Reputation: 585
I currently have a program setup to run two different ways. One way is to run over a specified time frame, and the other way is to run everyday. However, when I have it set to run everyday, I only want it to continue if its a business day. Now from research I've seen that you can iterate through business days using Pandas like so:
start = 2016-08-05
end = datetime.date.today().strftime("%Y-%m-%d")
for day in pd.bdate_range(start, end):
print str(day) + " is a business Day"
And this works great when I run my program over the specified period.
But when I want to have the program ran everyday, I can't quite figure out how to test one specific day for being a business day. Basically I want to do something like this:
start = datetime.date.today().strftime("%Y-%m-%d")
end = datetime.date.today().strftime("%Y-%m-%d")
if start == end:
if not Bdate(start)
print "Not a Business day"
I know I could probably setup pd.bdate_range() to do what I'm looking for, but in my opinion would be sloppy and not intuitive. I feel like there must be a simpler solution to this. Any advice?
Upvotes: 18
Views: 35245
Reputation: 967
There is builtin method to do this in pandas.
For Pandas version <1.0
from pandas.tseries.offsets import Day, BDay
from datetime import datetime
bday=BDay()
is_business_day = bday.onOffset(datetime(2020,8,20))
For Pandas version >=1.1.0 (onOffset
is deprecated)
from pandas.tseries.offsets import Day, BDay
from datetime import datetime
bday=BDay()
is_business_day = bday.is_on_offset(datetime(2020,8,20))
Upvotes: 7
Reputation: 5437
I just found a different solution to this. This might be interesting if you want to find the next business day if your date is not a business day.
bdays=BDay()
def is_business_day(date):
return date == date + 0*bdays
adding 0*bdays
rolls forward on the next business day including the current one. Unfortunately, subtracting 0*bdays
does not roll backwards (at least with the pandas version I was using).
Moreover, due to this behavior, you also need to be careful since not necessarily
0*bdays + 1*bdays != 1*bdays
Upvotes: 7
Reputation: 553
for me I use an old trick from Excel:
from pandas.tseries.offsets import Day, BDay
def is_bday(x):
return x == x + Day(1) - BDay(1)
Upvotes: 2
Reputation: 3285
Using at least numpy
version 1.7.0., try np.is_busday()
start = datetime.date.today().strftime("%Y-%m-%d")
end = datetime.date.today().strftime("%Y-%m-%d")
if start == end:
# added code here
if not np.is_busday(start):
print("Not a Business day")
Upvotes: 2
Reputation: 4196
Please check this module - bdateutil
Please check the below code using above module :
from bdateutil import isbday
from datetime import datetime,date
now = datetime.now()
val = isbday(date(now.year, now.month, now.day))
print val
Please let me know if this help.
Upvotes: 0
Reputation: 1134
Since len
of pd.bdate_range()
tells us how many business days are in the supplied range of dates, we can cast this to a bool
to determine if a range of a single day is a business day:
def is_business_day(date):
return bool(len(pd.bdate_range(date, date)))
Upvotes: 30