I have the following query:
>>> z = Restaurant.objects.values_list('city',flat=True).order_by('city').distinct()
>>> z
[u'ELURU', u'Eluru', u'Hyderabad']
As you can see, it is not completely distinct because of the case sensitivity. How do i correct this issue?
You can use annotate in conjunction with Lower (or Upper, etc...) to normalize your values and return truly distinct values like this...
from django.db.models.functions import Lower
z = Restaurant.objects.annotate(
city_lower=Lower('city')).values_list(
'city_lower',flat=True).order_by('city_lower').distinct()
Note: Make sure order_by is set to 'city_lower' and not 'city' to avoid duplicates.
I'm not sure you're going to find a solution to this since django doesn't offer a case-insensitive distinct method (currently). But then maybe it would be better to fix the values in your database anyway since you don't really want your end users to see their city in capitals since it will look ugly.
I'd suggest thinking about making a simple method that you could run either once in a data migration and stopping the city field from ever getting in this state again - or just running this periodically.
something similar to
for restaurant in Restaurant.objects.all():
if restaurant.city != restaurant.city.title():
restaurant.city = restaurant.city.title()
restaurant.save()
Try this;
z = Restaurant.objects.extra(select = {'tmp_city': lower('city')}).values_list('city',flat=True).order_by('city').distinct('tmp_city')
This works, although it is a little messy. I ended up having to use values, since distinct only works on database tables, regardless of whether or not you use annotate, extra, or rawSQL.
You end up creating an extra field with annotate, and then use that field in your list of dictionaries created by values. Once you have that list of dictionaries, you can use groupby to group dictionaries based on the Lower values key in the values list of dicts. Then, depending on how you want to select the object (in this case, just taking the first object of the group), you can select the version of the distinct that you want.
from django.db.models.functions import Lower
from itertools import groupby
restaurant = [g.next() for k, g in groupby(
list(
Restaurant.objects.annotate(city_lower=Lower('message_text')).values_list('city', flat=True)
).order_by('city').values('city_lower', 'city')
), lambda x: x['city_lower'])]
Related
I have a scenario that, i want a greatest value with the field name. I can get greatest value using Greatest db function which django provides. but i am not able to get its field name. for example:
emps = Employee.objects.annotate(my_max_value=Greatest('date_time_field_1', 'date_time_field_1'))
for e in emps:
print(e.my_max_value)
here i will get the value using e.my_max_value but i am unable to find out the field name of that value
You have to annotate a Conditional Expression using Case() and When().
from django.db.models import F, Case, When
emps = Employee.objects.annotate(
greatest_field=Case(
When(datetime_field_1__gt=F("datetime_field_2"),
then="datetime_field_1"),
When(datetime_field_2__gt=F("datetime_field_1"),
then="datetime_field_2"),
default="equal",
)
)
for e in emps:
print(e.greatest_field)
If you want the database query to tell you which of the fields was larger, you'll need to add another annotated column, using case/when logic to return one field name or the other. (See https://docs.djangoproject.com/en/4.0/ref/models/conditional-expressions/#when)
Unless you're really trying to offload work onto the database, it'll be much simpler to do the comparison work in Python.
I have a filter which should return a queryset with 2 objects, and should have one different field. for example:
obj_1 = (name='John', age='23', is_fielder=True)
obj_2 = (name='John', age='23', is_fielder=False)
Both the objects are of same model, but different primary key. I tried usign the below filter:
qs = Model.objects.filter(name='John', age='23').annotate(is_fielder=F('plays__outdoor_game_role')=='Fielder')
I used annotate first time, but it gave me the below error:
TypeError: QuerySet.annotate() received non-expression(s): False.
I am new to Django, so what am I doing wrong, and what should be the annotate to get the required objects as shown above?
The solution by #ktowen works well, quite straightforward.
Here is another solution I am using, hope it is helpful too.
queryset = queryset.annotate(is_fielder=ExpressionWrapper(
Q(plays__outdoor_game_role='Fielder'),
output_field=BooleanField(),
),)
Here are some explanations for those who are not familiar with Django ORM:
Annotate make a new column/field on the fly, in this case, is_fielder. This means you do not have a field named is_fielder in your model while you can use it like plays.outdor_game_role.is_fielder after you add this 'annotation'. Annotate is extremely useful and flexible, can be combined with almost every other expression, should be a MUST-KNOWN method in Django ORM.
ExpressionWrapper basically gives you space to wrap a more complecated combination of conditions, use in a format like ExpressionWrapper(expression, output_field). It is useful when you are combining different types of fields or want to specify an output type since Django cannot tell automatically.
Q object is a frequently used expression to specify a condition, I think the most powerful part is that it is possible to chain the conditions:
AND (&): filter(Q(condition1) & Q(condition2))
OR (|): filter(Q(condition1) | Q(condition2))
Negative(~): filter(~Q(condition))
It is possible to use Q with normal conditions like below:
(Q(condition1)|id__in=[list])
The point is Q object must come to the first or it will not work.
Case When(then) can be simply explained as if con1 elif con2 elif con3 .... It is quite powerful and personally, I love to use this to customize an ordering object for a queryset.
For example, you need to return a queryset of watch history items, and those must be in an order of watching by the user. You can do it with for loop to keep the order but this will generate plenty of similar queries. A more elegant way with Case When would be:
item_ids = [list]
ordering = Case(*[When(pk=pk, then=pos)
for pos, pk in enumerate(item_ids)])
watch_history = Item.objects.filter(id__in=item_ids)\
.order_by(ordering)
As you can see, by using Case When(then) it is possible to bind those very concrete relations, which could be considered as 1) a pinpoint/precise condition expression and 2) especially useful in a sequential multiple conditions case.
You can use Case/When with annotate
from django.db.models import Case, BooleanField, Value, When
Model.objects.filter(name='John', age='23').annotate(
is_fielder=Case(
When(plays__outdoor_game_role='Fielder', then=Value(True)),
default=Value(False),
output_field=BooleanField(),
),
)
Lets say if I have a model that has lots of fields, but I only care about a charfield. Lets say that charfield can be anything so I don't know the possible values, but I know that the values frequently overlap. So I could have 20 objects with "abc" and 10 objects with "xyz" or I could have 50 objects with "def" and 80 with "stu" and i have 40000 with no overlap which I really don't care about.
How do I count the objects efficiently? What I would like returned is something like:
{'abc': 20, 'xyz':10, 'other': 10,000}
or something like that, w/o making a ton of SQL calls.
EDIT:
I dont know if anyone will see this since I am editing it kind of late, but...
I have this model:
class Action(models.Model):
author = models.CharField(max_length=255)
purl = models.CharField(max_length=255, null=True)
and from the answers, I have done this:
groups = Action.objects.filter(author='James').values('purl').annotate(count=Count('purl'))
but...
this is what groups is:
{"purl": "waka"},{"purl": "waka"},{"purl": "waka"},{"purl": "waka"},{"purl": "mora"},{"purl": "mora"},{"purl": "mora"},{"purl": "mora"},{"purl": "mora"},{"purl": "lora"}
(I just filled purl with dummy values)
what I want is
{'waka': 4, 'mora': 5, 'lora': 1}
Hopefully someone will see this edit...
EDIT 2:
Apparently my database (BigTable) does not support the aggregate functions of Django and this is why I have been having all the problems.
You want something similar to "count ... group by". You can do this with the aggregation features of django's ORM:
from django.db.models import Count
fieldname = 'myCharField'
MyModel.objects.values(fieldname)
.order_by(fieldname)
.annotate(the_count=Count(fieldname))
Previous questions on this subject:
How to query as GROUP BY in django?
Django equivalent of COUNT with GROUP BY
This is called aggregation, and Django supports it directly.
You can get your exact output by filtering the values you want to count, getting the list of values, and counting them, all in one set of database calls:
from django.db.models import Count
MyModel.objects.filter(myfield__in=('abc', 'xyz')).\
values('myfield').annotate(Count('myfield'))
You can use Django's Count aggregation on a queryset to accomplish this. Something like this:
from django.db.models import Count
queryset = MyModel.objects.all().annotate(count = Count('my_charfield'))
for each in queryset:
print "%s: %s" % (each.my_charfield, each.count)
Unless your field value is always guaranteed to be in a specific case, it may be useful to transform it prior to performing a count, i.e. so 'apple' and 'Apple' would be treated as the same.
from django.db.models import Count
from django.db.models.functions import Lower
MyModel.objects.annotate(lower_title=Lower('title')).values('lower_title').annotate(num=Count('lower_title')).order_by('num')
I fetch the latest 5 rows from a Foo model which is ordered by a datetime field.
qs = Foo.objects.all()[:5]
In the following step, I want to reorder the queryset by some other criteria (actually, by the same datetime field in the opposite direction). But reordering after a slice is not permitted. reverse() undoes the first ordering, giving me a differet queryset. Is there a way to accomplish what I want without creating a list from the queryset and doing the ordering using it?
order_by gives you SQL in-database ordering. You're already using that, and then slicing on it. At that point, the results are retrieved into memory. If you want to change their order, you need to use Python in-memory sorting to do it, not the ORM in-database sorting.
In your case, Daniel has already given the best solution: since you simply want to sort by the same field, but in the other order, just reverse the list you have:
qs = Foo.objects.all()[:5]
objs = reversed(qs)
If you had wanted to sort by some other field, then you'd use the sorted() function with a custom key function:
qs = Foo.objects.all()[:5]
objs = sorted(qs, key=lambda o: o.some_other_field)
No, there's no way of doing that. order_by is an operation on the database, but when you slice a queryset it is evaluated and doesn't go back to the database after that.
Sounds like you already know the solution, though: run reversed() on the evaluated qs.
qs = reversed(Foo.objects.all()[:5])
Late answer, but this just worked for me:
import random
sorted(queryset[:10], key=lambda x: random.random())
Thank to this post I'm able to easily do count and group by queries in a Django view:
Django equivalent for count and group by
What I'm doing in my app is displaying a list of coin types and face values available in my database for a country, so coins from the UK might have a face value of "1 farthing" or "6 pence". The face_value is the 6, the currency_type is the "pence", stored in a related table.
I have the following code in my view that gets me 90% of the way there:
def coins_by_country(request, country_name):
country = Country.objects.get(name=country_name)
coin_values = Collectible.objects.filter(country=country.id, type=1).extra(select={'count': 'count(1)'},
order_by=['-count']).values('count', 'face_value', 'currency_type')
coin_values.query.group_by = ['currency_type_id', 'face_value']
return render_to_response('icollectit/coins_by_country.html', {'coin_values': coin_values, 'country': country } )
The currency_type_id comes across as the number stored in the foreign key field (i.e. 4). What I want to do is retrieve the actual object that it references as part of the query (the Currency model, so I can get the Currency.name field in my template).
What's the best way to do that?
You can't do it with values(). But there's no need to use that - you can just get the actual Collectible objects, and each one will have a currency_type attribute that will be the relevant linked object.
And as justinhamade suggests, using select_related() will help to cut down the number of database queries.
Putting it together, you get:
coin_values = Collectible.objects.filter(country=country.id,
type=1).extra(
select={'count': 'count(1)'},
order_by=['-count']
).select_related()
select_related() got me pretty close, but it wanted me to add every field that I've selected to the group_by clause.
So I tried appending values() after the select_related(). No go. Then I tried various permutations of each in different positions of the query. Close, but not quite.
I ended up "wimping out" and just using raw SQL, since I already knew how to write the SQL query.
def coins_by_country(request, country_name):
country = get_object_or_404(Country, name=country_name)
cursor = connection.cursor()
cursor.execute('SELECT count(*), face_value, collection_currency.name FROM collection_collectible, collection_currency WHERE collection_collectible.currency_type_id = collection_currency.id AND country_id=%s AND type=1 group by face_value, collection_currency.name', [country.id] )
coin_values = cursor.fetchall()
return render_to_response('icollectit/coins_by_country.html', {'coin_values': coin_values, 'country': country } )
If there's a way to phrase that exact query in the Django queryset language I'd be curious to know. I imagine that an SQL join with a count and grouping by two columns isn't super-rare, so I'd be surprised if there wasn't a clean way.
Have you tried select_related() http://docs.djangoproject.com/en/dev/ref/models/querysets/#id4
I use it a lot it seems to work well then you can go coin_values.currency.name.
Also I dont think you need to do country=country.id in your filter, just country=country but I am not sure what difference that makes other than less typing.