我试图找到一种非常干净的方法来计算两个日期之间的小时数，不包括周末和某些假期 。
我发现BusinessHours（https://pypi.python.org/pypi/BusinessHours/1.01）包可以做到这一点。但是我没有找到任何关于如何使用包（实际上是语法）的指令，特别是如何输入假期。 我找到了包的原始代码（https://github.com/dnel/BusinessHours/blob/master/BusinessHours.py），但仍不太确定。 我猜它可能是这样的：
date1 = pd.to_datetime('2017-01-01 00:00:00')
date2 = pd.to_datetime('2017-01-22 12:00:00')
import BusinessHour
gethours(date1, date2, worktiming=[8, 17], weekends=[6, 7])
不过，我在哪里可以输入假期？如果我不想要排除非办公时间，我会将worktiming调整为worktiming=[0,23]吗？
任何人都知道如何使用这个包请告诉我。我很感激。
P / s：我知道numpy中的命令可以获得两个日期（busday_count）之间的工作日数，但是没有命令可以在小时中获得结果。 pandas或numpy中可以完成任务的任何其他命令也受到欢迎。
谢谢
答案 0 :(得分：0)
此软件包1.2的最新pip安装在第51行有一个错误，“extraday”需要更改为“extradays”。
我也一直在网上搜索一些可行的代码来计算营业时间和工作日。这个软件包有一点点tweeking但是当你启动并运行它时工作得很好。
这就是我在笔记本中的内容：
#import BusinessHours
from BusinessHours import BusinessHours as bh
import numpy as np
import pandas as pd
from pandas import Series, DataFrame
date1 = pd.to_datetime('2017-01-01 00:00:00')
date2 = pd.to_datetime('2017-01-22 12:00:00')
bh(date1, date2, worktiming=[8, 17], weekends=[6, 7]).gethours()
这也在源代码中：
'''
holidayfile - A file consisting of the predetermined office holidays.
Each date starts in a new line and currently must only be in the format
dd-mm-yyyy
'''
希望这有帮助
答案 1 :(得分：0)
在PyPi中尝试这个名为business-hour的包
from business_duration import businessDuration
import pandas as pd
from datetime import time,datetime
import holidays as pyholidays
startdate = pd.to_datetime('2017-01-01 00:00:00')
enddate = pd.to_datetime('2017-01-22 12:00:00')
holidaylist = pyholidays.Australia()
unit='hour'
#By default Saturday and Sunday are excluded
print(businessDuration(startdate,enddate,holidaylist=holidaylist,unit=unit))
Output: 335.99611
holidaylist:
{datetime.date(2017, 1, 1): "New Year's Day",
datetime.date(2017, 1, 2): "New Year's Day (Observed)",
datetime.date(2017, 1, 26): 'Australia Day',
datetime.date(2017, 3, 6): 'Canberra Day',
datetime.date(2017, 4, 14): 'Good Friday',
datetime.date(2017, 4, 15): 'Easter Saturday',
datetime.date(2017, 4, 17): 'Easter Monday',
datetime.date(2017, 4, 25): 'Anzac Day',
datetime.date(2017, 6, 12): "Queen's Birthday",
datetime.date(2017, 9, 26): 'Family & Community Day',
datetime.date(2017, 10, 2): 'Labour Day',
datetime.date(2017, 12, 25): 'Christmas Day',
datetime.date(2017, 12, 26): 'Boxing Day'}
答案 2 :(得分：0)
我从源头重新使用了代码，我汇编了这段似乎可行的代码（适用于英国假期），但我希望就如何改进它发表意见。 我知道这不是特别优雅，但可以帮助某人。 顺便说一句，我想找到一种方法将假日库中的日历插入此库中。
无论如何，目前它不需要太多的库，只需要熊猫和日期时间，这可能是一个加分。
import pandas as pd
import datetime
from pandas.tseries.offsets import CDay
from pandas.tseries.holiday import (
AbstractHolidayCalendar, DateOffset, EasterMonday,
GoodFriday, Holiday, MO,
next_monday, next_monday_or_tuesday)
# This function will calculate the number of working minutes by first
# generating a time series of business days. Then it will calculate the
# precise working minutes for the start and end date, and use the total
# working hours for each day in-between.
def count_mins(starttime,endtime, bus_day_series, bus_start_time,bus_end_time):
mins_in_working_day=(bus_end_time-bus_start_time)*60
# now we are going to take the series of business days (pre-calculated)
# and sub select the period provided as argument of the function
# we could do the calculation of that "calendar" in the function itself
# but to improve performance, we calculate it separately and then we c
# call the function with that series as argument, provided the dates
# fall within the calculated range, of course
days = bus_day_series[starttime.date():endtime.date()]
daycount = len(days)
if len(days)==0:
return 0
else:
first_day_start = days[0].replace(hour=bus_start_time, minute=0)
first_day_end = days[0].replace(hour=bus_end_time, minute=0)
first_period_start = max(first_day_start, starttime)
first_period_end = min(first_day_end, endtime)
if first_period_end<=first_period_start:
first_day_mins=0
else:
first_day_sec=first_period_end - first_period_start
first_day_mins=first_day_sec.seconds/60
if daycount == 1:
return first_day_mins
else:
last_period_start = days[-1].replace(hour=bus_start_time, minute=0)
#we know the last day will always start in the bus_start_time
last_day_end = days[-1].replace(hour=bus_end_time, minute=0)
last_period_end = min(last_day_end, endtime)
if last_period_end<=last_period_start:
last_day_mins=0
else:
last_day_sec=last_period_end - last_period_start
last_day_mins=last_day_sec.seconds/60
middle_days_mins=0
if daycount>2:
middle_days_mins=(daycount-2)*mins_in_working_day
return first_day_mins + last_day_mins + middle_days_mins
# Calculates the date series with all the business days
# of the period we are interested on
class EnglandAndWalesHolidayCalendar(AbstractHolidayCalendar):
rules = [
Holiday('New Years Day', month=1, day=1, observance=next_monday),
GoodFriday,
EasterMonday,
Holiday('Early May bank holiday',
month=5, day=1, offset=DateOffset(weekday=MO(1))),
Holiday('Spring bank holiday',
month=5, day=31, offset=DateOffset(weekday=MO(-1))),
Holiday('Summer bank holiday',
month=8, day=31, offset=DateOffset(weekday=MO(-1))),
Holiday('Christmas Day', month=12, day=25, observance=next_monday),
Holiday('Boxing Day',
month=12, day=26, observance=next_monday_or_tuesday)
]
# From this point its how we use the function
# Here we hardcode a start/end date to create the list of business days
cal = EnglandAndWalesHolidayCalendar()
dayindex = pd.bdate_range(datetime.date(2019,1,1),datetime.date.today(),freq=CDay(calendar=cal))
day_series = dayindex.to_series()
# Convenience function to simplify how we call the main function
# It will take a pre calculated day_series.
def bus_hr(ts_start, ts_end, day_series ):
BUS_START=8
BUS_END=20
minutes = count_mins(ts_start, ts_end, day_series, BUS_START, BUS_END)
return int(round(minutes/60,0))
#A set of checks that the function is working properly
assert bus_hr( pd.Timestamp(2019,9,30,6,1,0) , pd.Timestamp(2019,10,1,9,0,0),day_series) == 13
assert bus_hr( pd.Timestamp(2019,10,3,10,30,0) , pd.Timestamp(2019,10,3,23,30,0),day_series)==10
assert bus_hr( pd.Timestamp(2019,8,25,10,30,0) , pd.Timestamp(2019,8,27,10,0,0),day_series) ==2
assert bus_hr( pd.Timestamp(2019,12,25,8,0,0) , pd.Timestamp(2019,12,25,17,0,0),day_series) ==0
assert bus_hr( pd.Timestamp(2019,12,26,8,0,0) , pd.Timestamp(2019,12,26,17,0,0),day_series) ==0
assert bus_hr( pd.Timestamp(2019,12,27,8,0,0) , pd.Timestamp(2019,12,27,17,0,0),day_series) ==9
assert bus_hr( pd.Timestamp(2019,6,24,5,10,44) , pd.Timestamp(2019,6,24,7,39,17),day_series)==0
assert bus_hr( pd.Timestamp(2019,6,24,5,10,44) , pd.Timestamp(2019,6,24,8,29,17),day_series)==0
assert bus_hr( pd.Timestamp(2019,6,24,5,10,44) , pd.Timestamp(2019,6,24,10,0,0),day_series)==2
assert bus_hr(pd.Timestamp(2019,4,30,21,19,0) , pd.Timestamp(2019,5,1,16,17,56),day_series)==8
assert bus_hr(pd.Timestamp(2019,4,30,21,19,0) , pd.Timestamp(2019,5,1,20,17,56),day_series)==12