#first and last day of every month
s_january, e_january = ("1/1/2017"), ("1/31/2017")
s_february, e_february = ("2/1/2017"), ("2/28/2017")
s_march, e_march = ("3/1/2017"), ("3/31/2017")
s_april, e_april = ("4/1/2017"), ("4/30/2017")
s_may, e_may = ("5/1/2017"), ("5/31/2017")
s_june, e_june = ("6/1/2017"), ("6/30/2017")
s_july, e_july = ("7/1/2017"), ("7/31/2017")
s_august, e_august = ("8/1/2017"), ("8/31/2017")
s_September, e_September = ("9/1/2017"), ("9/30/2017")
s_october, e_october = ("10/1/2017"), ("10/31/2017")
s_november, e_november = ("11/1/2017"), ("11/30/2017")
s_december, e_december = ("12/1/2017"), ("12/31/2017")
def foo(s_date, e_date):
does stuff
foo(s_january, e_january)
foo(s_february, e_february)
foo(s_march, e_march)
foo(s_april, e_april)
foo(s_may, e_may)
foo(s_june, e_june)
foo(s_july, e_july)
foo(s_august, e_august)
foo(s_september, e_september)
foo(s_october, e_october)
foo(s_november, e_november)
foo(s_december, e_december)
I have a function that on a random date does stuff, but I have to call the function for every month, if I put the range for year I don't get the result that I want.
Is there any better way to avoid running it 12 times?
Set up your dates in a dictionary rather than 24 variables, and make life easier for yourself by computing the first and last day of each month. It would be useful also to represent your dates as datetimes not strings, since it's clear from your question header that you want to do computation on them.
import datetime
from dateutil import relativedelta
year = 2017
dates = {}
for month in range(1,13):
dates[(year,month)] = (
datetime.date(year,month,1),
datetime.date(year,month,1)
+ relativedelta.relativedelta(months=1)
- relativedelta.relativedelta(days=1))
The first element in each tuple is computed straightforwardly as the first day of the month. The second date is the same date, but with one month added (first day of the next month) and then one day subtracted, to get the last day of the month.
Then you can do:
for (year,month),(start,end) in dates.items():
print(year, month, foo (start,end))
You could use a dictionary to keep all start end end dates:
import calendar
import datetime as dt
def foo(s_date, e_date):
print ("Doing something between {} and {}".format(s_date.strftime('%d/%m/%Y'), e_date.strftime('%d/%m/%Y')))
def getMonths(year):
result = {}
for month in range(1, 13):
lastDayOfMonth = calendar.monthrange(year, month)[1]
result[month] = (dt.datetime(year, month, 1), dt.datetime(year, month, lastDayOfMonth))
return result
for month, start_end_dates in getMonths(2018).items():
foo(*start_end_dates)
Prints:
Doing something between 01/01/2018 and 31/01/2018
Doing something between 01/02/2018 and 28/02/2018
Doing something between 01/03/2018 and 31/03/2018
...
What do you mean by putting the range for year?
You could consider putting your dates to a dictionary or nested lists.
Related
I'm trying to build a list of "pay days" for a given month in the future knowing only when the pay days started months ago. For example:
Starting date - When the paychecks started: 1/6/2023
Frequency is every two weeks
So if I want to know which dates are pay days in March, I have to start at the 1/6/2023 and add two weeks until I get to March to know that the first pay day in March is 3/3/2/2023.
Then I want my final list of dates to be only those March dates of:
(3/3/2023, 3/17/2023, 3/31/2023)
I know I can use pandas to do something like:
pd.date_range(starting_date, starting_date+relativedelta(months=1), freq='14d')
but it would include every date back to 1/6/2023.
The easiest thing to do here would be to just update the starting_date parameter to be the first pay day in the month you're interested in.
To do this, you can use this function that finds the first pay day in a given month by first finding the difference between your start date and the desired month.
# month is the number of the month (1-12)
def get_first_pay_day_in_month(month=datetime.datetime.now().month,
year=datetime.datetime.now().year,
start_date=datetime.datetime(2023, 1, 6),
):
diff = datetime.datetime(year, month, 1) - start_date
freq = 14
if diff.days % freq == 0:
print(f'Difference: {diff.days/freq} weeks')
return datetime.datetime(year,month,1)
else:
print(f'Difference: {diff.days} days')
print(f'Days: {diff.days % freq} extra')
return datetime.datetime(year,month,1 + 14 - (diff.days % freq))
Then you can use this function to get the first pay day of a specific month and plug it into the date_range method.
from dateutil import relativedelta
starting_date = get_first_pay_day_in_month(month=3)
pay_days = pd.date_range(starting_date, starting_date+relativedelta.relativedelta(months=1), freq='14d')
print(pay_days)
Which function can I use to extract day, month and year from dates written in this manner 4.5.6 where 4 is the day, 5 is the month and 6 is the year (presumably 2006). I have already tried using dateparser.parse but it is not working.
day, month, year = map(int, '4.5.6'.split('.'))
And then add 2000 as necessary to the year.
You can then construct a datetime object with
from datetime import datetime
dt = datetime(year, month, day)
While it would be logical to use datetime.strptime, the one-digit year messes things up, and the above will just work fine.
Here is how you can use the datetime.datetime.strptime() method:
import datetime
s = "4.5.6"
i = s.rindex('.') + 1
s = s[:i] + s[i:].rjust(2, '0') # Add padding to year
dt = datetime.datetime.strptime(s, "%d.%m.%y")
print(dt)
Output:
2006-05-04 00:00:00
With the resulting datetime.datetime object, you can access plenty of information about the date, for example you can get the year by printing dt.year (outputs 2006).
I am trying to find the first and last date of the previous five months from now using python.
today = datetime.today()
first = today.replace(day=1)
lastMonth = first - timedelta(days=153)
Now how do i find the first and last date of each of the previous five months in python?
Can anyone help me with this please?
The below is based off an two answers already on SO. It adds in the ability to respond with the first and last day over a range of dates.
Get year month for last X months
How to get the first and last day of the month
import calendar
import datetime
from dateutil.relativedelta import relativedelta
def get_last_months(start_date, months):
for i in range(months):
_, num_days = calendar.monthrange(start_date.year,start_date.month)
first_day = datetime.date(start_date.year, start_date.month, 1).strftime('%Y-%m-%d')
last_day = datetime.date(start_date.year, start_date.month, num_days).strftime('%Y-%m-%d')
yield (first_day, last_day)
start_date += relativedelta(months = -1)
months_back = 5
print([i for i in get_last_months(datetime.datetime.today(), months_back)])
Output:
[('2020-10-01', '2020-10-31'), ('2020-09-01', '2020-09-30'), ('2020-08-01', '2020-08-31'), ('2020-07-01', '2020-07-31'), ('2020-06-01', '2020-06-30')]
I've written this function to get the last Thursday of the month
def last_thurs_date(date):
month=date.dt.month
year=date.dt.year
cal = calendar.monthcalendar(year, month)
last_thurs_date = cal[4][4]
if month < 10:
thurday_date = str(year)+'-0'+ str(month)+'-' + str(last_thurs_date)
else:
thurday_date = str(year) + '-' + str(month) + '-' + str(last_thurs_date)
return thurday_date
But its not working with the lambda function.
datelist['Date'].map(lambda x: last_thurs_date(x))
Where datelist is
datelist = pd.DataFrame(pd.date_range(start = pd.to_datetime('01-01-2014',format='%d-%m-%Y')
, end = pd.to_datetime('06-03-2019',format='%d-%m-%Y'),freq='D').tolist()).rename(columns={0:'Date'})
datelist['Date']=pd.to_datetime(datelist['Date'])
Jpp already added the solution, but just to add a slightly more readable formatted string - see this awesome website.
import calendar
def last_thurs_date(date):
year, month = date.year, date.month
cal = calendar.monthcalendar(year, month)
# the last (4th week -> row) thursday (4th day -> column) of the calendar
# except when 0, then take the 3rd week (February exception)
last_thurs_date = cal[4][4] if cal[4][4] > 0 else cal[3][4]
return f'{year}-{month:02d}-{last_thurs_date}'
Also added a bit of logic - e.g. you got 2019-02-0 as February doesn't have 4 full weeks.
Scalar datetime objects don't have a dt accessor, series do: see pd.Series.dt. If you remove this, your function works fine. The key is understanding that pd.Series.apply passes scalars to your custom function via a loop, not an entire series.
def last_thurs_date(date):
month = date.month
year = date.year
cal = calendar.monthcalendar(year, month)
last_thurs_date = cal[4][4]
if month < 10:
thurday_date = str(year)+'-0'+ str(month)+'-' + str(last_thurs_date)
else:
thurday_date = str(year) + '-' + str(month) + '-' + str(last_thurs_date)
return thurday_date
You can rewrite your logic more succinctly via f-strings (Python 3.6+) and a ternary statement:
def last_thurs_date(date):
month = date.month
year = date.year
last_thurs_date = calendar.monthcalendar(year, month)[4][4]
return f'{year}{"-0" if month < 10 else "-"}{month}-{last_thurs_date}'
I know that a lot of time has passed since the date of this post, but I think it would be worth adding another option if someone came across this thread
Even though I use pandas every day at work, in that case my suggestion would be to just use the datetutil library. The solution is a simple one-liner, without unnecessary combinations.
from dateutil.rrule import rrule, MONTHLY, FR, SA
from datetime import datetime as dt
import pandas as pd
# monthly options expiration dates calculated for 2022
monthly_options = list(rrule(MONTHLY, count=12, byweekday=FR, bysetpos=3, dtstart=dt(2022,1,1)))
# last satruday of the month
last_saturday = list(rrule(MONTHLY, count=12, byweekday=SA, bysetpos=-1, dtstart=dt(2022,1,1)))
and then of course:
pd.DataFrame({'LAST_ST':last_saturdays}) #or whatever you need
This question answer Calculate Last Friday of Month in Pandas
This can be modified by selecting the appropriate day of the week, here freq='W-FRI'
I think the easiest way is to create a pandas.DataFrame using pandas.date_range and specifying freq='W-FRI.
W-FRI is Weekly Fridays
pd.date_range(df.Date.min(), df.Date.max(), freq='W-FRI')
Creates all the Fridays in the date range between the min and max of the dates in df
Use a .groupby on year and month, and select .last(), to get the last Friday of every month for every year in the date range.
Because this method finds all the Fridays for every month in the range and then chooses .last() for each month, there's not an issue with trying to figure out which week of the month has the last Friday.
With this, use pandas: Boolean Indexing to find values in the Date column of the dataframe that are in last_fridays_in_daterange.
Use the .isin method to determine containment.
pandas: DateOffset objects
import pandas as pd
# test data: given a dataframe with a datetime column
df = pd.DataFrame({'Date': pd.date_range(start=pd.to_datetime('2014-01-01'), end=pd.to_datetime('2020-08-31'), freq='D')})
# create a dateframe with all Fridays in the daterange for min and max of df.Date
fridays = pd.DataFrame({'datetime': pd.date_range(df.Date.min(), df.Date.max(), freq='W-FRI')})
# use groubpy and last, to get the last Friday of each month into a list
last_fridays_in_daterange = fridays.groupby([fridays.datetime.dt.year, fridays.datetime.dt.month]).last()['datetime'].tolist()
# find the data for the last Friday of the month
df[df.Date.isin(last_fridays_in_daterange)]
If I want to add a loop to constrain days as well, what is the easiest way to do it, considering different length of month, leap years etc.
This is the script with years and months:
yearStart = 2010
yearEnd = 2017
monthStart = 1
monthEnd = 12
for year in list(range(yearStart, yearEnd + 1)):
for month in list(range(monthStart, monthEnd + 1)):
startDate = '%04d%02d%02d' % (year, month, 1)
numberOfDays = calendar.monthrange(year, month)[1]
lastDate = '%04d%02d%02d' % (year, month, numberOfDays)
If you want only the days then this code, using the pendulum library, is probably the easiest.
>>> import pendulum
>>> first_date = pendulum.Pendulum(2010, 1, 1)
>>> end_date = pendulum.Pendulum(2018, 1, 1)
>>> for day in pendulum.period(first_date, end_date).range('days'):
... print (day)
... break
...
2010-01-01T00:00:00+00:00
pendulum has many other nice features. For one thing, it's a drop-in replacement for datetime. Therefore, many of the properties and methods that you are familiar with using for that class will also be available to you.
You may want to use datetime in addition to calendar library. I am exactly not sure on requirements. But it appears you want the first date and last date of a given month and year. And, then loop through those dates. The following function will give you the first day and last day of each month. Then, you can loop between those two dates in whichever way you want.
import datetime
import calendar
def get_first_last_day(month, year):
date = datetime.datetime(year=year, month=month, day=1)
first_day = date.replace(day = 1)
last_day = date.replace(day = calendar.monthrange(date.year, date.month)[1])
return first_day, last_day
Adding the logic for looping through 2 dates as well.
d = first_day
delta = datetime.timedelta(days=1)
while d <= last_day:
print d.strftime("%Y-%m-%d")
d += delta