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Calculation of business working hour in python

I would like to write a function that calculate working business hours in python, to do that I don't like to define a class and use python ready function to calculate.
I tried with following code but the code is not working well. I need to modify the code and change it for the hour instead of minutes too.
Do you have any suggestion?
def getminutes(datetime1,datetime2,worktiming=[9, 17]):
day_hours = (worktiming[1]-worktiming[0])
day_minutes = day_hours * 60 # minutes in a work day
weekends=[6, 7]
# Set initial default variables
dt_start = datetime1.datetime # datetime of start
dt_end = datetime2.datetime # datetime of end
worktime_in_seconds = 0
if dt_start.date() == dt_end.date():
# starts and ends on same workday
full_days = 0
if dt_start in [6, 7]:
return 0
else:
if dt_start.hour < worktiming[0]:
# set start time to opening hour
dt_start = datetime.datetime(
year=dt_start.year,
month=dt_start.month,
day=dt_start.day,
hour=worktiming[0],
minute=0)
if dt_start.hour >= worktiming[1] or \
dt_end.hour < worktiming[0]:
return 0
if dt_end.hour >= worktiming[1]:
dt_end = datetime.datetime(
year=dt_end.year,
month=dt_end.month,
day=dt_end.day,
hour=worktiming[1],
minute=0)
worktime_in_seconds = (dt_end-dt_start).total_seconds()
elif (dt_end-dt_start).days < 0:
# ends before start
return 0
else:
# start and ends on different days
current_day = dt_start # marker for counting workdays
while not current_day.date() == dt_end.date():
if not is_weekend(current_day):
if current_day == dt_start:
# increment hours of first day
if current_day.hour < worktiming[0]:
# starts before the work day
worktime_in_seconds += day_minutes*60 # add 1 full work day
elif current_day.hour >= worktiming[1]:
pass # no time on first day
else:
# starts during the working day
dt_currentday_close = datetime.datetime(
year=dt_start.year,
month=dt_start.month,
day=dt_start.day,
hour= worktiming[1],
minute=0)
worktime_in_seconds += (dt_currentday_close
- dt_start).total_seconds()
else:
# increment one full day
worktime_in_seconds += day_minutes*60
current_day += datetime.timedelta(days=1) # next day
# Time on the last day
if not is_weekend(dt_end):
if dt_end.hour >= worktiming[1]: # finish after close
# Add a full day
worktime_in_seconds += day_minutes*60
elif dt_end.hour < worktiming[0]: # close before opening
pass # no time added
else:
# Add time since opening
dt_end_open = datetime.datetime(
year=dt_end.year,
month=dt_end.month,
day=dt_end.day,
hour=worktiming[0],
minute=0)
worktime_in_seconds += (dt_end-dt_end_open).total_seconds()
return int(worktime_in_seconds / 60)
How can I modify the code that works with the following input ?
getminutes(2019-12-02 09:30:00,2019-12-07 12:15:00,worktiming=[9, 17])

You can use pd.bdate_range(datetime1, datetime2) to compute the number of working days. When converting worktiming to a pandas datetime, it is easy to compute the difference (in seconds) between the two datetimes:
import pandas as pd
datetime1 = "2019-12-02 09:30:00"
datetime2 = "2019-12-07 12:15:00"
def getminutes(datetime1, datetime2, worktiming=[9, 17]):
d1 = pd.to_datetime(datetime1)
d2 = pd.to_datetime(datetime2)
wd = pd.bdate_range(d1, d2) # working days
day_hours = (worktiming[1] - worktiming[0])
day_minutes = day_hours * 60 # minutes in a work day
day_seconds = day_minutes * 60 # seconds in a work day
full_days = len(wd)
day1 = datetime1[:10]
day2 = datetime2[:10]
dt1 = pd.to_datetime(day1 + " " + str(worktiming[0]) + ":00")
dt2 = pd.to_datetime(day2 + " " + str(worktiming[1]) + ":00")
ex1, ex2 = 0, 0
if day1 in wd:
ex1 = max(pd.Timedelta(d1 - dt1).seconds, 0)
if day2 in wd:
ex2 = max(pd.Timedelta(dt2 - d2).seconds, 0)
total_seconds = full_days * day_seconds - ex1 - ex2
total_minutes = total_seconds / 60
total_hours = total_minutes / 60
return int(total_minutes)
print(getminutes(datetime1, datetime2))
Output: 2370

datetime.datetime.strptime("%d/%m/%Y %H:%M:%S").timetuple().tm_hour

Here, you can see my code :
import datetime
DateS = datetime.datetime.strptime('30/03/2019 00:00:00',"%d/%m/%Y %H:%M:%S").timetuple().tm_hour
DateR = datetime.datetime.strptime('15/09/2019 00:00:00',"%d/%m/%Y %H:%M:%S").timetuple().tm_hour
ETP=ET0.copy()
for i in range(8760):
if i >= (DateS - 1) and i <= (DateR - 1) :
ETP[i] = ET0[i] * DB2.Kc[0]
else:
ETP[i] = ET0[i] * DB2.SolNu[0]
ETP[0] = 0
I want to get:
Date S = 2136 (day89 * 24hour = 2135)
Date R = 6192 (day258 * 24hour = 6192)
But I get:
Date S = 0
Date R = 0
When I use:
DateS = datetime.datetime.strptime('30/03/2019-00:00:00','%d/%m/%Y-%H:%M:%S').timetuple().tm_yday
DateR = datetime.datetime.strptime('15/09/2019-23:00:00','%d/%m/%Y-%H:%M:%S').timetuple().tm_yday
I get:
Date S = 89
Date R = 258
But I would like to get the 'hour step time'...
I try different datetime with strptime and timetupl but I never success.

try this code with tm_yday instead of tm_hour
DateS = datetime.datetime.strptime('30/03/2019-00:00:00','%d/%m/%Y-%H:%M:%S').timetuple().tm_yday * 24
DateR = datetime.datetime.strptime('15/09/2019-23:00:00','%d/%m/%Y-%H:%M:%S').timetuple().tm_yday * 24
and you will have the output you ask for.

Python - Generating url for date range - Outputs 0 result if year is lower than 2019

I had a struggle for few days now and I really hope that someone here can help me.
I have tried everything to make this work for dates lower than year 2019 and it doesn't want to generate any date in range that is lower than year 2019.
You can try to change
tmpTo = '2018-05-31'
to
tmpTo = '2019-05-31'
and it will give output
Here is Demo code i made for preview that goes together with much more code:
from datetime import timedelta, date, datetime
import time
dateFrom = datetime.strptime('2019-01-01', "%Y-%m-%d")
dateTo = datetime.strptime('2019-01-10', "%Y-%m-%d")
plusDAYS = 0
todayPlus1 = dateFrom + timedelta(days=plusDAYS)
datesFromTo = []
visitUrl = []
def autoInput():
global dateTo
global dateFrom
tmpFrom = '2018-05-02'
tmpTo = '2018-05-31'
dateFrom = datetime.strptime(tmpFrom, "%Y-%m-%d")
dateTo = datetime.strptime(tmpTo, "%Y-%m-%d")
separator = ' * '
pausa0 = 2
pausa = 3
autoInput()
print(dateFrom)
print(dateTo)
print('----------------------------')
print('----------------------------')
def createDatesFromTo():
global datesFromTo
global todayPlus1
global plusDAYS
while dateTo >= todayPlus1:
todayPlus1 = dateFrom + timedelta(days=plusDAYS)
datesFromTo.append(todayPlus1.strftime('%Y-%m-%d'))
plusDAYS += 1
todayPlus1 = dateFrom + timedelta(days=plusDAYS)
createDatesFromTo()
print('PRINTS EVERYTHING LIST IS CONTAINING')
print(datesFromTo)
print('----------------------------')
print('----------------------------')
def createUrlForDates():
global visitUrl
#for i in range(len(datesFromTo)): # ORIGINAL
for i in range(5): # DEMO 5 ITEMS
tmp = datesFromTo[i]+"&tom="+datesFromTo[i]+""
visitUrl.append(tmp)
createUrlForDates()
for i in visitUrl:
print('URL RESULT ---> www.TEST.com/?from='+i)

Change to the previous day date in python

I'm working on my python script to get the strings from the button objects so I can use it to set the date formats with the time that I got from the strings to store it in the lists. When I get the strings from the button objects, I want to set the date for each string, example: 29/08/2017 11:30PM, 30/08/2017 12:00AM, 30/08/2017 12:30AM.
When I try this:
if day_date >= 0 and day_date <= 6:
if epg_time3 == '12:00AM':
if day_date > 0:
currentday = datetime.datetime.now() + datetime.timedelta(days = 0)
nextday = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
if currentday != nextday:
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
elif currentday == nextday:
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day - 1)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day - 1)
It will show the output:
self.epg_time_1
['29/08/2017 11:00PM']
self.epg_time_2
['29/08/2017 11:30PM']
self.epg_time_3
['29/08/2017 12:00AM']
When I'm calling the EPG_Times function again, it will show the output like this:
self.epg_time_1
['30/08/2017 11:00PM']
self.epg_time_2
['30/08/2017 11:30PM']
self.epg_time_3
['30/08/2017 12:00AM']
It should be:
self.epg_time_1
['30/08/2017 11:00PM']
self.epg_time_2
['30/08/2017 11:30PM']
self.epg_time_3
['31/08/2017 12:00AM']
As you can see the time 12:00AM is the next day so I want to set it to 31 not 30. I have changed from days = self.program_day + 1 to days = self.program_day - 1, but when the strings show 11:00PM, 11:30PM and 12:00AM from the variables epg_time_1, epg_time_2 and epg_time_3, it will show the output like this:
self.epg_time_1
['30/08/2017 11:00PM']
self.epg_time_2
['30/08/2017 11:30PM']
self.epg_time_3
['30/08/2017 12:00AM']
Here is the full code:
self.program_day = list()
def EPG_Times(self):
self.epg_time_1 = list()
self.epg_time_2 = list()
self.epg_time_3 = list()
epg_time1 = str(self.getControl(344).getLabel())
epg_time2 = str(self.getControl(345).getLabel())
epg_time3 = str(self.getControl(346).getLabel())
day_date = self.program_day
day = ''
month = ''
year = ''
if day_date >= 0 and day_date <= 6:
if epg_time3 == '12:00AM':
if day_date == 0:
if epg_time1 == '12:00AM':
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
else:
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
else:
currentday = datetime.datetime.now() + datetime.timedelta(days = 0)
nextday = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
if currentday != nextday:
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day + 1)
elif currentday == nextday:
epg_time_1 = datetime.datetime.now() + datetime.timedelta(days = self.program_day)
epg_time_2 = datetime.datetime.now() + datetime.timedelta(days = self.program_day - 1)
epg_time_3 = datetime.datetime.now() + datetime.timedelta(days = self.program_day - 1)
epg1_day = epg_time_1.strftime("%d")
epg1_month = epg_time_1.strftime("%m")
epg1_year = epg_time_1.strftime("%Y")
epg2_day = epg_time_2.strftime("%d")
epg2_month = epg_time_2.strftime("%m")
epg2_year = epg_time_2.strftime("%Y")
epg3_day = epg_time_2.strftime("%d")
epg3_month = epg_time_2.strftime("%m")
epg3_year = epg_time_2.strftime("%Y")
half_hour = str(epg1_day + "/" + epg1_month + "/" + epg1_year + " " + epg_time1)
one_hour = str(epg2_day + "/" + epg2_month + "/" + epg2_year + " " + epg_time2)
one_hour_half = str(epg3_day + "/" + epg3_month + "/" + epg3_year + " " + epg_time3)
#Store the times and date in the list....
self.epg_time_1.append(half_hour)
self.epg_time_2.append(one_hour)
self.epg_time_3.append(one_hour_half)
What I'm expected the code to do is to change to the previous day date for each string that I get from the button objects when I call the EPG_time(self) function. If the epg_time_1 and epg_time_2 show the strings 11:00PM and 11:30PM, I want to set the time and date to 29/08/2017 11:00PM for epg_time_1 and 29/08/2017 11:30PM for the epg_time_2. If the epg_time_3 show the string 12:00AM then I want to add it to the next day date with the time 30/08/2017 12:00AM.
In the next 24 hours if the epg_time_1 and epg_time_2 show the strings 11:00PM and 11:30PM, I want to set the time and date to 30/08/2017 11:00PM for epg_time_1 and 30/08/2017 11:30PM for the epg_time_2. If the epg_time_3 show the string 12:00AM then I want to set to the next day date with the time 1/09/2017 12:00AM
If the epg_time_1 and epg_time_2 show the strings 11:30PM and 12:00AM, I want to change to the previous date for epg_time_1 which it is 29/08/2017 11:30PM and 30/08/2017 12:00AM. It will be depends on the time and date when I have stored the strings in the list.
Can you please tell me an example how I could use to change the date to the previous date and add to the next day date using in python?

There's a lot of text in your question that makes it hard to pinpoint the issue exactly. However, it appears to boil down to adding a variable number of days to a particular date and ensuring that the month is also updated (if necessary).
You should use the datetime.datetime.strptime() method to convert your dates to datetimes, which makes it trivial to add timedelta (you use both timedelta and strftime but miss this crucial method) and then just convert back to a string.
import datetime as dt
def add_days(datelist, days_to_add):
# Convert the strings to datetime objects
datelist = [dt.datetime.strptime(item, '%d/%m/%Y %I:%M%p')
for item in datelist]
# Add a variable number of days (can be negative) to the datetimes
mod_dates = [item + dt.timedelta(days=days_to_add) for item in datelist]
# Convert back to strings
str_dates = [dt.datetime.strftime(item, '%d/%m/%Y %I:%M%p')
for item in mod_dates]
return str_dates
# This is the list right at the top of your question
a = ['29/08/2017 11:30PM', '30/08/2017 12:00AM', '30/08/2017 12:30AM']
print("Start:")
print(a)
b = add_days(a, 1)
print("After 1 call:")
print(b)
c = add_days(b, 1)
print("After 2 calls:")
print(c)

Python difference between two dates

For some reason I'm really stumped with this (relativity) question.
How to calculate the difference between two dates. I want to do this without using modules. But for some reason my code isn't outputting the correct answer.
This is my thought process:
If asked calculate the # of days between Dec 10th 2014 and Feb 2nd 2015.
First find the number of days left in Dec from the 10th on (31 - 10)
= 21 days
Find the number of months between Dec and Feb ( aka Jan) add the
number days in that month = 31 days
Add the Days left in Dec (21) + the days in between the months (31) + the days in the last month (2) = 54 days.
Then check for anomalies ie Leap Year etc.
This is my function:
def Calculate_Date (year1, month1, day1, year2, month2, day2):
"""
This function takes to dates (year/month/day) and returned the
difference between the dates
"""
#Create a dict for the # of days in each month
month_days = {1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}
days_left_in_month1 = month_days[month1] - day1
days_left_in_year1 =0
days_into_year2 =0
days_between_year1_and_year2= 0
difference_in_days = 0
# Find the number days left in year one
i = month1
days_left_in_year = []
while i <= 12:
days = month_days[i]
days_left_in_year.append(days)
i = i + 1
days_left_in_year1 = (sum(days_left_in_year)) - day1
# Find the number days into year two
i = 1
days_into_year = []
while i <= month2:
days = month_days[i]
days_into_year.append(days)
i = i + 1
days_into_year2 = sum(days_into_year) - day2
#find the differernce in years
days_between_year1_and_year2 = (year2 - year1) * 365
#Check if its a leap year
leap_year = False
while True:
if float(year1 % 4) == 0:
if float(year1 % 100) != 0:
leap_year = True
break
if float(year1 % 100) == 0:
if float(year1 % 400) ==0:
leap_year = True
break
else:
break
#test output
print "The number of days left in the year One are %r " % days_left_in_year1
print "The number of days into the year Two are %r " % days_into_year2
print "The number of days between the years are %r " % days_between_year1_and_year2
#add an increment if leap year was true
if leap_year == True:
difference_in_days = days_left_in_year1 + days_into_year2 + days_between_year1_and_year2 + 1
else:
difference_in_days = days_left_in_year1 + days_into_year2 + days_between_year1_and_year2
return difference_in_days
print Calculate_Date(2011,6,30,2012,06,30)

Instead of doing date2 - date1, you might find it simpler to do (date2 - x) - (date1 - x) where x is an easy-to-handle date, ie "Jan 0" of year1.
Let's define a couple of functions:
def days_in_month(year, month):
"""
Return number of days in the specified month (28, 29, 30, or 31)
"""
if month == 2: # February
if not (year % 400):
return 29
elif not (year % 100):
return 28
elif not (year % 4):
return 29
else:
return 28
elif month in {1, 3, 5, 7, 8, 10, 12}:
return 31
else:
return 30
def days_in_year(year):
"""
Return the number of days in the specified year (365 or 366)
"""
return 337 + days_in_month(year, 2)
def days_this_year(year, month, day):
"""
Return the number of days so far this year
"""
return sum(days_in_month(year, m) for m in range(1, month)) + day
def year_days_since(base_year, this_year):
"""
Return the number of days from the start of base_year to the start of this_year
"""
if base_year > this_year:
raise ValueError("base_year must be <= this_year")
elif base_year == this_year:
return 0
else:
return sum(days_in_year(y) for y in range(base_year, this_year))
then the difference between two dates becomes:
def date_diff(y1, m1, d1, y2, m2, d2):
x = min(y1, y2) # base date
days1 = year_days_since(x, y1) + days_this_year(y1, m1, d1)
days2 = year_days_since(x, y2) + days_this_year(y2, m2, d2)
return days2 - days1
and because of the symmetry in this answer it will also happily do negative differences:
date_diff(2001, 1, 3, 2002, 2, 5) # => 398 == 365 + 31 + 2
date_diff(2002, 2, 5, 2001, 1, 3) # => -398

In case this is a real code, and not a school assignment, this is the way I'd do it:
from datetime import date
def date_diff(y1, m1, d1, y2, m2, d2):
return (date(y2, m2, d2) - date(y1, m1, d1)).days