I am writing code that lets users write down dates and times for things they have on. It takes in the date on which it starts, the start time and finish time. It also allows the user to specify if they want it to carry over into multiple weeks (every Monday for a month for example)
I am using a for loop to do this, and because of the different months having different days I obviously want (if the next Monday for example is in the next month) it to have the correct date.
This is my code:
for i in range(0 , times):
day = day
month = month
fulldateadd = datetime.date(year, month, day)
day = day + 7
if month == ( '01' or '03' or '05' or '07' or '10'or '12'):
if day > 31:
print(day)
day = day - 31
print(day)
month = month + 1
elif month == ( '04' or '06'or '09' or '11'):
if day > 30:
print(day)
day = day - 30
print(day)
month = month + 1
elif month == '02':
if day > 29:
print(day)
day = day - 29
print(day)
month = month + 1
When running this and testing to see if it goes correctly into the new month I get the error:
File "C:\Users\dansi\AppData\Local\Programs\Python\Python36-32\gui test 3.py", line 73, in addtimeslot
fulldateadd = datetime.date(year, month, day)
ValueError: day is out of range for month
Where have I gone wrong?
It's hard to be completely accurate without seeing some of the previous code (for example, where do day, month, year, and times come from?), but here's how you might be able to use timedelta in your code:
fulldateadd = datetime.date(year, month, day)
for i in range(times):
fulldateadd = fulldateadd + datetime.timedelta(7)
A timedelta instance represents a period of time, rather than a specific absolute time. By default, a single integer passed to the constructor represents a number of days. So timedelta(7) gives you an object that represents 7 days.
timedelta instances can then be used with datetime or date instances using basic arithmetic. For example, date(2016, 12, 31) + timedelta(1) would give you date(2017, 1, 1) without you needing to do anything special.
Related
Problem Statement:
Am developing a custom job scheduler that needs to be run on given days. It takes start date and end date as string and third param is list of week days on which job should run.
Start day can be different with given days but first job should run on next valid day
Let suppose Start date is 2022-09-07 (so day name is Wednesday) but given frequency days are ["Monday", "Friday", "Saturday"] so i need to run my first job on coming Friday and for this i need to calculate difference between start date and first valid day (in this case it's Friday)
So how can i do this python to run my first job on valid day (that can be in any position of given frequency days list) and also after one job complete i need to also get next valid day. I did some work but unfortunately its not working. Here is what i did
sorted_week_days_list = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
start_date = "2022-09-07"
valid_frequency_days = ["Monday", "Tuesday", "Friday"] # It can be any days in sorted order
start_date_object = datetime.datetime.strptime(start_date, "%Y-%m-%d")
given_start_day = start_date_object.strftime("%A")
if given_start_day not in valid_frequency_days:
# Need help to implement logic to get date for valid day
You should use the datetime.weekday() method to pull out the day of the week for days of interest. Assuming that you have dates similar to the format you show above, it is easy to convert, and also just use the day index for your "allowable start days" (Monday=0).
Then you can jig up a little function to look for the next start date in your sorted list and figure out how many days you need to wait.
Example below does that and also "rolls over" the weekend as needed.
Code:
from datetime import datetime, timedelta
from bisect import bisect_left
start_date = "2022-09-09"
valid_start_dates = [1, 4] # It can be any days in sorted order
start_date_object = datetime.strptime(start_date, "%Y-%m-%d")
d=start_date_object.weekday()
print(f'the numbered day of the week is: {d}')
def days_till_start(day, valid_start_days):
idx = bisect_left(valid_start_days, day)
if idx >= len(valid_start_days): # wrap around to next start
return valid_start_days[0] + 7 - day
elif day == valid_start_days[idx]:
return 0
else:
return valid_start_days[idx] - day
print(days_till_start(d, valid_start_dates))
start_dates = ['2022-09-05', '2022-09-06', '2022-09-07', '2022-09-08', '2022-09-09', '2022-09-10']
start_wkdys = [datetime.strptime(d, "%Y-%m-%d").weekday() for d in start_dates]
for d in start_wkdys:
print(f'day index is: {d}')
print(f'next start date is {days_till_start(d, valid_start_dates)} away')
print()
Output:
the numbered day of the week is: 4
0
day index is: 0
next start date is 1 away
day index is: 1
next start date is 0 away
day index is: 2
next start date is 2 away
day index is: 3
next start date is 1 away
day index is: 4
next start date is 0 away
day index is: 5
next start date is 3 away
Given a month and a year, such as 03 2022, I need to get the epoch time of the first day of the month and the last day of the month. I am not sure how to do that. Any help would be appreciated. Thank you.
You can get the beginning of the month easily by setting the day to 1.
To get the end of the month conveniently, you can calculate the first day of the next month, then go back one day.
Then set the time zone (tzinfo) to UTC to prevent Python using local time.
Finally a call to .timestamp()
import datetime
def date_to_endofmonth(
dt: datetime.datetime, offset: datetime.timedelta = datetime.timedelta(days=1)
) -> datetime.datetime:
"""
Roll a datetime to the end of the month.
Parameters
----------
dt : datetime.datetime
datetime object to use.
offset : datetime.timedelta, optional.
Offset to the next month. The default is 1 day; datetime.timedelta(days=1).
Returns
-------
datetime.datetime
End of month datetime.
"""
# reset day to first day of month, add one month and subtract offset duration
return (
datetime.datetime(
dt.year + ((dt.month + 1) // 12), ((dt.month + 1) % 12) or 12, 1
)
- offset
)
year, month = 2022, 11
# make datetime objects; make sure to set UTC
dt_month_begin = datetime.datetime(year, month, 1, tzinfo=datetime.timezone.utc)
dt_month_end = date_to_endofmonth(dt_month_begin).replace(tzinfo=datetime.timezone.utc)
ts_month_begin = dt_month_begin.timestamp()
ts_month_end = dt_month_end.timestamp()
print(ts_month_begin, ts_month_end)
# 1667260800.0 1701302400.0
Unable to comment due to reputation but #FObersteiner is excellent just I would recommend a small change.
For example running the current code it would produce this for Nov 2022
print(dt_month_begin.timestamp())
print(dt_month_begin)
print(dt_month_end.timestamp())
print(dt_month_end)
--->
1667260800.0
2022-11-01 00:00:00+00:00
1701302400.0
2023-11-30 00:00:00+00:00
Note the year field
I'd suggest the following
import datetime
def date_to_endofmonth(
dt: datetime.datetime, offset: datetime.timedelta = datetime.timedelta(seconds=1)
) -> datetime.datetime:
"""
Roll a datetime to the end of the month.
Parameters
----------
dt : datetime.datetime
datetime object to use.
offset : datetime.timedelta, optional.
Offset to the next month. The default is 1 second; datetime.timedelta(seconds=1).
Returns
-------
datetime.datetime
End of month datetime.
"""
# reset day to first day of month, add one month and subtract offset duration
return (
datetime.datetime(
dt.year + ((dt.month + 1) // 13), ((dt.month + 1) % 12) or 12, 1
)
- offset
)
year, month = 2022, 11
# make datetime objects; make sure to set UTC
dt_month_begin = datetime.datetime(year, month, 1, tzinfo=datetime.timezone.utc)
dt_month_end = date_to_endofmonth(dt_month_begin).replace(tzinfo=datetime.timezone.utc)
Differences being floor division by 13 instead of 12 to handle the month of November.
Changing the offset to seconds delta because I felt the user ( and myself who came looking for the answer) wanted the starting epoch time and the ending epoch time so
Nov 1st 00:00:00 --> Nov 30th 23:59:59 would be better than
Nov 1st 00:00:00 --> Nov 30th 00:00:00 ( Losing a day worth of seconds)
Output of the above would be
:
1667260800.0
2022-11-01 00:00:00+00:00
1669852799.0
2022-11-30 23:59:59+00:00
I need to print out “Your birthday is 31 March 2001 (a years, b days, c hours, d minutes and e seconds ago).”
I create input
birth_day = int(input("your birth day?"))
birth_month = int(input("your birth month?"))
birth_year = int(input("your birth year?"))
and I understand
print("your birthday is"+(birth_day)+(birth_month)+(birth_year)) to print out first sentence. but I faced problem with second one which is this part (a years, b days, c hours, d minutes and e seconds ago)
I guess I have to use “the epoch”
and use some of various just like below
year_sec=365*60*60*24
day_sec=60*60*24
hour_sec=60*60
min_sec=60
calculate how many seconds of the date since 1 January 1970 00:00:00 UTC:
import datetime, time
t = datetime.datetime(2001, 3, 31, 0, 0)
time.mktime(t.timetuple())
985960800.0
can anyone, could you solve my problem please?
Thank a lot
EDIT: See this answer in the thread kaya3 mentioned above for a more consistently reliable way of doing the same thing. I'm leaving my original answer below since it's useful to understand how to think about the problem, but just be aware that my answer below might mess up in tricky situations due to the quirks of the Gregorian calendar, in particular:
Every year that is exactly divisible by four is a leap year, except for years that are exactly divisible by 100, but these centurial years are leap years if they are exactly divisible by 400. For example, the years 1700, 1800, and 1900 are not leap years, but the years 1600 and 2000 are.
ORIGINAL ANSWER:
You can try using the time module:
import time
import datetime
def main(ask_for_hour_and_minute, convert_to_integers):
year, month, day, hour, minute = ask_for_birthday_info(ask_for_hour_and_minute)
calculate_time_since_birth(year, month, day, hour, minute, convert_to_integers)
def ask_for_birthday_info(ask_for_hour_and_minute):
birthday_year = int(input('What year were you born in?\n'))
birthday_month = int(input('What month were you born in?\n'))
birthday_day = int(input('What day were you born on?\n'))
if ask_for_hour_and_minute is True:
birthday_hour = int(input('What hour were you born?\n'))
birthday_minute = int(input('What minute were you born?\n'))
else:
birthday_hour = 0 # set to 0 as default
birthday_minute = 0 # set to 0 as default
return (birthday_year, birthday_month, birthday_day, birthday_hour, birthday_minute)
def calculate_time_since_birth(birthday_year, birthday_month, birthday_day, birthday_hour, birthday_minute, convert_to_integers):
year = 31557600 # seconds in a year
day = 86400 # seconds in a day
hour = 3600 # seconds in a hour
minute = 60 # seconds in a minute
# provide user info to datetime.datetime()
birthdate = datetime.datetime(birthday_year, birthday_month, birthday_day, birthday_hour, birthday_minute)
birthdate_tuple = time.mktime(birthdate.timetuple())
# figure out how many seconds ago birth was
seconds_since_birthday = time.time() - birthdate_tuple
# start calculations
years_ago = seconds_since_birthday // year
days_ago = seconds_since_birthday // day % 365
hours_ago = seconds_since_birthday // hour % 24
minutes_ago = seconds_since_birthday // minute % 60
seconds_ago = seconds_since_birthday % minute
# convert calculated values to integers if convert_to_integers is True
if convert_to_integers is True:
years_ago = int(years_ago)
days_ago = int(days_ago)
hours_ago = int(hours_ago)
minutes_ago = int(minutes_ago)
seconds_ago = int(seconds_ago)
# print calculations
print(f'Your birthday was {years_ago} years, {days_ago}, days, {hours_ago} hours, {minutes_ago} minutes, {seconds_ago} seconds ago.')
# to ask for just the year, month, and day
main(False, False)
# to ask for just the year, month, and day AND convert the answer to integer values
main(False, True)
# to ask for just the year, month, day, hour, and minute
main(True, False)
# to ask for just the year, month, day, hour, and minute AND convert the answer to integer values
main(True, True)
Tried to use descriptive variable names so the variables should make sense, but the operators might need some explaining:
10 // 3 # the // operator divides the numerator by the denominator and REMOVES the remainder, so answer is 3
10 % 3 # the % operator divides the numerator by the denominator and RETURNS the remainder, so the answer is 1
After understanding the operators, the rest of the code should make sense. For clarity, let's walk through it
Create birthdate by asking user for their information in the ask_for_birthday_info() function
Provide the information the user provided to the calculate_time_since_birth() function
Convert birthdate to a tuple and store it in birthdate_tuple
Figure out how many seconds have passed since the birthday and store it in seconds_since_birthday
Figure out how many years have passed since the birthday by dividing seconds_since_birthday by the number of seconds in a year
Figure out how many days have passed since the birthday by dividing seconds_since_birthday by the number of seconds in a day and keeping only the most recent 365 days (that's the % 365 in days_ago)
Figure out how many hours have passed since the birthday by dividing seconds_since_birthday by the number of seconds in a hour and keeping only the most recent 24 hours (that's the % 24 in hours_ago)
Figure out how many minutes have passed since the birthday by dividing seconds_since_birthday by the number of seconds in a minute and keeping only the most recent 60 minutes (that's the % 60 in minutes_ago)
Figure out how many seconds have passed since the birthday by dividing seconds_since_birthday and keeping only the most recent 60 seconds (that's the % 60 in seconds_ago)
Then, we just need to print the results:
print(f'Your birthday was {years_ago} years, {days_ago}, days, {hours_ago} hours, {minutes_ago} minutes, {seconds_ago} seconds ago.')
# if you're using a version of python before 3.6, use something like
print('Your birthday was ' + str(years_ago) + ' years, ' + str(days_ago) + ' days, ' + str(hours_ago) + ' hours, ' + str(minutes_ago) + ' minutes, ' + str(seconds_ago) + ' seconds ago.')
Finally, you can add some error checking to make sure that the user enters valid information, so that if they say they were born in month 15 or month -2, your program would tell the user they provided an invalid answer. For example, you could do something like this AFTER getting the birthday information from the user, but BEFORE calling the calculate_time_since_birth() function:
if not (1 <= month <= 12):
print('ERROR! You provided an invalid month!')
return
if not (1 <= day <= 31):
# note this isn't a robust check, if user provides February 30 or April 31, that should be an error - but this won't catch that
# you'll need to make it more robust to catch those errors
print('ERROR! You provided an invalid day!')
return
if not (0 <= hour <= 23):
print('ERROR! You provided an invalid hour!')
return
if not (0 <= minute <= 59):
print('ERROR! You provided an invalid minute!')
return
if not (0 <= second <= 59):
print('ERROR! You provided an invalid second!')
return
I'm new to python and I'm looking to work with datetime. I have some files generated every Sunday and I like to move the furthest Sunday out of the current folder eg: 2020-04-12, 2020-04-19, 2020-04-26.
I have found some examples on getting a specific date from today's date and I was able to modify it a tab bit. Eg. I can go back and get last week's Sunday with a specific date:
from datetime import date
from datetime import timedelta
import datetime
today = datetime.datetime(2020,4,13)
offset = (today.weekday() + 1) % 7
sunday = today - timedelta(days=offset)
#print (offset)
print(sunday)
I am confused by the offset variable. What is (today.weekday() + 1) % 7 doing? I have read the Python doc and not quite wrapping my head around it. With +1, I get the date 2020-04-12, which is a Sunday, great. When I do -1 (the other thing is if I set it to (today.weekday() - 1) % 7), I get 2020-04-07, a Tuesday. How did it jump from Sunday the 12th to Tuesday the 7th?
Additionally, how do I get it to jump back 3 weeks? that's where I'm also stuck.
Alright, so if today's Wednesday, then today.weekday() is 2, because it starts counting from 0 on Monday. Not sure why, but that's life.
So (2 + 1) % 7) = 3. That means that 3 days ago was Sunday. Hence your code:
offset = (today.weekday() + 1) % 7 # How many days ago was sunday
sunday = today - timedelta(days=offset) # Go backwards from today that many days
You'll notice that if you subtract one instead of add one, that means we're going backwards (because we're sutracting the timedelta object) by two fewer days than before (because 2 - 1 is equivalent to (2 + 1) - 2, that is, two fewer days). If you started by going backwards enough days to get to Sunday, and now you're going backwards two fewer days, you'll end up on Tuesday, which is two days later than Sunday.
The easiest way to shift which week you're headed to is to set the weeks argument in timedelta:
n_weeks = 3
sunday = today - timedelta(days=offset, weeks=n_weeks)
that's equivalent to, but much prettier than:
sunday = today - timedelta(days=offset + n_weeks * 7)
I want to find the last business day for the month of June and Dec for 2019. my below code gives me the last business day for last month and the current month. Ideally i want a code which allows me to input a year and month and it will tell me the last business day for the month and year i input.
i want my output to be like this for June 2019 and Dec 2019
2906
3112
hope someone can help me here, thanks
from pandas.tseries.offsets import BMonthEnd
from datetime import date
d=date.today()
offset = BMonthEnd()
#Last day of current month
current_month = offset.rollforward(d)
print(current_month)
#Last day of previous month
last_month = offset.rollback(d)
print(last_month)
Here's a function that will find the last weekday in a given month and format it as a string in the format given
import datetime
def last_weekday(year, month):
# Add a month
if month == 12:
month = 1
year += 1
else:
month += 1
d = datetime.date(year, month, 1)
# Subtract a day
d -= datetime.timedelta(days=1)
# Subtract more days if we have a weekend
while d.weekday() >= 5:
d -= datetime.timedelta(days=1)
return '{:02d}{:02d}'.format(d.month, d.day)
# Examples:
last_weekday(2019, 6) # -> '0628'
last_weekday(2019, 12) # -> '1231'