• 4

A PHP Error was encountered

Severity: Notice

Message: Undefined index: userid

Filename: views/question.php

Line Number: 191


File: /home/prodcxja/public_html/questions/application/views/question.php
Line: 191
Function: _error_handler

File: /home/prodcxja/public_html/questions/application/controllers/Questions.php
Line: 433
Function: view

File: /home/prodcxja/public_html/questions/index.php
Line: 315
Function: require_once

name Punditsdkoslkdosdkoskdo

python date of the previous month

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'
    if int(time.strftime('%m')) < 10:
        return '0'+str(time.strftime('%m')-1)
        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!

datetime and the datetime.timedelta classes are your friend.

  1. find today.
  2. use that to find the first day of this month.
  3. use timedelta to backup a single day, to the last day of the previous month.
  4. print the YYYYMM string you're looking for.

Like this:

 import datetime
 today = datetime.date.today()
 first = today.replace(day=1)
 lastMonth = first - datetime.timedelta(days=1)

201202 is printed.

  • 299
Reply Report
      • 1
    • You can also chain the .replace() function. Do it once to get the last month, then do it again to get the day you want. First: d = date.today() then one_month_ago = (d.replace(day=1) - timedelta(days=1)).replace(day=d.day)
      • 1
    • @J.F.Sebastian You are correct -- thanks for pointing that out. There doesn't seem to be an elegant one-liner for this as a "month" is not a constant time period. You can do something ugly by importing calendar and using calendar.mdays[d.month-1] and more ugliness in a min() function in the 2nd replace, but it seems un-Pythonic and doesn't account for corner cases. I've updated my answer below using try - except which accounts for all cases, although a I hate using exceptions as part of an algorithm.

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
  • 69
Reply Report
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


from datetime import date, timedelta

prev = date.today().replace(day=1) - timedelta(days=1)
print prev.month
  • 45
Reply Report
      • 2
    • part of the solution... to find the day of last month, add something like this: day_previous_month = min(today.day, last_day_of_previous_month.day) to avoid exceeding the number of days.

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'))
  • 9
Reply Report

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
  • 5
Reply Report

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)


First day of prev month: 2019-02-01
Last day of prev month: 2019-02-28
  • 4
Reply Report

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
  • 1
Reply Report

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 to go back to same day last month
    one_month_ago = one_month_ago.replace(day=d.day)
except ValueError:
print("one_month_ago: {0}".format(one_month_ago))


one_month_ago: 2012-02-29 00:00:00
  • 0
Reply Report

If you want to look at the ASCII letters in a EXE type file in a LINUX/UNIX Environment, try "od -c 'filename' |more"

You will likely get a lot of unrecognizable items, but they will all be presented, and the HEX representations will be displayed, and the ASCII equivalent characters (if appropriate) will follow the line of hex codes. Try it on a compiled piece of code that you know. You might see things in it you recognize.

  • 0
Reply Report

Trending Tags