python date of the previous month

Question

I am trying to get the date of the previous month with python. Here is what i\'ve tried:

str( time.strftime(\'%Y\') ) + str( int(time.strftime(\'%m\'))-1 )
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Answer 1

Building off the comment of @J.F. Sebastian, you can chain the replace() function to go back one "month". Since a month is not a constant time period, this solution tries to go back to the same date the previous month, which of course does not work for all months. In such a case, this algorithm defaults to the last day of the prior month.

from datetime import datetime, timedelta

d = datetime(2012, 3, 31) # A problem date as an example

# last day of last month
one_month_ago = (d.replace(day=1) - timedelta(days=1))
try:
    # try to go back to same day last month
    one_month_ago = one_month_ago.replace(day=d.day)
except ValueError:
    pass
print("one_month_ago: {0}".format(one_month_ago))

Output:

one_month_ago: 2012-02-29 00:00:00

Answer 2

Building on bgporter's answer.

def prev_month_range(when = None): 
    """Return (previous month's start date, previous month's end date)."""
    if not when:
        # Default to today.
        when = datetime.datetime.today()
    # Find previous month: https://stackoverflow.com/a/9725093/564514
    # Find today.
    first = datetime.date(day=1, month=when.month, year=when.year)
    # Use that to find the first day of this month.
    prev_month_end = first - datetime.timedelta(days=1)
    prev_month_start = datetime.date(day=1, month= prev_month_end.month, year= prev_month_end.year)
    # Return previous month's start and end dates in YY-MM-DD format.
    return (prev_month_start.strftime('%Y-%m-%d'), prev_month_end.strftime('%Y-%m-%d'))

Answer 3

from datetime import date, timedelta

first_day_of_current_month = date.today().replace(day=1)
last_day_of_previous_month = first_day_of_current_month - timedelta(days=1)

print "Previous month:", last_day_of_previous_month.month

Or:

from datetime import date, timedelta

prev = date.today().replace(day=1) - timedelta(days=1)
print prev.month

Answer 4

For someone who got here and looking to get both the first and last day of the previous month:

from datetime import date, timedelta

last_day_of_prev_month = date.today().replace(day=1) - timedelta(days=1)

start_day_of_prev_month = date.today().replace(day=1) - timedelta(days=last_day_of_prev_month.day)

# For printing results
print("First day of prev month:", start_day_of_prev_month)
print("Last day of prev month:", last_day_of_prev_month)

Output:

First day of prev month: 2019-02-01
Last day of prev month: 2019-02-28

Answer 5

Its very easy and simple. Do this

from dateutil.relativedelta import relativedelta
from datetime import datetime

today_date = datetime.today()
print "todays date time: %s" %today_date

one_month_ago = today_date - relativedelta(months=1)
print "one month ago date time: %s" % one_month_ago
print "one month ago date: %s" % one_month_ago.date()

Here is the output: $python2.7 main.py

todays date time: 2016-09-06 02:13:01.937121
one month ago date time: 2016-08-06 02:13:01.937121
one month ago date: 2016-08-06

Answer 6

With the Pendulum very complete library, we have the subtract method (and not "subStract"):

import pendulum
today = pendulum.datetime.today()  # 2020, january
lastmonth = today.subtract(months=1)
lastmonth.strftime('%Y%m')
# '201912'

We see that it handles jumping years.

The reverse equivalent is add.

https://pendulum.eustace.io/docs/#addition-and-subtraction