I've got a working SQL query that I'm trying to write in Django (without resorting to RAW) and was hoping you might be able to help.
Broadly, I'm looking to next two queries - the first calculates a COUNT, and then I'm looking to calculate an AVERAGE of the COUNTS. (this'll give you the average number of items on a ticket, per location)
The SQL that works is:
SELECT location_name, Avg(subq.num_tickets) FROM (
SELECT Count(ticketitem.id) AS num_tickets, location.name AS location_name
FROM ticketitem
JOIN ticket ON ticket.id = ticketitem.ticket_id
JOIN location ON location.id = ticket.location_id
JOIN location ON location.id = location.app_location_id
GROUP BY ticket_id, location.name) AS subq
GROUP BY subq.location_name;
For my Django code, I'm trying something like this:
# Get the first count
qs = TicketItem.objects.filter(<my complicated filter>).\
values('ticket__location__app_location__name','posticket').\
annotate(num_tickets=Count('id'))
# now get the average of the count
qs2 = qs.values('ticket__location__app_location__name').\
annotate(Avg('num_tickets')).\
order_by('location__app_location__name')
but that fails because num_tickets doesn't exist ... Anyway - suspect I'm being slow. Would love someone to enlighten me!
Check out the section on aggregating annotations from the Django docs. Their example takes an average of a count.
I was playing around with this a bit in a manage.py shell, and I think the django ORM might not be able to do that kind of annotation. Honestly you're probably going to have to resort to doing a raw query or bind in something like https://github.com/Deepwalker/aldjemy which would let you do that via SQLAlchemy.
When I playing with this I tried
(my_model.objects.filter(...)
.values('parent_id', 'parent__name', 'thing')
.annotate(Count('thing'))
.values('name', 'thing__count')
.annotate(Avg('thing__count')))
Which gave a lovely traceback about FieldError: Cannot compute Avg('thing__count'): 'thing__count' is an aggregate, which makes sense since I doubt the ORM is trying to convert that first group by to a nested query.
Related
I've this query. Orders post records by last comment on post. This query works well with small tables. However, I've filled database with random data approximately 2M rows on comment table. Analyzed query with explain and saw that sequential scan is performed on Post table.
Post.objects.extra(select={'last_update': 'select max(c.create_date) from comment_comment c where c.post_id = post_post.id'}).order_by('-last_update')
I've rewritten same query which is faster than current one. But I could not find a way to fit the query on django's orm. How can I rewrite it? If it is possible, I want to write it not using raw query as much as possible.
Regards. Thanks for any help.
select
p.*,
t.last_update
from
post_post p
join
( select c.post_id as pid, max(c.create_date) as last_update from comment_comment c group by pid) t
on p.id = t.pid
order by t.last_update desc
limit 50;
If I make some assumptions about your Django model, it will look something like this:
posts.objects
.annotate(last_update=Max('comments__create_date'))
.order_by('-last_update')[:50]
In Django, annotate is your friend.
This is a bleeding-edge feature that I'm currently skewered upon and quickly bleeding out. I want to annotate a subquery-aggregate onto an existing queryset. Doing this before 1.11 either meant custom SQL or hammering the database. Here's the documentation for this, and the example from it:
from django.db.models import OuterRef, Subquery, Sum
comments = Comment.objects.filter(post=OuterRef('pk')).values('post')
total_comments = comments.annotate(total=Sum('length')).values('total')
Post.objects.filter(length__gt=Subquery(total_comments))
They're annotating on the aggregate, which seems weird to me, but whatever.
I'm struggling with this so I'm boiling it right back to the simplest real-world example I have data for. I have Carparks which contain many Spaces. Use Book→Author if that makes you happier but —for now— I just want to annotate on a count of the related model using Subquery*.
spaces = Space.objects.filter(carpark=OuterRef('pk')).values('carpark')
count_spaces = spaces.annotate(c=Count('*')).values('c')
Carpark.objects.annotate(space_count=Subquery(count_spaces))
This gives me a lovely ProgrammingError: more than one row returned by a subquery used as an expression and in my head, this error makes perfect sense. The subquery is returning a list of spaces with the annotated-on total.
The example suggested that some sort of magic would happen and I'd end up with a number I could use. But that's not happening here? How do I annotate on aggregate Subquery data?
Hmm, something's being added to my query's SQL...
I built a new Carpark/Space model and it worked. So the next step is working out what's poisoning my SQL. On Laurent's advice, I took a look at the SQL and tried to make it more like the version they posted in their answer. And this is where I found the real problem:
SELECT "bookings_carpark".*, (SELECT COUNT(U0."id") AS "c"
FROM "bookings_space" U0
WHERE U0."carpark_id" = ("bookings_carpark"."id")
GROUP BY U0."carpark_id", U0."space"
)
AS "space_count" FROM "bookings_carpark";
I've highlighted it but it's that subquery's GROUP BY ... U0."space". It's retuning both for some reason. Investigations continue.
Edit 2: Okay, just looking at the subquery SQL I can see that second group by coming through ☹
In [12]: print(Space.objects_standard.filter().values('carpark').annotate(c=Count('*')).values('c').query)
SELECT COUNT(*) AS "c" FROM "bookings_space" GROUP BY "bookings_space"."carpark_id", "bookings_space"."space" ORDER BY "bookings_space"."carpark_id" ASC, "bookings_space"."space" ASC
Edit 3: Okay! Both these models have sort orders. These are being carried through to the subquery. It's these orders that are bloating out my query and breaking it.
I guess this might be a bug in Django but short of removing the Meta-order_by on both these models, is there any way I can unsort a query at querytime?
*I know I could just annotate a Count for this example. My real purpose for using this is a much more complex filter-count but I can't even get this working.
Shazaam! Per my edits, an additional column was being output from my subquery. This was to facilitate ordering (which just isn't required in a COUNT).
I just needed to remove the prescribed meta-order from the model. You can do this by just adding an empty .order_by() to the subquery. In my code terms that meant:
from django.db.models import Count, OuterRef, Subquery
spaces = Space.objects.filter(carpark=OuterRef('pk')).order_by().values('carpark')
count_spaces = spaces.annotate(c=Count('*')).values('c')
Carpark.objects.annotate(space_count=Subquery(count_spaces))
And that works. Superbly. So annoying.
It's also possible to create a subclass of Subquery, that changes the SQL it outputs. For instance, you can use:
class SQCount(Subquery):
template = "(SELECT count(*) FROM (%(subquery)s) _count)"
output_field = models.IntegerField()
You then use this as you would the original Subquery class:
spaces = Space.objects.filter(carpark=OuterRef('pk')).values('pk')
Carpark.objects.annotate(space_count=SQCount(spaces))
You can use this trick (at least in postgres) with a range of aggregating functions: I often use it to build up an array of values, or sum them.
I just bumped into a VERY similar case, where I had to get seat reservations for events where the reservation status is not cancelled. After trying to figure the problem out for hours, here's what I've seen as the root cause of the problem:
Preface: this is MariaDB, Django 1.11.
When you annotate a query, it gets a GROUP BY clause with the fields you select (basically what's in your values() query selection). After investigating with the MariaDB command line tool why I'm getting NULLs or Nones on the query results, I've came to the conclusion that the GROUP BY clause will cause the COUNT() to return NULLs.
Then, I started diving into the QuerySet interface to see how can I manually, forcibly remove the GROUP BY from the DB queries, and came up with the following code:
from django.db.models.fields import PositiveIntegerField
reserved_seats_qs = SeatReservation.objects.filter(
performance=OuterRef(name='pk'), status__in=TAKEN_TYPES
).values('id').annotate(
count=Count('id')).values('count')
# Query workaround: remove GROUP BY from subquery. Test this
# vigorously!
reserved_seats_qs.query.group_by = []
performances_qs = Performance.objects.annotate(
reserved_seats=Subquery(
queryset=reserved_seats_qs,
output_field=PositiveIntegerField()))
print(performances_qs[0].reserved_seats)
So basically, you have to manually remove/update the group_by field on the subquery's queryset in order for it to not have a GROUP BY appended on it on execution time. Also, you'll have to specify what output field the subquery will have, as it seems that Django fails to recognize it automatically, and raises exceptions on the first evaluation of the queryset. Interestingly, the second evaluation succeeds without it.
I believe this is a Django bug, or an inefficiency in subqueries. I'll create a bug report about it.
Edit: the bug report is here.
Problem
The problem is that Django adds GROUP BY as soon as it sees using an aggregate function.
Solution
So you can just create your own aggregate function but so that Django thinks it is not aggregate. Just like this:
total_comments = Comment.objects.filter(
post=OuterRef('pk')
).order_by().annotate(
total=Func(F('length'), function='SUM')
).values('total')
Post.objects.filter(length__gt=Subquery(total_comments))
This way you get the SQL query like this:
SELECT "testapp_post"."id", "testapp_post"."length"
FROM "testapp_post"
WHERE "testapp_post"."length" > (SELECT SUM(U0."length") AS "total"
FROM "testapp_comment" U0
WHERE U0."post_id" = "testapp_post"."id")
So you can even use aggregate subqueries in aggregate functions.
Example
You can count the number of workdays between two dates, excluding weekends and holidays, and aggregate and summarize them by employee:
class NonWorkDay(models.Model):
date = DateField()
class WorkPeriod(models.Model):
employee = models.ForeignKey(User, on_delete=models.CASCADE)
start_date = DateField()
end_date = DateField()
number_of_non_work_days = NonWorkDay.objects.filter(
date__gte=OuterRef('start_date'),
date__lte=OuterRef('end_date'),
).annotate(
cnt=Func('id', function='COUNT')
).values('cnt')
WorkPeriod.objects.values('employee').order_by().annotate(
number_of_word_days=Sum(F('end_date__year') - F('start_date__year') - number_of_non_work_days)
)
Hope this will help!
A solution which would work for any general aggregation could be implemented using Window classes from Django 2.0. I have added this to the Django tracker ticket as well.
This allows the aggregation of annotated values by calculating the aggregate over partitions based on the outer query model (in the GROUP BY clause), then annotating that data to every row in the subquery queryset. The subquery can then use the aggregated data from the first row returned and ignore the other rows.
Performance.objects.annotate(
reserved_seats=Subquery(
SeatReservation.objects.filter(
performance=OuterRef(name='pk'),
status__in=TAKEN_TYPES,
).annotate(
reserved_seat_count=Window(
expression=Count('pk'),
partition_by=[F('performance')]
),
).values('reserved_seat_count')[:1],
output_field=FloatField()
)
)
If I understand correctly, you are trying to count Spaces available in a Carpark. Subquery seems overkill for this, the good old annotate alone should do the trick:
Carpark.objects.annotate(Count('spaces'))
This will include a spaces__count value in your results.
OK, I have seen your note...
I was also able to run your same query with other models I had at hand. The results are the same, so the query in your example seems to be OK (tested with Django 1.11b1):
activities = Activity.objects.filter(event=OuterRef('pk')).values('event')
count_activities = activities.annotate(c=Count('*')).values('c')
Event.objects.annotate(spaces__count=Subquery(count_activities))
Maybe your "simplest real-world example" is too simple... can you share the models or other information?
"works for me" doesn't help very much. But.
I tried your example on some models I had handy (the Book -> Author type), it works fine for me in django 1.11b1.
Are you sure you're running this in the right version of Django? Is this the actual code you're running? Are you actually testing this not on carpark but some more complex model?
Maybe try to print(thequery.query) to see what SQL it's trying to run in the database. Below is what I got with my models (edited to fit your question):
SELECT (SELECT COUNT(U0."id") AS "c"
FROM "carparks_spaces" U0
WHERE U0."carpark_id" = ("carparks_carpark"."id")
GROUP BY U0."carpark_id") AS "space_count" FROM "carparks_carpark"
Not really an answer, but hopefully it helps.
The short of it is, the table names of all queries that are inside a filter get renamed to u0, u1, ..., so my extra where clauses won't know what table to point to. I would love to not have to hand-make all the queries for every way I might subselect on this data, and my current workaround is to turn my extra'd queries into pk values_lists, but those are really slow and something of an abomination.
Here's what this all looks like. You can mostly ignore the details of what goes in the extra of this manager method, except the first sql line which points to products_product.id:
def by_status(self, *statii):
return self.extra(where=["""products_product.id IN
(SELECT recent.product_id
FROM (
SELECT product_id, MAX(start_date) AS latest
FROM products_productstatus
GROUP BY product_id
) AS recent
JOIN products_productstatus AS ps ON ps.product_id = recent.product_id
WHERE ps.start_date = recent.latest
AND ps.status IN (%s))""" % (', '.join([str(stat) for stat in statii]),)])
Which works wonderfully for all the situations involving only the products_product table.
When I want these products as a subselect, i do:
Piece.objects.filter(
product__in=Product.objects.filter(
pk__in=list(
Product.objects.by_status(FEATURED).values_list('id', flat=True))))
How can I keep the generalized abilities of a query set, yet still use an extra where clause?
At first: the issue is not totally clear to me. Is the second code block in your question the actual code you want to execute? If this is the case the query should work as expected since there is no subselect performed.
I assume so that you want to use the second code block without the list() around the subselect to prevent a second query being performed.
The django documentation refers to this issue in the documentation about the extra method. However its not very easy to overcome this issue.
The easiest but most "hakish" solution is to observe which table alias is produced by django for the table you want to query in the extra method. You can rely on the persistent naming of this alias as long as you construct the query always in the same fashion (you don't change the order of multiple extra methods or filter calls that cause a join).
You can inspect a query that will be execute in the DB queryset by using:
print Model.objects.filter(...).query
This will reveal the aliases that are used for the tables you want to query.
As of Django 1.11, you should be able to use Subquery and OuterRef to generate an equivalent query to your extra (using a correlated subquery rather than a join):
def by_status(self, *statii):
return self.filter(
id__in=Subquery(ProductStatus.values("product_id").filter(
status__in=statii,
product__in=Subquery(ProductStatus.objects.values(
"product_id",
).annotate(
latest=Max("start_date"),
).filter(
latest=OuterRef("start_date"),
).values("product_id"),
),
)
You could probably do it with Window expressions as well (as of Django 2.0).
Note that this is untested, so may need some tweaks.
I've a model called Valor. Valor has a Robot. I'm querying like this:
Valor.objects.filter(robot=r).reverse()[0]
to get the last Valor the the r robot. Valor.objects.filter(robot=r).count() is about 200000 and getting the last items takes about 4 seconds in my PC.
How can I speed it up? I'm querying the wrong way?
The optimal mysql syntax for this problem would be something along the lines of:
SELECT * FROM table WHERE x=y ORDER BY z DESC LIMIT 1
The django equivalent of this would be:
Valor.objects.filter(robot=r).order_by('-id')[:1][0]
Notice how this solution utilizes django's slicing method to limit the queryset before compiling the list of objects.
If none of the earlier suggestions are working, I'd suggest taking Django out of the equation and run this raw sql against your database. I'm guessing at your table names, so you may have to adjust accordingly:
SELECT * FROM valor v WHERE v.robot_id = [robot_id] ORDER BY id DESC LIMIT 1;
Is that slow? If so, make your RDBMS (MySQL?) explain the query plan to you. This will tell you if it's doing any full table scans, which you obviously don't want with a table that large. You might also edit your question and include the schema for the valor table for us to see.
Also, you can see the SQL that Django is generating by doing this (using the query set provided by Peter Rowell):
qs = Valor.objects.filter(robot=r).order_by('-id')[0]
print qs.query
Make sure that SQL is similar to the 'raw' query I posted above. You can also make your RDBMS explain that query plan to you.
It sounds like your data set is going to be big enough that you may want to denormalize things a little bit. Have you tried keeping track of the last Valor object in the Robot object?
class Robot(models.Model):
# ...
last_valor = models.ForeignKey('Valor', null=True, blank=True)
And then use a post_save signal to make the update.
from django.db.models.signals import post_save
def record_last_valor(sender, **kwargs):
if kwargs.get('created', False):
instance = kwargs.get('instance')
instance.robot.last_valor = instance
post_save.connect(record_last_valor, sender=Valor)
You will pay the cost of an extra db transaction when you create the Valor objects but the last_valor lookup will be blazing fast. Play with it and see if the tradeoff is worth it for your app.
Well, there's no order_by clause so I'm wondering about what you mean by 'last'. Assuming you meant 'last added',
Valor.objects.filter(robot=r).order_by('-id')[0]
might do the job for you.
django 1.6 introduces .first() and .last():
https://docs.djangoproject.com/en/1.6/ref/models/querysets/#last
So you could simply do:
Valor.objects.filter(robot=r).last()
Quite fast should also be:
qs = Valor.objects.filter(robot=r) # <-- it doesn't hit the database
count = qs.count() # <-- first hit the database, compute a count
last_item = qs[ count-1 ] # <-- second hit the database, get specified rownum
So, in practice you execute only 2 SQL queries ;)
Model_Name.objects.first()
//To get the first element
Model_name.objects.last()
//For get last()
in my case, the last is not work because there is only one row in the database
maybe help full for you too :)
Is there a limit clause in django? This way you can have the db, simply return a single record.
mysql
select * from table where x = y limit 1
sql server
select top 1 * from table where x = y
oracle
select * from table where x = y and rownum = 1
I realize this isn't translated into django, but someone can come back and clean this up.
The correct way of doing this, is to use the built-in QuerySet method latest() and feeding it whichever column (field name) it should sort by. The drawback is that it can only sort by a single db column.
The current implementation looks like this and is optimized in the same sense as #Aaron's suggestion.
def latest(self, field_name=None):
"""
Returns the latest object, according to the model's 'get_latest_by'
option or optional given field_name.
"""
latest_by = field_name or self.model._meta.get_latest_by
assert bool(latest_by), "latest() requires either a field_name parameter or 'get_latest_by' in the model"
assert self.query.can_filter(), \
"Cannot change a query once a slice has been taken."
obj = self._clone()
obj.query.set_limits(high=1)
obj.query.clear_ordering()
obj.query.add_ordering('-%s' % latest_by)
return obj.get()
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.