I want to calculate difference between two time columns without considering non-business hours. I have used pyholidays, which worked totally fine. But even when i define starttime and endtime for Business-duration, Result still includes Non-Business Hours as you shown in attached photos.
for index, row in df.iterrows():
first=row['New']
second=row['Assigned']
third=row['In Progress']
if(pd.notnull(second)):
starttime = (8,0,0)
endtime = (17,0,0)
holidaylist = pyholidays.Germany()
unit='hour'
row['AP'] = businessDuration(first,second,holidaylist=holidaylist,unit=unit)
else:
starttime = (8,0,0)
endtime = (17,0,0)
holidaylist = pyholidays.Germany()
unit='hour'
row['AP'] = businessDuration(first,third,holidaylist=holidaylist,unit=unit)
ap.append(row['AP'])
CodePudding user response：
I don't know if this works, but try this:
# == Imports needed ===========================
from __future__ import annotations
import pandas as pd
import holidays as pyholidays
from datetime import time
from bizdays import Calendar
# == Function ==================================
def compute_bizhours_diff(
start_date: str | pd.Timestamp,
end_date: str | pd.Timestamp,
biz_open_time: datetime.time | None = None,
biz_close_time: datetime.time | None = None,
cal: bizdays.Calendar | None = None,
) -> float:
"""Compute the number of business hours between two dates.
Parameters
----------
start_date : str | pd.Timestamp
The initial date.
end_date : str | pd.Timestamp
The final date.
biz_open_time : datetime.time | None
The beginning hour/minute of a business day.
biz_close_time : datetime.time | None
The ending hour/minute of a business day.
cal : bizdays.Calendar | None
Calendar object, used to determine compute the number of days between `start_date`
and `end_date` that aren't holidays. If None, considers every day as a business day,
except Saturdays, or Sundays.
Returns
-------
float
The total number of business hours between `start_date`, and `end_date`.
Examples
--------
>>> import holidays as pyholidays
>>> from datetime import time
>>> from bizdays import Calendar
>>> # 2022-09-07 is a national holiday in Brazil, therefore only
>>> # the hours between 2022-09-08 09:00:00, and 2022-09-08 15:48:00
>>> # should be considered. This should equal 6.8 hours.
>>> start_date = pd.to_datetime('2022-09-07 15:55:00')
>>> end_date = pd.to_datetime('2022-09-08 15:48:00')
>>> BR_holiday_list = pyholidays.BR(years={start_date.year, end_date.year}, state='RJ')
>>> cal = Calendar(holidays=BR_holiday_list, weekdays=['Saturday', 'Sunday'])
>>> print(compute_bizhours_diff(start_date, end_date, cal=cal))
6.8
>>> # Both dates in the next example are holidays, therefore, the result should be 0.0
>>> start_date = pd.to_datetime('2022-09-07 15:55:00')
>>> end_date = pd.to_datetime('2022-09-07 15:48:00')
>>> print(compute_bizhours_diff(start_date, end_date, cal=cal))
0.0
>>> # What if the end_date preceeds start_date by mistake?
>>> # In such cases, we switch start_date to end_date, and vice-versa.
>>> start_date = pd.to_datetime('2022-09-02 00:00:00')
>>> end_date = pd.to_datetime('2022-09-01 15:55:00')
>>> print(compute_bizhours_diff(start_date, end_date, cal=cal))
2.0833333333333335
>>> # What if the start_date, and end_date begin and finish on the same day, but they both have timestamps that end before
>>> # or after the bussiness hours?
>>> # In such cases, the total number of hours is equal to 0.0
>>> start_date = pd.to_datetime('2022-09-02 00:00:00')
>>> end_date = pd.to_datetime('2022-09-02 8:00:00')
>>> print(compute_bizhours_diff(start_date, end_date, cal=cal))
0.0
"""
if biz_open_time is None:
biz_open_time = time(9, 0, 0)
if biz_close_time is None:
biz_close_time = time(18, 0, 0)
if cal is None:
cal = Calendar(weekdays=['Saturday', 'Sunday'])
open_delta = relativedelta(hour=biz_open_time.hour, minute=biz_open_time.minute)
end_delta = relativedelta(hour=biz_close_time.hour, minute=biz_close_time.minute)
start_date = pd.to_datetime(start_date)
end_date = pd.to_datetime(end_date)
_end_date = max(start_date, end_date)
_start_date = min(start_date, end_date)
start_date = _start_date
end_date = _end_date
start_date = (
start_date if cal.isbizday(start_date) else cal.following(start_date) open_delta
)
end_date = (
end_date if cal.isbizday(end_date) else cal.preceding(end_date) end_delta
)
if end_date < start_date:
return 0.00
start_date_biz = max(start_date, start_date open_delta)
end_first_day = start_date_biz end_delta
end_date_biz = min(
end_date,
end_date end_delta
)
start_last_day = end_date_biz open_delta
if start_last_day > end_date:
end_date_biz = start_last_day
if end_first_day < start_date:
end_first_day = start_date_biz
if end_first_day.date() == end_date_biz.date():
return (end_date_biz - start_date_biz).seconds / 3600
return (
(end_first_day - start_date_biz).seconds
(end_date_biz - start_last_day).seconds
(
max((len(list(cal.seq(start_date, end_date))) - 2), 0)
* (end_first_day - (start_date open_delta)).seconds
)
) / 3600
Before running the preceding code, you need to install the following packages, if you don't already have them:
pip install holidays bizdays
Link to both packages' documentation:
Examples
Here's how you can use compute_bizhours_diff:
import pandas as pd
import holidays as pyholidays
from datetime import time
from bizdays import Calendar
# OPTIONAL: define custom start, and end to your business hours.
biz_open_time = time(9, 0, 0)
biz_close_time = time(18, 0, 0)
# Define your start, and end dates.
start_date = pd.to_datetime('2022-09-07 04:48:00')
end_date = pd.to_datetime('2022-09-10 15:55:00')
# Create a list of holidays, and create a Calendar instance.
BR_holiday_list = pyholidays.BR(years={start_date.year, end_date.year}, state='RJ')
# For German holidays, you can use something like:
German_holiday_list = pyholidays.Germany(years={start_date.year, end_date.year})
# Define the Calendar instance. Here, we're using the German holidays, and excluding Saturday, and Sunday from weekdays.
cal = Calendar(holidays=German_holiday_list, weekdays=['Saturday', 'Sunday'])
# Finally, compute the total number of working hours between your two dates:
compute_bizhours_diff(start_date, end_date, cal=cal)
# Returns: 27.0
You can also use with a pandas dataframe, using apply:
df['working_hours_delta'] = df.apply(lambda row: compute_bizhours_diff(row[START_DATE_COLNAME], row[END_DATE_COLNAME], cal=cal), axis=1)
Notes
The function compute_bizhours_diff is far from perfect. Before using it in any production environment, or for any serious use-case, I strongly recommend refactoring it.
CodePudding user response：
Thank you for your suggestion. I have tried your method, i have also defined calendar instance. Later i was getting 'relativedelta' error which i have somehow solved by 'dateutil'. Now i am at final stage to compute business-hour difference between two columns.
`de_holidays = pyholidays.Germany()
cal = Calendar(holidays=de_holidays, weekdays=['Saturday', 'Sunday'])
df['rp'] = df.apply(lambda row: compute_bizhours_diff(row['Resolved'], row['Pending'], cal=cal, biz_open_time = time(8, 0, 0), biz_close_time = time(17, 0, 0)), axis=1)`
Now i am getting error about month number, which can not be nan. I have also attached photo of errors.
