I know this question is a repeated one. But what I am trying to do is, I want to iterate through a date range and for each iteration i need to set the fromDate and toDate.
for ex:
If I give the date range as startDate = '2022-10-31'
and endDate = '2022-11-04'
and for each iteration fromDate = '2022-10-31' and toDate = '2022-11-01'
next iteration fromDate = '2022-11-01' and endDate = '2022-11-02' and so on.
I did some research and got to know how to iterate through dateRange.
sample code:
import datetime
start_date = datetime.date(2022, 10, 31)
end_date = datetime.date(2022, 11, 04)
dates_2011_2013 = [ start_date + datetime.timedelta(n) for n in range(int ((end_date - start_date).days))]
This just prints the incremented dates in the date Range. Am new to Python language. Any help is appreciated.
Thank you.
You can change the code slightly to,
import datetime
start_date = datetime.date(2022, 10, 31)
end_date = datetime.date(2022, 11, 4)
dates_2011_2013 = [ (start_date + datetime.timedelta(n), start_date + datetime.timedelta(n+1)) for n in range(int ((end_date - start_date).days))]
[(datetime.date(2022, 10, 31), datetime.date(2022, 11, 1)),
(datetime.date(2022, 11, 1), datetime.date(2022, 11, 2)),
(datetime.date(2022, 11, 2), datetime.date(2022, 11, 3)),
(datetime.date(2022, 11, 3), datetime.date(2022, 11, 4))]
First item of the tuple is start date and the second item is end date.
I want to iterate through a date range
another option is using while:
while start_date < end_date:
print(start_date, start_date + datetime.timedelta(1))
start_date += datetime.timedelta(1)
'''
2022-10-31 2022-11-01
2022-11-01 2022-11-02
2022-11-02 2022-11-03
2022-11-03 2022-11-04
I tried to exclude specific dates from a range of date and not getting valid output and below is the code. Could you please help to fix it.
from datetime import timedelta, date
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
delta = edate - sdate
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
for i in range(delta.days + 1):
day = sdate + timedelta(days=i)
print(day)
for j in range(len(toRemoveDate)):
if day != toRemoveDate[j]:
print(day)
Thanks,
Anand
try this,
from datetime import timedelta, date
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
delta = edate - sdate
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
for i in range(delta.days + 1):
day = sdate + timedelta(days=i)
if day not in toRemoveDate: # <-- check if day not in list
print(day)
Try .drop()
from datetime import timedelta, date
import pandas as pd
sdate = date(2020, 7, 1)
edate = date(2020, 7, 7)
range = pd.date_range(start=str(sdate), end=str(edate))
toRemoveDate = [date(2020, 7, 6), date(2020, 7, 2)]
days = range.drop(toRemoveDate)
print(days)
I'm trying to specify business days in a foreign country, but I can't get the pandas function pd.bdate_range() to recognize holidays. My code is as follows:
import pandas as pd
import datetime
weekmask = "Mon Tue Wed Thu Fri"
holidays = [datetime.datetime(2017, 1, 9), datetime.datetime(2017, 3, 20),
datetime.datetime(2017, 4, 13)]
BdaysCol2017 = pd.bdate_range(start = pd.datetime(2017, 1, 1),
end = pd.datetime(2017, 12, 31),
weekmask = weekmask,
holidays = holidays)
But I get the following error on the holidays parameter:
ValueError: a custom frequency string is required when holidays or weekmask are passed, got frequency B
Why is this? How can I specify custom holidays? Is there a better way to do this?
Thank you
as the docs specify about weekmask and holidays:
only used when custom frequency strings are passed
so you need:
BdaysCol2017 = pd.bdate_range(start = pd.datetime(2017, 1, 1),
end = pd.datetime(2017, 12, 31),
freq='C',
weekmask = weekmask,
holidays=holidays)
I would do that:
import pandas as pd
from datetime import datetime
weekmask = 'Sun Mon Tue Wed Thu'
exclude = [pd.datetime(2017, 3, 20),
pd.datetime(2017, 4, 13),
pd.datetime(2017, 5, 3)]
pd.bdate_range('2017/1/1','2017/12/31',
freq='C',
weekmask = weekmask,
holidays=exclude)
I have an ISO 8601 week number, say 201801 - corresponding to week 1 of year 2018.
How do I extract the dates within the week using python?
I would get the sunday of that week, and then give 7 days starting from that. Depending on what you want to happen in week 52, you might want to remove the break
from datetime import datetime, timedelta
def days_from_weekstr(weekstr):
sunday = datetime.strptime(weekstr + "-0", "%Y%W-%w")
year = sunday.year
for i in range(7):
day = sunday + timedelta(1) * i
if day.year != year:
break
yield day.date()
Usage:
>>> list(days_from_weekstr("201801"))
[datetime.date(2018, 1, 7), datetime.date(2018, 1, 8), datetime.date(2018, 1, 9), datetime.date(2018, 1, 10), datetime.date(2018, 1, 11), datetime.date(2018, 1, 12), datetime.date(2018, 1, 13)]
I'm looking for an elegant and pythonic way to get the date of the end of the previous quarter.
Something like this:
def previous_quarter(reference_date):
...
>>> previous_quarter(datetime.date(2013, 5, 31))
datetime.date(2013, 3, 31)
>>> previous_quarter(datetime.date(2013, 2, 1))
datetime.date(2012, 12, 31)
>>> previous_quarter(datetime.date(2013, 3, 31))
datetime.date(2012, 12, 31)
>>> previous_quarter(datetime.date(2013, 11, 1))
datetime.date(2013, 9, 30)
Edit: Have I tried anything?
Yes, this seems to work:
def previous_quarter(ref_date):
current_date = ref_date - timedelta(days=1)
while current_date.month % 3:
current_date -= timedelta(days=1)
return current_date
But it seems unnecessarily iterative.
You can do it the "hard way" by just looking at the month you receive:
def previous_quarter(ref):
if ref.month < 4:
return datetime.date(ref.year - 1, 12, 31)
elif ref.month < 7:
return datetime.date(ref.year, 3, 31)
elif ref.month < 10:
return datetime.date(ref.year, 6, 30)
return datetime.date(ref.year, 9, 30)
Using dateutil:
import datetime as DT
import dateutil.rrule as rrule
def previous_quarter(date):
date = DT.datetime(date.year, date.month, date.day)
rr = rrule.rrule(
rrule.DAILY,
bymonth=(3,6,9,12), # the month must be one of these
bymonthday=-1, # the day has to be the last of the month
dtstart = date-DT.timedelta(days=100))
result = rr.before(date, inc=False) # inc=False ensures result < date
return result.date()
print(previous_quarter(DT.date(2013, 5, 31)))
# 2013-03-31
print(previous_quarter(DT.date(2013, 2, 1)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 3, 31)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 11, 1)))
# 2013-09-30
Exploit the data pattern involved and turn the problem into a table-lookup - your classic space-time tradeff:
from datetime import date
PQTBL = (((12,31,-1),)*3 + ((3,31,0),)*3 + ((6,30,0),)*3 + ((9,30,0),)*3)
def previous_quarter(ref):
entry = PQTBL[ref.month-1]
return date(ref.year+entry[2], entry[0], entry[1])
Find the first day and month of the quarter, then use relativedelta to subtract a day.
from dateutil.relativedelta import relativedelta
def previous_quarter(ref):
first_month_of_quarter = ((ref.month - 1) // 3) * 3 + 1
return ref.replace(month=first_month_of_quarter, day=1) - relativedelta(days=1)
It's almost certain you would be happier using pandas (a python library), it has many functions for "business time" data.
http://pandas.pydata.org/pandas-docs/dev/timeseries.html
Reworked Justin Ethier's code for a "next quarter" version. Also added timezone via pytz and strftime formatting. #justin-ethier
import pytz
from datetime import datetime, timedelta
import datetime as dt
def nextQuarter():
ref = datetime.now(pytz.timezone('America/New_York'))
if ref.month < 4:
next = dt.datetime(ref.year, 3, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
elif ref.month < 7:
next = dt.datetime(ref.year, 6, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
elif ref.month < 10:
next = dt.datetime(ref.year, 9, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
else:
next = dt.datetime(ref.year + 1, 12, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
return next
next = nextQuarter()
import datetime
def previous_quarter(ref):
quarter = (ref.month - 1) // 3
prev_quarter = (quarter - 1) % 4
return datetime.datetime(ref.year if quarter>0 else ref.year-1, prev_quarter*3+1, 1)
Solution using only python's datetime library -
import datetime
def get_quarter_end(dt):
'''
given a datetime object, find the end of the quarter
'''
quarter_of_month = int((dt.month-1)/3 + 1)
#======================================================
# find the first day of the next quarter
#======================================================
# if in last quarter then go to the next year
year = dt.year + 1 if quarter_of_month==4 else dt.year
# if in last quarter then month is january (or 1)
month = 1 if quarter_of_month==4 else (quarter_of_month*3) + 1
first_of_next_quarter = datetime.datetime(year = year,
month = month,
day = 1
)
# last day of quarter for dt will be minus 1 day of first of next quarter
quarter_end_dt = first_of_next_quarter - datetime.timedelta(days=1)
return quarter_end_dt
if __name__=='__main__':
dt = datetime.datetime.strptime('2016-07-15', '%Y-%m-%d')
target_dt = get_quarter_end(dt)
and if you want to retreive the last fours quarter you can do this
if ref.month < 4:
list1 = [datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30),
datetime.date(ref.year - 1, 6, 30),
datetime.date(ref.year - 1, 3, 31)]
list1 = [i.strftime('%Y%m%d') for i in list1]
return list1
elif ref.month < 7:
return [datetime.date(ref.year, 3, 31),
datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30),
datetime.date(ref.year - 1, 6, 30)]
elif ref.month < 10:
return [datetime.date(ref.year, 6, 30),
datetime.date(ref.year, 3, 31),
datetime.date(ref.year - 1, 12, 31),
datetime.date(ref.year - 1, 9, 30)]
return [datetime.date(ref.year, 9, 30),
datetime.date(ref.year, 6, 30),
datetime.date(ref.year, 3, 30),
datetime.date(ref.year - 1, 12, 31)]