What I am trying to do is see if date is in 1 week from currdate
from datetime import datetime, timedelta
import yagmail
year = datetime.now().year
month = datetime.now().month
day = datetime.now().day
currdate = '{}-{}-{}'.format(year, month, day)
currdate = datetime.strptime(currdate, '%Y-%m-%d')
date = '2018-04-01'
days = currdate - timedelta(int(date[-2:]))
days = str(days)
print(days)
if days[8:11] == '07':
yag = yagmail.SMTP("##########gmail.com", "######")
content = ['One Of Your Homework\'s Is Due In 1 Week!']
yag.send('###########gmail.com', 'Homework Due Soon!', content)
else:
print('It Isn\'t')
But it prints:
2018-04-07 00:00:00
It Isnt't
And I'm not sure why. Because days[8:11] is 07.
It is not 07. It's 07 (note the trailing space).
The following change will work:
if int(days[8:11]) == 7:
I'd create a function that you pass the date as a string. Something like this:
import datetime
def check_if_less_than_seven_days(x):
d = datetime.datetime.strptime(x, "%Y-%m-%d") # Add .date() if hour doesn't matter
now = datetime.datetime.now() # Add .date() if hour doesn't matter
return (d - now).days < 7
if check_if_less_than_seven_days("2018-04-18"):
print('Do something') # This will not print
if check_if_less_than_seven_days("2018-04-14"):
print('Do something') # This will print
Will print:
'Do something'
I suppose your first line when you initiate datetime.now() three times is just for testing purposes but dont do this as it could end up over different days (if you run this exactly at the milliseconds around midnight..) this will work better in that regard.
now = datetime.datetime.now()
year = now.year
month = now.month
day = now.day
Anyway, read up on datetime timedelta. Just make you logic around that.
https://docs.python.org/3/library/datetime.html#timedelta-objects
import datetime
test_date_string = "2018-04-10"
d = datetime.datetime.strptime(test_date_string, "%Y-%m-%d")
now = datetime.datetime.now()
delta = d - now
elif delta.days < 7:
print("You have less then 7 days to go")
For days[8:11] you get the following output
>>> days[8:11]
'08 '
So you should use days[8:10]=='07' in case you want to use the same method,as it wont have extra space at the end.
>>> days[8:10]
'08'
so you should use
if days[8:10] == '07':
Related
How to remove 0:00:00 from the output of datetime.date() function? For example:
import datetime
now = datetime.datetime.now()
year1 = now.strftime("%Y")
month2 = now.strftime("%m")
day3 = now.strftime("%d")
year = int(year1)
month = int(month2)
day = int(day3)
first_day = datetime.date(2021,8,1)
second_day = datetime.date(year,month,day)
daysleft = first_day - second_day
print(daysleft)
I get the output:
9 days, 0:00:00
If you didn't understand the question title, My main goal is to remove the 0:00:00 before the period (.). I've seen many other questions like this (in stack overflow/exchange and other websites), but it was nothing in python coding language.
You can get just the days by asking for the .days attribute of a datetime.timedelta object. E.g.:
print('{} days'.format(daysleft.days))
If your only goal is printing the output and you don't care about maintaining daysleft as a datetime object, you can always convert it to a string and .split() it like this:
print(str(daysleft).split(",")[0])
datetime has many useful functions and values
To get only days you can daysleft.days
print(f'{daysleft.days} days')
but you can also reduce other code to
now = datetime.datetime.now()
daysleft = first_day - now.date()
or
today = datetime.date.today()
daysleft = first_day - today
BTW:
If you need year, month, day as numbers then you can get it as
year = now.year
month = now.month
day = now.day
hour = now.hour
minute = now.minute
second = now.second
ms = now.microsecond
weekday = now.weekday() # 0 = monday, ..., 6 = sunday
weekday = now.isoweekday() # 1 = monday, ..., 7 = sunday
Other useful function timedelta
one_day = datetime.timedelta(days=1)
today = datetime.date.today()
yesterday = today - one_day
tomorrow = today + one_day
day_after_tomorrow = today + 2*one_day
Minimal working code with tkinter
import datetime
first_day = datetime.date(2021, 8, 1)
now = datetime.datetime.now()
daysleft = first_day - now.date()
today = datetime.date.today()
daysleft = first_day - today
print(f'{daysleft.days} days')
# ---
import tkinter as tk
root = tk.Tk()
lbl = tk.Label(root, text=f'{daysleft.days} days till {first_day}')
lbl.pack()
root.mainloop()
I tried to develop a Python function that determines the difference between two datetime objects. I need an algorithm that calculates the number of hours per day. Is there a built-in function for this?
from datetime import datetime, timedelta, date
def getHoursByDay(dateA, dateB):
...
dateA = datetime.strptime('2018-09-01 09:00:00', '%Y-%m-%d %H:%M:%S')
dateB = datetime.strptime('2018-09-03 11:30:00', '%Y-%m-%d %H:%M:%S')
hours = getHoursByDay(dateA, dateB)
print hours
# {
# '2018-09-01': 15,
# '2018-09-02': 24,
# '2018-09-03': 11.5,
# }
There is no built-in function, though it is very simple to build one.
from datetime import datetime, timedelta, time
def deltaByDay(dateA, dateB):
dateAstart = datetime.combine(dateA, time())
dateBstart = datetime.combine(dateB, time())
result = {}
oneday = timedelta(1)
if dateAstart == dateBstart:
result[dateA.date()] = dateB - dateA
else:
nextDate = dateAstart + oneday
result[dateA.date()] = nextDate - dateA
while nextDate < dateBstart:
result[nextDate.date()] = oneday
nextDate += oneday
result[dateB.date()] = dateB - dateBstart
return result
def deltaToHours(delta, ndigits=None):
return delta.days * 24 + round(delta.seconds / 3600.0, ndigits)
dateA = datetime.strptime('2018-09-01 09:00:00', '%Y-%m-%d %H:%M:%S')
dateB = datetime.strptime('2018-09-03 11:30:00', '%Y-%m-%d %H:%M:%S')
deltas = deltaByDay(dateA, dateB);
output = {k.strftime('%Y-%m-%d'): deltaToHours(v, 1) for k, v in deltas.items()}
print(output)
# => {'2018-09-01': 15.0, '2018-09-02': 24.0, '2018-09-03': 11.5}
The built in timedelta functions would be able to get you the total days, and the remaining hours difference. If you want the output specifically in that dictionary format posted you would have to create it manually like this:
from datetime import datetime, time, timedelta
def getHoursByDay(dateA, dateB):
if dateA.strftime("%Y-%m-%d") == dateB.strftime("%Y-%m-%d"):
return {dateA.strftime("%Y-%m-%d"): abs(b-a).seconds / 3600}
result = {}
delta = dateB - dateA
tomorrow = dateA + timedelta(days=1)
day1 = datetime.combine(tomorrow, time.min) - dateA
result[dateA.strftime("%Y-%m-%d")] = day1.seconds / 3600
for day in range(1, delta.days):
result[(dateA + timedelta(days=day)).strftime("%Y-%m-%d")] = 24
priorday = dateB - timedelta(days1)
lastday = dateB - datetime.combine(priorday, time.min)
result[dateB.strftime("%Y-%m-%d")] = lastday.seconds / 3600
return result
Essentially this function calculates the first day and the last day values, then populates all the days in between with 24.
There is a kind of simple way to do this.
hours = (dateA - dateB).hours
I've used this to caclulate a difference in days.
https://docs.python.org/2/library/datetime.html#datetime.timedelta
I have a method which returns the number of days within a specified range and excluding some specific days like Friday. here is an example if you take out friday and thursday, from 2016-8-6 to 2016-9-6 the result will be 8 days holiday and 24 working day. in case i want to do the reverse operation how do i find the end date (2016-9-6) if i have only working days and start date.
from datetime import datetime, timedelta
def measure_workingdays(start_date, end_date, off_days):
format = "%Y-%m-%d"
if not isinstance(start_date, datetime):
start_date = datetime.strptime(start_date, format)
if not isinstance(end_date, datetime):
end_date = datetime.strptime(end_date, format)
total_days = (end_date - start_date).days + 1 # + 1 Because it count one day less
holiday = 0
start = start_date
for rec in range(total_days):
day = start.strftime("%a")
if day in off_days:
holiday += 1
start += timedelta(days=1)
print(holiday) # 8
working_days = total_days - holiday
print(working_days) # 24
start_date = "2016-8-6"
start_date = datetime.strptime(start_date, "%Y-%m-%d")
end_date = "2016-9-6"
end_date = datetime.strptime(end_date, "%Y-%m-%d")
off_day = ['Fri','Thu']
working_days = measure_weekdays(start_date, end_date, off_day)
Example of Reverse operation
def measure_weekdays_reverse(start_date, paid, off_days):
format = "%Y-%m-%d"
if not isinstance(start_date, datetime):
start_date = datetime.strptime(start_date, format)
holiday = 0
start = start_date
for rec in range(paid):
day = start.strftime("%a")
if day in off_days:
holiday += 1
start += timedelta(days=1)
print(holiday) # Output 6 instead of 8
last_paid_date = start + timedelta(days=holiday)
print(last_paid_date) # output 2016-09-05 insteaad of 2016-09-06
total_days = measure_weekdays_reverse(start_date, 24, ["Fri","Thu"])
The error is that you only loop a fix number of times (the number of paid days), so if you encounter holidays, you will in fact not iterate enough to find all the true paid days, which may still hide some holidays.
You can fix this by adding an inner loop on the holidays. Change this:
for rec in range(paid):
day = start.strftime("%a")
if day in off_days:
holiday += 1
start += timedelta(days=1)
print(holiday) # Output 6 instead of 8
last_paid_date = start + timedelta(days=holiday)
to this:
for rec in range(paid):
day = start.strftime("%a")
while day in off_days:
holiday += 1
start += timedelta(days=1)
day = start.strftime("%a")
last_paid_date = start
start += timedelta(days=1)
print(holiday)
I have time string 11:15am or 11:15pm.
I am trying to convert this string into UTC timezone with 24 hour format.
FROM EST to UTC
For example: When I pass 11:15am It should convert into 15:15 and when I pass 11:15pm then it should convert to 3:15.
I have this code which I am trying:
def appointment_time_string(time_str):
import datetime
a = time_str.split()[0]
# b = re.findall(r"[^\W\d_]+|\d+",a)
# c = str(int(b[0]) + 4) + ":" + b[1]
# print("c", c)
in_time = datetime.datetime.strptime(a,'%I:%M%p')
print("In Time", in_time)
start_time = str(datetime.datetime.strftime(in_time, "%H:%M:%S"))
print("Start TIme", start_time)
if time_str.split()[3] == 'Today,':
start_date = datetime.datetime.utcnow().strftime("%Y-%m-%dT")
elif time_str.split()[3] == 'Tomorrow,':
today = datetime.date.today( )
start_date = (today + datetime.timedelta(days=1)).strftime("%Y-%m-%dT")
appointment_time = str(start_date) + str(start_time)
return appointment_time
x = appointment_time_string(time_str)
print("x", x)
But this is just converting to 24 hour not to UTC.
To convert the time from 12 hours to 24 hours format, you may use below code:
from datetime import datetime
new_time = datetime.strptime('11:15pm', '%I:%M%p').strftime("%H:%M")
# new_time: '23:15'
In order to convert time from EST to UTC, the most reliable way is to use third party library pytz. Refer How to convert EST/EDT to GMT? for more details
Developed the following script using provided options/solutions to satisfy my requirement.
def appointment_time_string(time_str):
import datetime
import pytz
a = time_str.split()[0]
in_time = datetime.datetime.strptime(a,'%I:%M%p')
start_time = str(datetime.datetime.strftime(in_time, "%H:%M:%S"))
if time_str.split()[3] == 'Today,':
start_date = datetime.datetime.utcnow().strftime("%Y-%m-%d")
elif time_str.split()[3] == 'Tomorrow,':
today = datetime.date.today( )
start_date = (today + datetime.timedelta(days=1)).strftime("%Y-%m-%d")
appointment_time = str(start_date) + " " + str(start_time)
# print("Provided Time", appointment_time)
utc=pytz.utc
eastern=pytz.timezone('US/Eastern')
fmt='%Y-%m-%dT%H:%M:%SZ'
# testeddate = '2016-09-14 22:30:00'
test_date = appointment_time
dt_obj = datetime.datetime.strptime(test_date,'%Y-%m-%d %H:%M:%S')
dt_str = datetime.datetime.strftime(dt_obj, '%m/%d/%Y %H:%M:%S')
date=datetime.datetime.strptime(dt_str,"%m/%d/%Y %H:%M:%S")
date_eastern=eastern.localize(date,is_dst=None)
date_utc=date_eastern.astimezone(utc)
# print("Required Time", date_utc.strftime(fmt))
return date_utc.strftime(fmt)
Following this answer I tried to get the date for last Thursday of the current month. But my code doesn't get out of loop.
from datetime import datetime
from dateutil.relativedelta import relativedelta, TH
todayte = datetime.today()
cmon = todayte.month
nthu = todayte
while nthu.month == cmon:
nthu += relativedelta(weekday=TH(1))
#print nthu.strftime('%d%b%Y').upper()
Looking at the documentation of relativedelta
Notice that if the calculated date is already Monday, for example, using (0, 1) or (0, -1) won’t change the day.
If nthu is already Thursday then adding TH(1) or TH(-1) won't have any effect but result in the same date and that's why your loop is running infinitely.
I will assume maximum 5 weeks in a month and do it like following:
todayte = datetime.today()
cmon = todayte.month
for i in range(1, 6):
t = todayte + relativedelta(weekday=TH(i))
if t.month != cmon:
# since t is exceeded we need last one which we can get by subtracting -2 since it is already a Thursday.
t = t + relativedelta(weekday=TH(-2))
break
Based on Adam Smith's answer on How can I get the 3rd Friday of a month in Python?, you can get the date of the last Thursday of the current month as follows:
import calendar
import datetime
def get_thursday(cal,year,month,thursday_number):
'''
For example, get_thursday(cal, 2017,8,0) returns (2017,8,3)
because the first thursday of August 2017 is 2017-08-03
'''
monthcal = cal.monthdatescalendar(year, month)
selected_thursday = [day for week in monthcal for day in week if \
day.weekday() == calendar.THURSDAY and \
day.month == month][thursday_number]
return selected_thursday
def main():
'''
Show the use of get_thursday()
'''
cal = calendar.Calendar(firstweekday=calendar.MONDAY)
today = datetime.datetime.today()
year = today.year
month = today.month
date = get_thursday(cal,year,month,-1) # -1 because we want the last Thursday
print('date: {0}'.format(date)) # date: 2017-08-31
if __name__ == "__main__":
main()
You should pass 2 to TH instead of 1, as 1 doesn't change anything. Modify your code to:
while (nthu + relativedelta(weekday=TH(2))).month == cmon:
nthu += relativedelta(weekday=TH(2))
print nthu.strftime('%d-%b-%Y').upper()
# prints 26-MAY-2016
Note that I modified the loop's condition in order to break in the last occurrence of the day on the month, otherwise it'll break in the next month (in this case, June).
from datetime import datetime , timedelta
todayte = datetime.today()
cmon = todayte.month
nthu = todayte
while todayte.month == cmon:
todayte += timedelta(days=1)
if todayte.weekday()==3: #this is Thursday
nthu = todayte
print nthu
You can also use calendar package.
Access the calendar in the form of monthcalendar. and notice that, the Friday is the last day of a week.
import calendar
import datetime
now = datetime.datetime.now()
last_sunday = max(week[-1] for week in calendar.monthcalendar(now.year,
now.month))
print('{}-{}-{:2}'.format(now.year, calendar.month_abbr[now.month],
last_sunday))
I think this will be fastest perhaps:
end_of_month = datetime.datetime.today() + relativedelta(day=31)
last_thursday = end_of_month + relativedelta(weekday=TH(-1))
This code can be used in python 3.x for finding last Thursday of current month.
import datetime
dt = datetime.datetime.today()
def lastThurs(dt):
currDate, currMth, currYr = dt, dt.month, dt.year
for i in range(31):
if currDate.month == currMth and currDate.year == currYr and currDate.weekday() == 3:
#print('dt:'+ str(currDate))
lastThuDate = currDate
currDate += datetime.timedelta(1)
return lastThuDate
You can do something like this:
import pandas as pd
from dateutil.relativedelta import relativedelta, TH
expiry_type = 0
today = pd.datetime.today()
expiry_dates = []
if expiry_type == 0:
# Weekly expiry
for i in range(1,13):
expiry_dates.append((today + relativedelta(weekday=TH(i))).date())
else:
# Monthly expiry
for i in range(1,13):
x = (today + relativedelta(weekday=TH(i))).date()
y = (today + relativedelta(weekday=TH(i+1))).date()
if x.month != y.month :
if x.day > y.day :
expiry_dates.append(x)
print(expiry_dates)
import datetime
def get_thursday(_month,_year):
for _i in range(1,32):
if _i > 9:
_dateStr = str(_i)
else:
_dateStr = '0' + str(_i)
_date = str(_year) + '-' + str(_month) + '-' + _dateStr
try:
a = datetime.datetime.strptime(_date, "%Y-%m-%d").strftime('%a')
except:
continue
if a == 'Thu':
_lastThurs = _date
return _lastThurs
x = get_thursday('05','2017')
print(x)
It is straightforward fast and easy to understand, we take the first of every month and then subtract it by 3 coz 3 is Thursday weekday number and then multiply it by either 4 and check if it is in the same month if it is then that is last Thursday otherwise we multiply it with 3 and we get our last Thursday
import datetime as dt
from datetime import timedelta
#start is the first of every month
start = dt.datetime.fromisoformat('2022-08-01')
if start.month == (start + timedelta((3 - start.weekday()) + 4*7)).month:
exp = start + timedelta((3 - start.weekday()) + 4*7)
else:
exp = start + timedelta((3 - start.weekday()) + 3*7)
You can use Calendar to achieve your result. I find it simple.
import calendar
import datetime
testdate= datetime.datetime.now()
weekly_thursday=[]
for week in calendar.monthcalendar(testdate.year,testdate.month):
if week[3] != 0:
weekly_thursday.append(week[3])
weekly_thursday
The list weekly_thursday will have all the Thursdays from the month.
weekly_thursday[-1] will give you the last Thursday of the month.
testdate : datetime.datetime(2022, 9, 9, 6, 35, 16, 752465)
weekly_thursday : [1, 8, 15, 22, 29]
weekly_thursday [-1]:29