Assume
I am using Django 2.x and use default settings. i.e, TIME_ZONE = 'UTC' and USE_TZ = True
I am recording data in Honolulu, Hawaii, which means 2019-4-9 9pm (user time) in Honolulu is 2019-4-10 in UTC (server time)
Now I want to filter by 2019-4-9 Honolulu time (user time)
Here is what a demo code
class TimelineTable(models.Model):
accessed = models.DateTimeField(auto_now_add=True)
def __str__(self):
return f'{self.accessed}'
Here is the data in TimelineTable (recorded in 2019-4-9 Honolulu time)
ID accessed
1 2019-04-10 07:19:30.319881
2 2019-04-10 07:19:35.004506
3 2019-04-10 07:19:37.612088
Without considering timezone, fetch data works fine
>>> TimelineTable.objects.filter(accessed__date=datetime.date(2019, 4, 9))
<QuerySet []>
>>> TimelineTable.objects.filter(accessed__date=datetime.date(2019, 4, 10))
<QuerySet [<TimelineTable: 2019-04-10 07:19:30.319881+00:00>, <TimelineTable: 2019-04-10 07:19:35.004506+00:00>, <TimelineTable: 2019-04-10 07:19:37.612088+00:00>]>
Now the problem is with timezone.
I am talking to server: Hey, give me those entries I recorded in 2019-4-9 at Honolulu.
>>> Honolulu = pytz.timezone("Pacific/Honolulu")
>>> t = datetime.datetime(2019, 4, 9, tzinfo=Honolulu)
>>> TimelineTable.objects.filter(accessed__date=t)
<QuerySet []>
I expected 3 entries, but nothing happens.
How to get these entries, without any bugs?
I think this should work:
t = datetime.datetime(2019, 4, 9, 14, 00, 00, tzinfo=Honolulu)
d = t.astimezone(pytz.UTC)
TimelineTable.objects.filter(accessed__date=d.date())
# OR
TimelineTable.objects.filter(accessed__gte=t)
Following code will not work:
t = datetime.datetime(2019, 4, 9, tzinfo=Honolulu)
d = t.astimezone(pytz.UTC)
TimelineTable.objects.filter(accessed__date=d.date())
Explanation
When you generate datetime.datetime(2019, 4, 9), it will return time datetime.time(0, 0). When you convert it to UTC, the date won't be changed because its not the time difference between UTC and Honolulu is not 24 hours.
Related
Context I get different values for datetime field when I access them differently. I am sure there is some utc edge magic going on here.
(Pdb++)
Foo.objects.all().values_list('gated_out__occurred__date')[0][0]
datetime.date(2021, 9, 9)
(Pdb++) Foo.objects.all()[0].gated_out.occurred.date()
datetime.date(2021, 9, 10)
Edit: They have the same PK
Foo.objects.all().order_by("pk")[0].gated_out.occurred.date()
datetime.date(2021, 9, 10)
(Pdb++) Foo.objects.all().order_by("pk").values_list('gated_out__occurred__date')[0][0]
datetime.date(2021, 9, 9)
How do I fix/figure out what is happening?
Take these models:
class Rocket(Model):
...
class Flight(Model):
rocket = ForeignKey(Rocket)
start_time = DateTimeField(...)
If I want to get start times of the latest flight for every rocket, that is simple:
>>> Flight.objects.values('rocket').annotate(max_start_time=Max('start_time'))
<QuerySet [
{'rocket': 3, 'max_start_time': datetime.datetime(2019, 6, 13, 6, 58, 46, 299013, tzinfo=<UTC>)},
{'rocket': 4, 'max_start_time': datetime.datetime(2019, 6, 13, 6, 59, 12, 759964, tzinfo=<UTC>)},
...]>
But what if instead of max_start_time I wanted to select IDs of those same Flights?
In other words, I want to get the ID of the latest Flight for every rocket.
What database backend are you using? If your database backend has support for DISTINCT ON this is most easily accomplished by:
Flight.objects.order_by("rocket", "-start_time").distinct("rocket").values("id", "rocket")
So to display a small bargraph using Django and Chart.js I constructed the following query on my model.
views.py
class BookingsView(TemplateView):
template_name = 'orders/bookings.html'
def get_context_data(self, **kwargs):
today = datetime.date.today()
seven_days = today + datetime.timedelta(days=7)
bookings = dict(Booking.objects.filter(start_date__range = [today, seven_days]) \
.order_by('start_date') \
.values_list('start_date') \
.annotate(Count('id')))
# Edit set default for missing dictonairy values
for dt in range(7):
bookings.setdefault(today+datetime.timedelta(dt), 0)
# Edit reorder the dictionary before using it in a template
context['bookings'] = OrderedDict(sorted(bookings.items()))
This led me to the following result;
# Edit; after setting the default on the dictionary and the reorder
{
datetime.date(2019, 8, 6): 12,
datetime.date(2019, 8, 7): 12,
datetime.date(2019, 8, 8): 0,
datetime.date(2019, 8, 9): 4,
datetime.date(2019, 8, 10): 7,
datetime.date(2019, 8, 11): 0,
datetime.date(2019, 8, 12): 7
}
To use the data in a chart I would like to add the missing start_dates into the dictionary but I'm not entirely sure how to do this.
So I want to update the dictionary with a value "0" for the 8th and 11th of August.
I tried to add the for statement but I got the error;
"'datetime.date' object is not iterable"
Like the error says, you can not iterate over a date object, so for start_date in seven_days will not work.
You can however use a for loop here like:
for dt in range(7):
bookings.setdefault(today+datetime.timedelta(dt), 0)
A dictionary has a .setdefault(..) function that allows you to set a value, given the key does not yet exists in the dicionary. This is thus shorter and more efficient than first checking if the key exists yourself since Python does not have to perform two lookups.
EDIT: Since python-3.7 dictionaries are ordered in insertion order (in the CPython version of python-3.6 that was already the case, but seen as an "implementation detail"). Since python-3.7, you can thus sort the dictionaries with:
bookings = dict(sorted(bookings.items()))
Prior to python-3.7, you can use an OrderedDict [Python-doc]:
from collections import OrderedDict
bookings = OrderedDict(sorted(bookings.items()))
I have a model with test data as below
id days
1, 30
1, 40
2, 10
2, 20
1, 90
I want output as
1, [30,40,90]
2, [10,20]
How can I get this in Django?
It's not much Django, it's pure python. To get the result as a mapping on 'id' as key:
result = {}
for obj in Mymodel.objects.all():
if result.has_key(obj.id):
result[obj.id].append(obj.days)
else:
result[obj.id] = [obj.days]
print result
>>> {1: [30, 40, 90], 2: [10, 20]}
The order of the elements in each list is not defined. If you require these to be ordered, best would be to append .order_by('days') on the Queryset.
A final remark: Your 'id' is not unique. I would consider a non-pk-column named 'id' a bad practice, since 'id' is Django's default name for the automatically created pk-field.
The following is a row that I have extracted from the web:
AIG $30 AIG is an international renowned insurance company listed on the NYSE. A period is required. Manual Auto Active 3 0.0510, 0.0500, 0.0300 [EXTRACT]
I will like to create 5 separate variables by parsing the text and retrieving the relevant data. However, i seriously don't understand the REGEX documentation! Can anyone guide me on how i can do it correctly with this example?
Name = AIG
CurrentPrice = $30
Status = Active
World_Ranking = 3
History = 0.0510, 0.0500, 0.0300
Not sure what do you want to achieve here. There's no need to use regexps, you could just use str.split:
>>> str = "AIG $30 AIG is an international renowned insurance company listed on the NYSE. A period is required. Manual Auto Active 3 0.0510, 0.0500, 0.0300 [EXTRACT]"
>>> list = str.split()
>>> dict = { "Name": list[0], "CurrentPrice": list[1], "Status": list[19], "WorldRanking": list[20], "History": ' '.join((list[21], list[22], list[23])) }
#output
>>> dict
{'Status': 'Active', 'CurrentPrice': '$30', 'Name': 'AIG', 'WorldRanking': '3', 'History': '0.0510, 0.0500, 0.0300'}
Instead of using list[19] and so on, you may want to change it to list[-n] to not depend to the company's description length. Like that:
>>> history = ' '.join(list[-4:-1])
>>> history
'0.0510, 0.0500, 0.0300'
For floating history indexes it could be easier to use re:
>>> import re
>>> history = re.findall("\d\.\d{4}", str)
>>> ['0.0510', '0.0500', '0.0300']
For identifying status, you could get the indexes of history values and then substract by one:
>>> [ i for i, substr in enumerate(list) if re.match("\d\.\d{4}", substr) ]
[21, 22, 23]
>>> list[21:24]
['0.0510,', '0.0500,', '0.0300,']
>>> status = list[20]
>>> status
'3'