I want to create a function that will return a date depending on each month. More specific I want it to return to me the 23d day of each month in the example format (YYYY, MM, DD):
2018, 1, 23 .
except for the occasion that this date coincides with a weekend, in which I want it to return the exact previous working day. For example for December 2017 the 23d was Saturday, so in this case I want my function to return to me
2017, 12, 22.
Right now I use:
import datetime
someday = datetime.date(2018, 1, 23)
print(someday)
with which in the someday variable I define the desired date manually for each month.
In short for the next seven months of 2018 (January, February, March, April, May, June, July), I want my function to return to me :
2018, 1, 23
2018, 2, 23
2018, 3, 23
2018, 4, 23
2018, 5, 23
2018, 6, 22
2018, 7, 23
and so on and so forth for the next months.
You can achieve this with weekday(), timedelta() and while loop:
from datetime import date, timedelta
def last_workday(somedate):
while somedate.weekday() > 4: #weekday returns values 0-6 (Mon-Sun)
somedate -= timedelta(days=1) #timedelta returns 1 day which you subtract from your date
return somedate
y = 2018
d = 23
for m in range(1,13):
print(last_workday(date(y, m, d)))
I wrote a small python script on how you could do it.
Use toordinal to check for the day of the week and if it's Saturday or Sunday, subtract 1 or 2 by using timedelta
import datetime
# Yes, from 1,13 because 13 is exclusive (Goes only to 12)
days_month = datetime.monthrange(2018,x)
for x in range(days_month[0],days_month[1]+1):
for x in range(1, monthrange):
someday = datetime.date(2018, x, 23) # Loop over every month
weekday = someday.weekday() # 0 = Monday, 6 = Sunday
if(weekday > 5 ): # If it's "more" than Friday
jumpback_days = weekday - 4; # Subtract 4 from the current weekday to get 1 for Saturday and 2 for Sunday
print(someday - datetime.timedelta(days=jumpback_days)) # Subtract days and print
else:
# Print without subtracting anything
print(someday)
Note: If you are using Python2, please replace range with xrange for it to work.
EDIT: If you want to print only and all workdays in a specific year, you can do it this way:
import datetime
from calendar import monthrange
year = 2018
for month in range(1, 13): # Yes, from 1,13 because 13 is exclusive (Goes only to 12)
days_month = monthrange(year, month) # Request the amount of days in that month
for day in range(1, days_month[1] + 1): # Iterate through all days of that month (+1 because it's exclusive)
someday = datetime.date(year, month, day) # Loop over every month
weekday = someday.weekday() # 0 = Monday, 6 = Sunday
if(weekday < 5 ): # If it's "less" than Saturday
# Print because if it's "less" than Saturday, then it's a workday
print(someday)
Related
I want to find the last friday date in each quater for the given year. For example, the output as follows
last friday in March
last friday in June
last friday in september
last friday in december
How can I find this intelligently based on given year as an input
I'm assuming you don't want the output to be literally
last friday in March
last friday in June
last friday in september
last friday in december
and you just didn't feel like looking up those dates for an example year to include in your question.
Since we know that quarters always end in those months, we can create the last dates of those months. Then, datetime.date.weekday() tells us which day of the week it is. Friday is 4, so we just need to find out how many days we need to go back to achieve this. Then, use datetime.timedelta can subtract that many days, and we should be good to go:
import datetime
def last_fridays_of_quarter(year):
last_days = [datetime.date(year, 3, 31), datetime.date(year, 6, 30), datetime.date(year, 9, 30), datetime.date(year, 12, 31)]
for day in last_days:
days_to_sub = (day.weekday() - 4) % 7
last_friday = day - datetime.timedelta(days=days_to_sub)
print(last_friday)
Testing this for 2022:
>>> last_fridays_of_quarter(2022)
2022-03-25
2022-06-24
2022-09-30
2022-12-30
Here's one way to solve it. You store last days of each quarter in a given year in an array. Then you iterate over said quarter ending dates, and calculate the offset that you would need to subtract in order to get to the previous friday.
from datetime import datetime, timedelta
year = 2022
quarter_ending_dates = [
datetime(year, 3, 31),
datetime(year, 6, 30),
datetime(year, 9, 30),
datetime(year, 12, 31)
]
for ending_date in quarter_ending_dates:
offset = (ending_date.weekday() - 4) % 7
last_friday = ending_date - timedelta(days=offset)
print(last_friday)
try:
import pandas as pd
def fridays(year):
df = pd.date_range(start=str(year), end=str(year+1),
freq='W-FRI').strftime('%m/%d/%Y')\
.to_frame().reset_index(drop = True)
df.columns = ['date']
df.index = pd.to_datetime(df['date']).dt.to_period('Q')
df = df.groupby(df.index).last()
df.index.name = 'quarter'
return df
fridays(2022)
output is:
quarter
date
2022Q1
03/25/2022
2022Q2
06/24/2022
2022Q3
09/30/2022
2022Q4
12/30/2022
How can i find all 'friday the 13th' due year 2000 to 2020 by python and have it print as below using a nested for loop:
print(i,k,13,"is friday the 13th!")
ex)
2000 10 13 is friday the 13th!
2001 4 13 is friday the 13th!
...
2020 11 13 is friday the 13th!
Based on your code in the comments, this should work for you:
from datetime import date, timedelta
def friday_13_in_year(y):
day = date(y, 1, 1)
end = date(y, 12, 31)
one_day = timedelta(days=1)
while day < end:
if day.weekday() == 4 and day.day == 13:
return str(day)+" is friday the 13th!"
day += one_day
print([friday_13_in_year(y) for y in range(2000, 2022+1)])
l=[i for i in ((datetime.datetime.strptime('20000101','%Y%m%d')+datetime.timedelta(days=i)) for i in range(7306)) if i.day == 13 and i.weekday() == 4]
l lists all Fridays 13th from 2000 to 2020. 20 years in days is 20*365.25+1 = 7306
I want last friday of each month for upcoming three months.
Friday_date = datetime.date.today()
while Friday_date.weekday() != 4:
Friday_date += datetime.timedelta(1)
This gives me the nearest friday. I want to make sure this is the last friday of this month so that i can add 28 days to get next friday.
The easiest way to do this is to use the module dateutil:
>>> from dateutil.relativedelta import FR, relativedelta
>>> datetime.date.today()+relativedelta(day=31, weekday=FR(-1))
datetime.date(2021, 6, 25)
Don't assume you can get the last Friday of subsequent months just by adding 28 days. It won't always work. Adding 28 days to the last Friday of February 2024 gives you this:
>>> datetime.date(2024,2,1)+relativedelta(day=31, weekday=FR(-1), days=28)
datetime.date(2024, 3, 22)
but the last Friday of that month is 29 March. Let dateutil do that correctly for you:
>>> datetime.date(2024,2,1)+relativedelta(day=31, weekday=FR(-1), months=1)
datetime.date(2024, 3, 29)
If needed with standard library only, here is with calendar and datetime:
import calendar
from datetime import date
today = date.today()
year, month = today.year, today.month
n_months = 4
friday = calendar.FRIDAY
for _ in range(n_months):
# get last friday
c = calendar.monthcalendar(year, month)
day_number = c[-1][friday] or c[-2][friday]
# display the found date
print(date(year, month, day_number))
# refine year and month
if month < 12:
month += 1
else:
month = 1
year += 1
where the line c[-1][friday] or c[-2][friday] first checks the last week of the month: is Friday nonzero there? if so take it, else look at the week before where there must be a Friday.
This prints
2021-06-25
2021-07-30
2021-08-27
2021-09-24
This formula gets you the day of the last Friday of any given month:
import calendar
year = 2021
month = 6
last_day = calendar.monthrange(year, month)[1]
last_weekday = calendar.weekday(year, month, last_day)
last_friday = last_day - ((7 - (4 - last_weekday)) % 7)
# ^ ^
# | Friday
# days in a week
This is my first coffee, so this can probably be condensed a bit, but it illustrates the logic. last_day is the last calendar day (30 for June), last_weekday is what weekday it is (2 for Wednesday), and based on that we simply calculate how many days to subtract to land on the last Friday (25).
If you want to know the last friday you can do this :
from datetime import date
from datetime import timedelta
today = date.today()
offset = (today.weekday() - 4) % 7
last_wednesday = today - timedelta(days=offset)
Write a function that accepts a string which contains a particular
date from the Gregorian calendar. Your function should return what day
of the week it was. Here are a few examples of what the input
string(Month Day Year) will look like. However, you should not
'hardcode' your program to work only for these input!
"June 12 2012"
"September 3 1955"
"August 4 1843"
Note that each item (Month Day Year) is separated by one space. For
example if the input string is:
"May 5 1992"
Then your function should return the day of the week (string) such as:
"Tuesday"
Algorithm with sample example:
# Assume that input was "May 5 1992"
day (d) = 5 # It is the 5th day
month (m) = 3 # (*** Count starts at March i.e March = 1, April = 2, ... January = 11, February = 12)
century (c) = 19 # the first two characters of the century
year (y) = 92 # Year is 1992 (*** if month is January or february decrease one year)
# Formula and calculation
day of the week (w) = (d + floor(2.6m - 0.2) - 2c + y + floor(y/4) + floor(c/4)) modulo 7
after calculation we get, (w) = 2
Count for the day of the week starts at Sunday, i.e Sunday = 0, Monday = 1, Tuesday = 2, ... Saturday = 6
Since we got 2, May 5 1992 was a Tuesday
My first question is how do I accept June 12 2012 as input from the user? Is there any method that allows me to do this? Any help would be appreciated.
user_input = input('Enter the date')
This question already has answers here:
How do I calculate the date six months from the current date using the datetime Python module?
(47 answers)
Closed 7 years ago.
So I am trying to find a way to increment a datetime object by one month. However, it seems this is not so simple, according to this question.
I was hoping for something like:
import datetime as dt
now = dt.datetime.now()
later = now + dt.timedelta(months=1)
But that doesn't work. I was also hoping to be able to go to the same day (or the closest alternative) in the next month if possible. For example, a datetime object set at January 1st would increment to Feb 1st whereas a datetime object set at February 28th would increment to March 31st as opposed to March 28th or something.
To be clear, February 28th would (typically) map to March 31st because it is the last day of the month, and thus it should go to the last day of the month for the next month. Otherwise it would be a direct link: the increment should go to the day in the next month with the same numbered day.
Is there a simple way to do this in the current release of Python?
Check out from dateutil.relativedelta import *
for adding a specific amount of time to a date, you can continue to use timedelta for the simple stuff i.e.
import datetime
from dateutil.relativedelta import *
use_date = datetime.datetime.now()
use_date = use_date + datetime.timedelta(minutes=+10)
use_date = use_date + datetime.timedelta(hours=+1)
use_date = use_date + datetime.timedelta(days=+1)
use_date = use_date + datetime.timedelta(weeks=+1)
or you can start using relativedelta
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(years=+1)
for the last day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
Right now this will provide 29/02/2016
for the penultimate day of next month:
use_date = use_date+relativedelta(months=+1)
use_date = use_date+relativedelta(day=31)
use_date = use_date+relativedelta(days=-1)
last Friday of the next month:
use_date = use_date+relativedelta(months=+1, day=31, weekday=FR(-1))
2nd Tuesday of next month:
new_date = use_date+relativedelta(months=+1, day=1, weekday=TU(2))
As #mrroot5 points out dateutil's rrule functions can be applied, giving you an extra bang for your buck, if you require date occurences.
for example:
Calculating the last day of the month for 9 months from the last day of last month.
Then, calculate the 2nd Tuesday for each of those months.
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import datetime
use_date = datetime(2020,11,21)
#Calculate the last day of last month
use_date = use_date+relativedelta(months=-1)
use_date = use_date+relativedelta(day=31)
#Generate a list of the last day for 9 months from the calculated date
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, bymonthday=(-1,)))
print("Last day")
for ld in x:
print(ld)
#Generate a list of the 2nd Tuesday in each of the next 9 months from the calculated date
print("\n2nd Tuesday")
x = list(rrule(freq=MONTHLY, count=9, dtstart=use_date, byweekday=TU(2)))
for tuesday in x:
print(tuesday)
Last day
2020-10-31 00:00:00
2020-11-30 00:00:00
2020-12-31 00:00:00
2021-01-31 00:00:00
2021-02-28 00:00:00
2021-03-31 00:00:00
2021-04-30 00:00:00
2021-05-31 00:00:00
2021-06-30 00:00:00
2nd Tuesday
2020-11-10 00:00:00
2020-12-08 00:00:00
2021-01-12 00:00:00
2021-02-09 00:00:00
2021-03-09 00:00:00
2021-04-13 00:00:00
2021-05-11 00:00:00
2021-06-08 00:00:00
2021-07-13 00:00:00
rrule could be used to find the next date occurring on a particular day.
e.g. the next 1st of January occurring on a Monday (Given today is the 4th November 2021)
from dateutil.relativedelta import *
from dateutil.rrule import *
from datetime import *
year = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0].year
year
2024
or the next 5 x 1st of January's occurring on a Monday
years = rrule(YEARLY,dtstart=datetime.now(),bymonth=1,bymonthday=1,byweekday=MO)[0:5]
for i in years:print(i.year)
...
2024
2029
2035
2046
2052
The first Month next Year that starts on a Monday:
>>> month = rrule(YEARLY,dtstart=datetime.date(2023, 1, 1),bymonthday=1,byweekday=MO)[0]
>>> month.strftime('%Y-%m-%d : %B')
'2023-05-01 : May'
If you need the months that start on a Monday between 2 dates:
months = rrule(YEARLY,dtstart=datetime.date(2025, 1, 1),until=datetime.date(2030, 1, 1),bymonthday=1,byweekday=MO)
>>> for m in months:
... print(m.strftime('%Y-%m-%d : %B'))
...
2025-09-01 : September
2025-12-01 : December
2026-06-01 : June
2027-02-01 : February
2027-03-01 : March
2027-11-01 : November
2028-05-01 : May
2029-01-01 : January
2029-10-01 : October
This is by no means an exhaustive list of what is available.
Documentation is available here: https://dateutil.readthedocs.org/en/latest/
Note: This answer shows how to achieve this using only the datetime and calendar standard library (stdlib) modules - which is what was explicitly asked for. The accepted answer shows how to better achieve this with one of the many dedicated non-stdlib libraries. If you can use non-stdlib libraries, by all means do so for these kinds of date/time manipulations!
How about this?
def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12
new_day = orig_date.day
# while day is out of range for month, reduce by one
while True:
try:
new_date = datetime.date(new_year, new_month, new_day)
except ValueError as e:
new_day -= 1
else:
break
return new_date
EDIT:
Improved version which:
keeps the time information if given a datetime.datetime object
doesn't use try/catch, instead using calendar.monthrange from the calendar module in the stdlib:
import datetime
import calendar
def add_one_month(orig_date):
# advance year and month by one month
new_year = orig_date.year
new_month = orig_date.month + 1
# note: in datetime.date, months go from 1 to 12
if new_month > 12:
new_year += 1
new_month -= 12
last_day_of_month = calendar.monthrange(new_year, new_month)[1]
new_day = min(orig_date.day, last_day_of_month)
return orig_date.replace(year=new_year, month=new_month, day=new_day)
Question: Is there a simple way to do this in the current release of Python?
Answer: There is no simple (direct) way to do this in the current release of Python.
Reference: Please refer to docs.python.org/2/library/datetime.html, section 8.1.2. timedelta Objects. As we may understand from that, we cannot increment month directly since it is not a uniform time unit.
Plus: If you want first day -> first day and last day -> last day mapping you should handle that separately for different months.
>>> now
datetime.datetime(2016, 1, 28, 18, 26, 12, 980861)
>>> later = now.replace(month=now.month+1)
>>> later
datetime.datetime(2016, 2, 28, 18, 26, 12, 980861)
EDIT: Fails on
y = datetime.date(2016, 1, 31); y.replace(month=2) results in ValueError: day is out of range for month
Ther is no simple way to do it, but you can use your own function like answered below.