How to use Django __time lookup in a queryset - python

I have a model with 2 datetime fields which looks like this:
class Booking(models.Model):
start_date = models.DateTimeField()
end_date = models.DateTimeField()
...
As test data I have 2 bookings with start_date before 17:30 and 2 bookings after 17:45, all on the same day (8 May 2018). I am trying to filter the bookings with the __time lookup to find all the bookings before (and including) 17:30. My queryset is:
bookings = Booking.objects.filter(date__time__lte=datetime.time())
Where datetime.time prints as
datetime.time(17, 30)
and where the date part of the datetime is the same as the bookings dates. The above query is returning an empty queryset but if I use the same query except filtering for times greater than datetime.time() i.e.
bookings = Booking.objects.filter(date__time__gte=datetime.time())
The queryset returns all the bookings (where it should only return the 2 bookings with start_date after 17:30). Can someone please explain to me how the __time lookup is meant to be used?
EDIT
I updated the filter to
bookings = Booking.objects.filter(start_date__time__lte=datetime.time())
and the result is the same. When I print the values of the bookings, the values are:
print Booking.objects.all().values('date', 'end_date')
[
{'start_date': datetime.datetime(2018, 5, 8, 16, 30, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>), 'end_date': datetime.datetime(2018, 5, 8, 17, 0, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>)},
{'start_date': datetime.datetime(2018, 5, 8, 17, 0, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>), 'end_date': datetime.datetime(2018, 5, 8, 17, 30, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>)},
{'start_date': datetime.datetime(2018, 5, 8, 17, 45, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>), 'end_date': datetime.datetime(2018, 5, 8, 18, 15, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>)},
{'start_date': datetime.datetime(2018, 5, 8, 17, 45, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>), 'end_date': datetime.datetime(2018, 5, 8, 18, 15, tzinfo=<DstTzInfo 'Africa/Harare' CAT+2:00:00 STD>)}
]
EDIT 2
I forgot to mention I need to get the bookings that are on the same date. As siddhant0905 suggested I filtered the queryset with datetimes instead and added an extra filter to make sure it was on the same date. The following worked for me:
bookings = Booking.objects.filter(Q(start_date__date=datetime.date()) & Q(start_date__lte=datetime))


I think you should compare the complete datetime object rather than just comparing the time part.
Debug the 'Type' of the time that the query returns and the type of time you are providing it to compare with. Both should be same.
Django shell can be of great help.

Related

Why isn't my code working to make datetime objects timezone-aware?

I have two python datetime objects that represent the same moment in time:
a = datetime.datetime(2018, 10, 28, 13, 26, 30)
b = datetime.datetime(2018, 10, 28, 7, 26, 30)
Both are coming from different sources.
I know that the first is in UTC, and the second is in "America/Edmonton" (MDT).
Neither initially have a timezone attached to them.
I need to add timezones to these objects and compare them in a way where a == b is True.
What I did was this:
import datetime
from pytz import timezone
a = datetime.datetime(2018, 10, 28, 13, 26, 30)
b = datetime.datetime(2018, 10, 28, 7, 26, 30)
a = a.replace(tzinfo=timezone("UTC"))
b = b.replace(tzinfo=timezone("America/Edmonton"))
a = a.astimezone(timezone("America/Edmonton"))
b = b.astimezone(timezone("America/Edmonton"))
print(repr(a))
# Result: datetime.datetime(2018, 10, 28, 7, 26, 30, tzinfo=<DstTzInfo 'America/Edmonton' MDT-1 day, 18:00:00 DST>)
print(repr(b))
# Result: datetime.datetime(2018, 10, 28, 7, 26, 30, tzinfo=<DstTzInfo 'America/Edmonton' LMT-1 day, 16:26:00 STD>)
a == b # Results in False for some reason
What is "MDT-1 day, 18:00:00 DST" vs "LMT-1 day, 16:26:00 STD"? Why are they different? What am I doing wrong?

The proper way to do this appears to be:
import datetime
from pytz import timezone
a = datetime.datetime(2018, 10, 28, 13, 26, 30)
b = datetime.datetime(2018, 10, 28, 7, 26, 30)
a = timezone('UTC').localize(a)
b = timezone('America/Edmonton').localize(b)
a == b
As demonstrated here. This does result in a being equal to b. Still not sure why it sounds like pytz is defaulting to using a system from before 1893.

Daylight time saving aware conversion of timestamps in python

Given a timestamp without time zone (e.g. 2018-03-12 09:30:00) AND the timezone EST5EDT, the goal is to parse the data returning a datetime object that is time zone AND daylight saving aware.
from datetime import datetime
import pytz
datetime(2018, 3, 8, 9, 30, tzinfo=pytz.timezone('EST5EDT')).astimezone(pytz.utc)
# returns:
# datetime.datetime(2018, 3, 8, 14, 30, tzinfo=<UTC>)
datetime(2018, 3, 12, 9, 30, tzinfo=pytz.timezone('EST5EDT')).astimezone(pytz.utc)
# returns:
# datetime.datetime(2018, 3, 12, 14, 30, tzinfo=<UTC>)
# BUT should return (second Sunday of march the daylight saving changes by 1 hour):
# datetime.datetime(2018, 3, 12, 13, 30, tzinfo=<UTC>)

Never set tzinfo directly when creating datetimes. Always use the localize() method of the timezone (see the note at the top of http://pytz.sourceforge.net/):
pytz.timezone('EST5EDT').localize(
datetime(2018, 3, 12, 9, 30)
).astimezone(pytz.utc)

Iterable Datetime - How to get continuos datetime objects with day name in Python?

I want to create a mapping of the everyday of the week with its datetime object. So my dictionary should be having keys as "Monday", "Tuesday", .. (so on) so that I can get a datetime object for every day on the next(!) week.
At the moment I have a dictionary with these values:
DAYS_DATETIME_RELATIONS = {
"today": datetime.datetime.now(),
"tomorrow": datetime.datetime.now() + datetime.timedelta(days=1),
"after_tomorrow": datetime.datetime.now() + datetime.timedelta(days=2)
}
Unfortunately I cannot find any algorithmic solution for this and hope anyone of you could help me.

This can be achieved by using 2 dictionaries in the following manner:
import calendar
import datetime
days = {i: calendar.day_name[i-1] for i in range(7)}
today = datetime.datetime.now()
# using i % 7 so an arbitrary range can be used, for example
# range(7, 15) to get the week after the next week
next_week = {days[i % 7]: (today + datetime.timedelta(days=i)).date()
for i in range(7)}
print(next_week)
# {'Tuesday': datetime.date(2018, 1, 9), 'Sunday': datetime.date(2018, 1, 7),
# 'Monday': datetime.date(2018, 1, 8), 'Thursday': datetime.date(2018, 1, 11),
# 'Wednesday': datetime.date(2018, 1, 10), 'Friday': datetime.date(2018, 1, 12),
# 'Saturday': datetime.date(2018, 1, 13)}
print(next_week['Saturday'])
# 2018-01-13

Here is another way to solve your question using datetime and timedelta from datetime module:
from datetime import datetime, timedelta
def generate_dict_relation(_time, _days=0):
keys = {'Yesterday': -1, 'Today': 0, 'Tomorow': 1, 'After_tomorrow': 2}
if not _days:
return {key: _time + timedelta(days=keys.get(key, 0)) for key in keys}
else:
return {(_time + timedelta(days=_days+k)).strftime('%A'): _time + timedelta(days=_days+k)
for k in range(0, 7)}
_date_now = datetime.now()
DAYS_DATETIME_RELATIONS = {}
# get dates: yesterday, today, tomorrow and after tomorrow
DAYS_DATETIME_RELATIONS.update(generate_dict_relation(_date_now, 0))
# get dates after 7 days = 1 week
DAYS_DATETIME_RELATIONS.update(generate_dict_relation(_date_now, 7))
next_tuesday = DAYS_DATETIME_RELATIONS.get('Tuesday')
next_monday = DAYS_DATETIME_RELATIONS.get('Monday')
yesterday = DAYS_DATETIME_RELATIONS.get('Yesterday')
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(next_tuesday, repr(next_tuesday)))
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(next_monday, repr(next_monday)))
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(yesterday, repr(yesterday)))
Output:
16/01/2018 10:56:26:1516096586 datetime.datetime(2018, 1, 16, 10, 56, 26, 659949)
15/01/2018 10:56:26:1516010186 datetime.datetime(2018, 1, 15, 10, 56, 26, 659949)
06/01/2018 10:56:26:1515232586 datetime.datetime(2018, 1, 6, 10, 56, 26, 659949)

One very generic way will be to create a custom iterator to return you the continuos datetime objects as:
from datetime import datetime, timedelta
class RepetetiveDate(object):
def __init__(self, day_range=7, datetime_obj=datetime.now(), jump_days=1):
self.day_range = day_range
self.day_counter = 0
self.datetime_obj = datetime_obj
self.jump_days = jump_days
self.time_deltadiff = timedelta(days=self.jump_days)
def __iter__(self):
return self
# If you are on Python 2.7
# define this function as `next(self)`
def __next__(self):
if self.day_counter >= self.day_range:
raise StopIteration
if self.day_counter != 0: # don't update for the first iteration
self.datetime_obj += self.time_deltadiff
self.day_counter += 1
return self.datetime_obj
Here, this iterator returns continuos datetime object starting from the datetime object you'll initially pass (default starts from current date).
It is using 3 optional params which you may customize as per your need:
day_range: Maximum allowed iteration for the RepetetiveDate iterator. Default value is 7.
jump_days: Integer value for jumping the number of days for the datetime object in next iteration. That means, if jump_days is equal to "2", will return datetime objects of every alternate date. To get the datetime objects of past, pass this value as negative. Default value is 1.
datetime_obj: Accepts the datetime from which date you want to start your iteration. Default value is current date.
If you are new to iterators, take a look at:
What exactly are Python's iterator, iterable, and iteration protocols?
Difference between Python's Generators and Iterators
Sample Run for upcoming dates:
>>> x = RepetetiveDate()
>>> next(x)
datetime.datetime(2018, 1, 8, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 9, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 10, 15, 55, 39, 124654)
Sample Run for previous dates:
>>> x = RepetetiveDate(jump_days=-1)
>>> next(x)
datetime.datetime(2018, 1, 6, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 5, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 4, 15, 55, 39, 124654)
How to get your desired dictionary?
Using this, you may create your dictionary using the dict comprehension as:
Dictionary of all days of week
>>> {d.strftime("%A"): d for d in RepetetiveDate(day_range=7)}
{
'Monday': datetime.datetime(2018, 1, 8, 15, 23, 16, 926364),
'Tuesday': datetime.datetime(2018, 1, 9, 15, 23, 16, 926364),
'Wednesday': datetime.datetime(2018, 1, 10, 15, 23, 16, 926364),
'Thursday': datetime.datetime(2018, 1, 11, 15, 23, 16, 926364),
'Friday': datetime.datetime(2018, 1, 12, 15, 23, 16, 926364),
'Saturday': datetime.datetime(2018, 1, 13, 15, 23, 16, 926364),
'Sunday': datetime.datetime(2018, 1, 14, 15, 23, 16, 926364)
}
Here I am using d.strftime("%A") to extract day name from the datetime object.
List of current days for next 4 weeks
>>> [d for d in RepetetiveDate(jump_days=7, day_range=4))]
[
datetime.datetime(2018, 1, 7, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 14, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 21, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 28, 16, 17, 45, 45005)
]

One very clean way to implement this is using rrule from the dateutil library. For example:
>>> from dateutil.rrule import rrule, DAILY
>>> from datetime import datetime
>>> start_date = datetime.now()
>>> {d.strftime("%A"): d for d in rrule(freq=DAILY, count=7, dtstart=start_date)}
which will return your desired dict object:
{
'Sunday': datetime.datetime(2018, 1, 7, 17, 2, 30),
'Monday': datetime.datetime(2018, 1, 8, 17, 2, 30),
'Tuesday': datetime.datetime(2018, 1, 9, 17, 2, 30),
'Wednesday': datetime.datetime(2018, 1, 10, 17, 2, 30),
'Thursday': datetime.datetime(2018, 1, 11, 17, 2, 30),
'Friday': datetime.datetime(2018, 1, 12, 17, 2, 30),
'Saturday': datetime.datetime(2018, 1, 13, 17, 2, 30)
}
(Special thanks to Jon Clements for telling me about rrule)

Django Queryset Datetime Filter

I have a column formatted as such in one of my models:
TEMP_START = models.DateTimeField(null=True)
And I am attempting to do an exact lookup using queryset syntax such as
x.filter(TEMP_START=my_datetime_object) # x can be thought of as y.objects.all()
This returns no objects, when it should do more than one. However,
x.filter(TEMP_START__date=my_datetime_object.date()).filter(TEMP_START__hour=my_datetime_object.hour)
Does return the proper objects (they're hourly). Are direct datetime filters not supported, and thus keywords must be used?
====== Edit with bad results:
Searching for: {'TEMP_START': datetime.datetime(2016, 3, 31, 2, 0)}
Values in column: [{'TEMP_START': datetime.datetime(2016, 3, 29, 8, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 29, 14, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 2, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 29, 20, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 8, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 20, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 31, 2, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 14, 0)}]
Values being returned: []
Code:
args_timeframe_start = {variables.temp_start: self.ranked_timeframes[rank][variables.temp_start]}
print(args_timeframe_start)
print(self.query_data.values(variables.temp_start).distinct())
query_data = self.query_data.filter(**args_timeframe_start)
print(query_data.values(variables.temp_start).distinct())

You need to find out what is my_datetime_object but most likely because DateTime fields contain python datetime.datetime objects, datetime.datetime objects is composed of year, month, date, hour, minute, second, microsecond. So if you merely compare date and hour, sure you could get results, but you couldn't guarantee that my_datetime_object matches one of the records in your database that has the same minute, second and microsecond.
Try this quickly and you could see what does datetime look like, also django doc about DateTimeField:
from datetime import datetime
date = datetime.now()
print date

Calculating a timewindow with Python datetime

I'd like to query my DB for all records posted on a particular day (e.g. today between 00:00 and 23:59:59) given a datetime.datetime timestamp such as datetime.datetime(2010, 12, 21, 17, 59, 43, 85335).
What's the best method to calculate the start and end datetime.datetime instances, please? I'd like to end up with something like this:
>>> timestamp = datetime.datetime(2010, 12, 21, 17, 59, 43, 85335)
>>> # do something with timestamp to get start_date and end_date
>>> start_date
datetime.datetime(2010, 12, 21, 0, 0, 0, 0)
>>> end_date
datetime.datetime(2010, 12, 21, 23, 59, 59, 0)
Thanks, HC

Ah, found what I want. I struggle with datetime every time.
>>> timestamp = datetime.datetime.now()
>>> timestamp
datetime.datetime(2010, 12, 21, 18, 31, 37, 900795)
>>> timestamp.replace(hour=0, minute=0, second=0, microsecond=0)
datetime.datetime(2010, 12, 21, 0, 0)

If it's just calculating a day, then what you have is fine. Just use the day and then put the hour as 0:00 and 23:59 respectively.
If you want to get the start and end times of a month, that gets a lot more complicated since months have different days. For something like that you could use this module
http://niemeyer.net/python-dateutil

timedelta sounds just like what you need.
http://docs.python.org/library/datetime.html#timedelta-objects
from datetime import datetime, date
date.today() - timedelta(days=7)
Will give you 7 days from today.

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