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Modify days counter on datetime [Python]

I have this code and I want to count the days between 2 dates.
from datetime import date, datetime
checkin= datetime(2022, 1, 30, 1, 15, 00)
checkout= datetime(2022, 1, 31, 0, 0, 00)
count_days= (checkout - checkin).days
In this case, the result of count_days result is 0, because in an operation with 2 datetimes, it takes into account hours, minutes and seconds.
I want the result to be 1 because is +1 day of difference. Type of variables must be datetimes. Thanks!

Convert them to dates first, with the date method.
from datetime import date, datetime
checkin = datetime(2022, 1, 30, 1, 15, 00)
checkout = datetime(2022, 1, 31, 0, 0, 00)
count_days = (checkout.date() - checkin.date()).days

Could you do something like this?
(assuming you want a minimum since the solution you have is similar)
from datetime import date, datetime
check_in= datetime(2022, 1, 30, 1, 15, 00)
check_out= datetime(2022, 1, 31, 0, 0, 00)
# Number of days between two dates (min 1 day)
days = (check_out - check_in).days + 1
print(days)

Why is only one date under DST influence?

I've been reading the pytz and datetime module documentation but I can't figure out why one date is under DST and the other is not.
import pytz
import datetime
mytz = pytz.timezone('America/New_York')
od = datetime.datetime(2021, 7, 1, 4, 0)
mytz.localize(od)
# Out: datetime.datetime(2021, 7, 1, 4, 0, tzinfo=<DstTzInfo 'America/New_York' EDT-1 day, 20:00:00 DST>)
mytz.localize(od).dst()
# Out: datetime.timedelta(0, 3600)
dt = datetime.datetime(2089, 7, 1, 4, 0)
mytz.localize(dt)
# Out: datetime.datetime(2089, 7, 1, 4, 0, tzinfo=<DstTzInfo 'America/New_York' EST-1 day, 19:00:00 STD>)
mytz.localize(dt).dst()
# Out: datetime.timedelta(0)

If you look at the source of the time zone rules, you find that they can have a keyword "max" specified that "is used to extend a rule’s application into the indefinite future" ref. For the US, you can find that here. Unless otherwise specified, DST just continues to be applied during the specified period of the year. But keep in mind that this does not mean that it will actually be the case in the future, since time zones are subject to political decisions.
As an addition to #balmy 's comment suggesting this is a deficiency of pytz, Python 3.9's zoneinfo gives the result to be expected from the above:
import datetime
from zoneinfo import ZoneInfo
od = datetime.datetime(2021, 7, 1, 4, 0, tzinfo=ZoneInfo('America/New_York'))
print(od.dst())
# 1:00:00
dt = datetime.datetime(2089, 7, 1, 4, 0, tzinfo=ZoneInfo('America/New_York'))
print(dt.dst())
# 1:00:00

How to create a datetime object with tzinfo set as 'UTC'?

I have a list of datetimes objects :
time_range = [datetime.datetime(2019, 7, 9, 0, 0, tzinfo=tzutc()),
datetime.datetime(2019, 7, 8, 0, 0, tzinfo=tzutc()),
datetime.datetime(2019, 7, 7, 0, 0, tzinfo=tzutc()),
datetime.datetime(2019, 7, 6, 0, 0, tzinfo=tzutc())
... ]
And I have an other datetime object :
time = datetime(2019, 7, 7)
I have to test if time is in time_range.
But each time I test :
time in time_range
I get the output False, because I don't have the tzinfo.
Here's what I've tried :
I tried to add the tzinfo :
time = datetime(2019, 7, 7, tzinfo=tzuct())
but I can't find where the tzutc() function is.
I also tried tu use pandas :
import pandas as pd
pd.to_datetime(str(time) + '+00:00')
I get the UTC :
Timestamp('2019-07-05 00:00:00+0000', tz='UTC')
But this is not a datetime.datetime object...
Do you have an idea how I could do ?
(Note : i'm compelled to use the form time in time_range, because of the rest of my program)

In datetime constructor, tzinfo parameter expects a type of timezone. It's not the clearest documentation. Try this:
from datetime import datetime, timezone
dt = datetime(2019, 7, 7, tzinfo=timezone.utc)

>>> from datetime import datetime, timezone
>>> time = datetime(2019, 7, 7, tzinfo=timezone.utc)
>>> print(time)
2019-07-07 00:00:00+00:00
>>> print(time.tzinfo)
UTC

After some research, I found an other solution, using pandas :
utc_time = pd.to_datetime(str(time) + '+00:00').to_pydatetime()
returns a datetime.datetime object :
datetime.datetime(2019, 7, 7, 0, 0, tzinfo=<UTC>)
However, to avoid importing pandas library, here's the solution I used :
from datetime import datetime, timezone
new_time = time.replace(tzinfo=timezone.utc)
new_time in time_range.
>>> True

rrule weekly for three weeks, then one week not, repeat

I'm using rrule from python dateutil and don't know how to create an rruleset for the following example:
Monday, three weeks in a row. Then a week not, then again three weeks in a row, one not, and so on.
Any advice on creating an rrule(set) for this?

One way to do this is to use an rruleset with a WEEKLY rrule and a corresponding exrule for every 4th week:
from dateutil.rrule import rrule, rruleset
from dateutil.rrule import WEEKLY
from dateutil.relativedelta import relativedelta
from datetime import datetime, timedelta
dtstart = datetime(2011, 1, 1)
rrset = rruleset()
weekly_rule = rrule(freq=WEEKLY, dtstart=dtstart)
every_4_weeks = rrule(freq=WEEKLY, interval=4,
dtstart=dtstart + relativedelta(weeks=4))
rrset.rrule(weekly_rule)
rrset.exrule(every_4_weeks)
rrset.between(dtstart, dtstart + timedelta(days=65))
The result:
[datetime.datetime(2011, 1, 8, 0, 0),
datetime.datetime(2011, 1, 15, 0, 0),
datetime.datetime(2011, 1, 22, 0, 0),
datetime.datetime(2011, 2, 5, 0, 0),
datetime.datetime(2011, 2, 12, 0, 0),
datetime.datetime(2011, 2, 19, 0, 0),
datetime.datetime(2011, 3, 5, 0, 0)]
The way it works is weekly_rule generates one date per week, and the every_4_weeks generates every 4th week, starting with the 4th week after dtstart. That gives you a 3-on 1-off schedule.

pandas representing datetime with midnight as hour 24

I am working in python pandas and I am doing the following:
StDt = datetime(2018, 1, 1, 1, 0)
EnDt = datetime(2020, 1, 1, 1, 0)
allHours = pd.date_range(StDt, EnDt, freq='H').to_pydatetime()
The midnight hours are represented as:
datetime(2018, 1, 3, 0, 0)
datetime(2018, 1, 5, 0, 0)
Is it possible to create the series in a way such that midnight is represented as hour 24 of previous day
i.e. the above two cases will look as:
datetime(2018, 1, 2, 24, 0)
datetime(2018, 1, 4, 24, 0)
i.e. I am looking for following:
datetime(2018, 1, 3, 0, 0) = datetime(2018, 1, 2, 24, 0)
datetime(2018, 1, 5, 0, 0) = datetime(2018, 1, 4, 24, 0)
Edit:
My particular situation requires working in hour ending world and that is how the convention is in what I am working in.

Using datetimes, this is not possible. Python simply doesn't accept datetime(2018, 1, 2, 24, 0) as a valid time.
There was a request in 2010 to allow for this time to be accepted
Issue 10427: 24:-00 Hour in DateTime
which was rejected.
My only suggestion would be to consider whether you really need this time depicted as you outlined. For actual data manipulation, it should not make any difference as any operations you'd like to do in Pandas with datetimes will conform to this same restriction anyways.

I was working with similar data, and found it useful to consider that Hour Ending data labeled 1-24 is the equivalent of Hour Beginning data labeled 0-23.
So you'll have to change your rule set notation, but it should be a straightforward change.