Getting error for December month.
ValueError: month must be in 1..12
def last_day_of_month(ds):
cur_ds = datetime.strptime(ds, '%Y-%m-%d')
next_month = datetime(year=cur_ds.year, month=cur_ds.month+1, day=1)
last_day_month = next_month - timedelta(days=1)
return datetime.strftime(last_day_month, '%Y-%m-%d')
print last_day_of_month('2016-12-01')
In line 3 month=cur_ds.month+1 you are giving 13th month which is not valid. If you want to calculate last day of a given month you could also use month range from calendar library.
>>import calendar
>>year, month = 2016, 12
>>calendar.monthrange(year, month)[1]
31
You can't make a datetime with a month of 13. So you have to find a way to fix it. A simple solution is to convert the incremented month to an extra year:
# Reduce 12 to 1, 0 and all other #s to 0, #
extrayear, month = divmod(cur_ds.month, 12)
# Add 1 or 0 to existing year, add one to month (which was reduced to 0-11)
next_month = datetime(year=cur_ds.year + extrayear, month=month + 1, day=1)
You're passing in 12 as current month, then adding one to get next_month, making it 13. Check for the 12 case and set month=1 instead.
this is how I did it.
from django.utils import timezone
from calendar import monthrange
from datetime import datetime
current = timezone.now()
firstdayofmonth = current.replace(day=1)
endmonth = monthrange(current.year, current.month)
lastdayofmonth = datetime(current.year, current.month, endmonth[1])
Related
I have a date in the format "YYYY-MM-DD", for example "2022-03-09". It is the 68th day of the year 2022. Is there a way to get the 68th day of 2021? More generally, if I have a date in "YYYY-MM-DD" format, which is the N th day of the year, is there a quick way to calculate the N th day of the previous year?
You could do:
from datetime import datetime, timedelta
date_string = '2020-03-09'
# Parse the string to a datetime.date object
date = datetime.strptime(date_string, '%Y-%m-%d')
# Get the number N of the day in the year
N = date.timetuple().tm_yday
# Take 1st of january of last year and add N-1 days
N_last_year = (
date.replace(year=date.year-1, month=1, day=1) +
timedelta(days=N-1)
)
print(N_last_year.date())
Or, another funny solution based on leap years. It is based on the fact that the Nth day of last year is the same as this year's, except if the date is after 29th february and there is a leap year somewhere.
from datetime import datetime, timedelta
def leap(year: int) -> bool:
# Returns True if year is a leap year, False otherwise
return (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
date_string = '2020-03-09'
# Parse the string to a datetime.date object
date = datetime.strptime(date_string, '%Y-%m-%d')
# Get the number N of the day in the year
N = date.timetuple().tm_yday
date_last_year = date.replace(year=date.year-1)
# Do the cool thing
if date.month >= 3:
if leap(date.year):
date_last_year += timedelta(days=1)
elif leap(date.year-1):
date_last_year += timedelta(days=-1)
print(date_last_year.date())
The datetime objects will you with that (https://docs.python.org/3/library/datetime.html).
Indeed, you can compute the time lap between your date and the beginning of the year. Then you can add this difference to the beginning of the next year.
import datetime
date_string = "2022-03-09"
date_object = datetime.datetime.strptime(date_string, "%Y-%m-%d").date()
beginning_of_year = datetime.date(date_object.year, 1, 1)
time_difference = date_object - beginning_of_year
beginning_of_last_year = datetime.date(date_object.year - 1, 1, 1)
same_day_last_year = beginning_of_last_year + time_difference
print(same_day_last_year)
However, the result is disappointing, since it is in fact the same date...
I think you can do something like
from datetime import timedelta, date
year = timedelta(days=365)
today = date.fromisoformat("2022-03-09")
print(today - year)
How can I get the first date of the next month in Python? For example, if it's now 2019-12-31, the first day of the next month is 2020-01-01. If it's now 2019-08-01, the first day of the next month is 2019-09-01.
I came up with this:
import datetime
def first_day_of_next_month(dt):
'''Get the first day of the next month. Preserves the timezone.
Args:
dt (datetime.datetime): The current datetime
Returns:
datetime.datetime: The first day of the next month at 00:00:00.
'''
if dt.month == 12:
return datetime.datetime(year=dt.year+1,
month=1,
day=1,
tzinfo=dt.tzinfo)
else:
return datetime.datetime(year=dt.year,
month=dt.month+1,
day=1,
tzinfo=dt.tzinfo)
# Example usage (assuming that today is 2021-01-28):
first_day_of_next_month(datetime.datetime.now())
# Returns: datetime.datetime(2021, 2, 1, 0, 0)
Is it correct? Is there a better way?
Here is a 1-line solution using nothing more than the standard datetime library:
(dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)
Examples:
>>> dt = datetime.datetime(2016, 2, 29)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2016-03-01 00:00:00
>>> dt = datetime.datetime(2019, 12, 31)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2020-01-01 00:00:00
>>> dt = datetime.datetime(2019, 12, 1)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2020-01-01 00:00:00
Using dateutil you can do it the most literally possible:
import datetime
from dateutil import relativedelta
today = datetime.date.today()
next_month = today + relativedelta.relativedelta(months=1, day=1)
In English: add 1 month(s) to the today's date and set the day (of the month) to 1. Note the usage of singular and plural forms of day(s) and month(s). Singular sets the attribute to a value, plural adds the number of periods.
You can store this relativedelta.relativedelta object to a variable and the pass it around. Other answers involve more programming logic.
EDIT You can do it with the standard datetime library as well, but it's not so beautiful:
next_month = (today.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)
sets the date to the 1st of the current month, adds 32 days (or any number between 31 and 59 which guarantees to jump into the next month) and then sets the date to the 1st of that month.
you can use calendar to get the number of days in a given month, then add timedelta(days=...), like this:
from datetime import date, timedelta
from calendar import monthrange
days_in_month = lambda dt: monthrange(dt.year, dt.month)[1]
today = date.today()
first_day = today.replace(day=1) + timedelta(days_in_month(today))
print(first_day)
if you're fine with external deps, you can use dateutil (which I love...)
from datetime import date
from dateutil.relativedelta import relativedelta
today = date.today()
first_day = today.replace(day=1) + relativedelta(months=1)
print(first_day)
Extract the year and month, add 1 and form a new date using the year, month and day=1:
from datetime import date
now = date(2020,12,18)
y,m = divmod(now.year*12+now.month,12)
nextMonth = date(y,m+1,1)
print(now,nextMonth)
# 2020-12-18 2021-01-01
Your way looks good yet I would have done it this way:
import datetime
from dateutil import relativedelta
dt = datetime.datetime(year=1998,
month=12,
day=12)
nextmonth = dt + relativedelta.relativedelta(months=1)
nextmonth.replace(day=1)
print(nextmonth)
Using only python standard libraries:
import datetime
today = datetime.date.today()
first_of_next_month = return date.replace(
day=1,
month=date.month % 12 + 1,
year=date.year + (date.month // 12)
)
could be generalized to...
def get_first_of_month(date, month_offset=0):
# zero based indexing of month to make math work
month_count = date.month - 1 + month_offset
return date.replace(
day=1, month=month_count % 12 + 1, year=date.year + (month_count // 12)
)
first_of_next_month = get_first_of_month(today, 1)
Other solutions that don't require 3rd party libraries include:
Toby Petty's answer is another good option.
If the exact timedelta is helpful to you,
a slight modification on Adam.Er8's answer might be convenient:
import calendar, datetime
today = datetime.date.today()
time_until_next_month = datetime.timedelta(
calendar.monthrange(today.year, today.month)[1] - today.day + 1
)
first_of_next_month = today + time_until_next_month
With Zope's DateTime library a very simple solution is possible
from DateTime.DateTime import DateTime
date = DateTime() # today
while date.day() != 1:
date += 1
print(date)
I see so many wonderful solutions to this problem I personally was looking for a solution for getting the first and last day of the previous month when I stmbled on this question.
But here is a solution I like to think is quite simple and elegant:
date = datetime.datetime.now().date()
same_time_next_month = date + datetime.timedelta(days = date.day)
first_day_of_next_month_from_date = same_time_next_month - datetime.timedelta(days = same_time_next_month.day - 1)
Here we simply add the day of the target date to the date to get the same time of the next month, and then remove the number of days elapsed from the new date gotten.
Try this, for starting day of each month, change MonthEnd(1) to MonthBegin(1):
import pandas as pd
from pandas.tseries.offsets import MonthBegin, MonthEnd
date_list = (pd.date_range('2021-01-01', '2022-01-31',
freq='MS') + MonthEnd(1)).strftime('%Y-%m-%d').tolist()
date_list
Out:
['2021-01-31',
'2021-02-28',
'2021-03-31',
'2021-04-30',
'2021-05-31',
'2021-06-30',
'2021-07-31',
'2021-08-31',
'2021-09-30',
'2021-10-31',
'2021-11-30',
'2021-12-31',
'2022-01-31']
With python-dateutil:
from datetime import date
from dateutil.relativedelta import relativedelta
last day of current month:
date.today() + relativedelta(day=31)
first day of next month:
date.today() + relativedelta(day=31) + relativedelta(days=1)
I am trying to get the weeks of the current month. I can get the number of the week in that month(current_w) but I will need to get a list in the correct order. For example, for July, it will be w = [27, 28, 29, 30] because current week is in the third order of the 4 weeks of this month. Can somebody help me on that?
import datetime
kw = datetime.datetime.today().isocalendar()[1]
weeks = [7,14,21,28]
for idx, i in enumerate(weeks):
delta = datetime.timedelta(days=i)
if (a - delta).month != a.month:
current_w = idx
You could use datetime to get both the first and the last week and then just everything in between.
import datetime
import calendar
year = datetime.datetime.now().year
month = datetime.datetime.now().month
first_week = datetime.datetime(year, month, calendar.monthrange(year, month)[0]).isocalendar()[1]
last_week = datetime.datetime(year, month, calendar.monthrange(year, month)[1]).isocalendar()[1]
print(list(range(first_week, last_week + 1)))
from datetime import datetime would save you all the datetime.datetime usage btw.
If you only want weeks that are fully included in a month, the first week will be the one that contains the 7th of that month. And the last week will one before the one that contains the 1st of next month.
That is enough to give the following function:
def weeks_in_month(month, year=None):
"""Returns the (ISO) weeks fully contained in a month.
month is an integer between 1 and 12. Year is the current year
if not given"""
if year is None:
year = datetime.date.today().year
first = datetime.date(year, month, 7).isocalendar()[1]
nextd = datetime.date(year + 1, 1, 1) if month == 12 \
else datetime.date(year, month + 1, 1)
nextw = nextd.isocalendar()[1]
return list(range(first, nextw))
How can I print the date of the last Tuesday of each month for next year using Python.
For example the first line outputted would be: 30/Jan/2018
I do not want to have the full name of the month only the first 3 characters!
Currently I have figured out how to get the next year:
import datetime
now = datetime.datetime.now()
next_year = now.year + 1
Can anyone please help?
The calendar module is perfect for this:
You can use calendar.month_abbr which is an array of
abbreviated months just like you want.
week is an array representing the days of the week starting at Monday so Tuesday would be week[1].
import calendar
import datetime
now = datetime.datetime.now()
next_year = now.year + 1
for month in range(1, 13):
last_tuesday = max(week[1] for week in calendar.monthcalendar(next_year, month))
print('{}/{}/{}'.format(last_tuesday, calendar.month_abbr[month], next_year))
Output:
30/Jan/2018
27/Feb/2018
27/Mar/2018
24/Apr/2018
29/May/2018
26/Jun/2018
31/Jul/2018
28/Aug/2018
25/Sep/2018
30/Oct/2018
27/Nov/2018
25/Dec/2018
I would also suggest the pandas DateOffset object LastWeekOfMonth.
Describes monthly dates in last week of month like "the last Tuesday
of each month"
from pandas.tseries.offsets import LastWeekOfMonth
def last_tues(year):
return (pd.date_range('1/1/' + str(year), periods=12, freq='M')
- LastWeekOfMonth(n=1, weekday=1)).strftime('%d/%b/%Y'))
last_tues(2018)
Out[31]:
array(['30/Jan/2018', '27/Feb/2018', '27/Mar/2018', '24/Apr/2018',
'29/May/2018', '26/Jun/2018', '26/Jun/2018', '28/Aug/2018',
'25/Sep/2018', '30/Oct/2018', '27/Nov/2018', '25/Dec/2018'],
dtype='<U11')
I want to get the value of datetime using python code
ex. 20141104
that is example what I want to get
then, How can I get the datetime like that.
import calendar
for month in range(1, 13):
year = 2014
make_calendar = calendar.monthcalendar(year, month)
for weekend in make_calendar:
for day in weekend:
if (day != 0):
parameter = str(year) + str(month) + str(day)
print parameter
-> I try to get value like example but, the result is 201442.
I want to 20140402 not 201442.
I'm in need of help.
user1153551 has shown how to do what you want using the calendar module, but you should consider using the datetime module instead, with its powerful strftime method. The calendar module is great when you need to manipulate and/or format calendar at the month or year level, but for lower level manipulation at the level of individual dates, datetime is probably more suitable.
For example:
#! /usr/bin/env python
from datetime import date, timedelta
#A timedelta object of 1 day
oneday = timedelta(days=1)
year = 2014
#A date object of the start of the year
current_day = date(year, 1, 1)
#Print all the days of the given year in YYYYmmdd format
while current_day.year == year:
print current_day.strftime("%Y%m%d")
current_day += oneday
You can use following code to get desired output:
from time import gmtime, strftime,time, sleep
date = strftime("%Y%m%d")
print date
Use '%02d' % month to archive day, month with leading zero for Example,
>>> import datetime
>>> '%02d' % datetime.date.today().month
'11'
Python Code
import calendar
for month in range(1, 13):
year = 2014
make_calendar = calendar.monthcalendar(year, month)
for weekend in make_calendar:
for day in weekend:
if (day != 0):
parameter = '%02d%02d%02d' % (year, month, day)
print parameter