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 )
However, this way is bad for 2 reasons: First it returns 20122 for the February of 2012 (instead of 201202) and secondly it will return 0 instead of 12 on January.
I have solved this trouble in bash with
echo $(date -d"3 month ago" "+%G%m%d")
I think that if bash has a built-in way for this purpose, then python, much more equipped, should provide something better than forcing writing one's own script to achieve this goal. Of course i could do something like:
if int(time.strftime('%m')) == 1:
return '12'
else:
if int(time.strftime('%m')) < 10:
return '0'+str(time.strftime('%m')-1)
else:
return str(time.strftime('%m') -1)
I have not tested this code and i don't want to use it anyway (unless I can't find any other way:/)
Thanks for your help!
Simple, one liner:
import datetime as dt
previous_month = (dt.date.today().replace(day=1) - dt.timedelta(days=1)).month
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
.
You should use dateutil. With that, you can use relativedelta, it's an improved version of timedelta.
>>> import datetime
>>> import dateutil.relativedelta
>>> now = datetime.datetime.now()
>>> print now
2012-03-15 12:33:04.281248
>>> print now + dateutil.relativedelta.relativedelta(months=-1)
2012-02-15 12:33:04.281248
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'))
def prev_month(date=datetime.datetime.today()):
if date.month == 1:
return date.replace(month=12,year=date.year-1)
else:
try:
return date.replace(month=date.month-1)
except ValueError:
return prev_month(date=date.replace(day=date.day-1))
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
import pandas as pd
lastmonth = int(pd.to_datetime("today").strftime("%Y%m"))-1
print(lastmonth)
202101
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
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
There is a high level library dateparser
that can determine the past date given natural language, and return the corresponding Python datetime
object
from dateparser import parse
parse('4 months ago')
Just for fun, a pure math answer using divmod. Pretty inneficient because of the multiplication, could do just as well a simple check on the number of month (if equal to 12, increase year, etc)
year = today.year
month = today.month
nm = list(divmod(year * 12 + month + 1, 12))
if nm[1] == 0:
nm[1] = 12
nm[0] -= 1
pm = list(divmod(year * 12 + month - 1, 12))
if pm[1] == 0:
pm[1] = 12
pm[0] -= 1
next_month = nm
previous_month = pm
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
Source: Stackoverflow.com