1).strftime('%B %Y')
u'July 2016'

Answer 5

Python .

:

• 1, .
• Doing - timedelta (days = 1) .
• format and use "% B% Y" to convert to the desired format.

import datetime as dt
format(dt.date.today().replace(day=1) - dt.timedelta(days=1), '%B %Y')
>>>'June-2019'