Can anyone "translate" this MySQL query to Django chain or Q(). This is just an example (but valid) query, but I want to see it with my own eyes, because Django documentation doesn't look very noob friendly in this aspect and I couldn't get those chains and stuff to work.
mysql -> SELECT position, COUNT(position) FROM
-> (SELECT * FROM log WHERE (aspect LIKE 'es%' OR brand LIKE '%pj%')
-> AND tag IN ('in','out')) AS list
-> GROUP BY position ORDER BY COUNT(position) DESC;
While I think chaining filters would be more convenient for me in the future, this below just seems way more straightforward at the moment.
query = "the query from above"
cursor.execute(query)
[new_list.append([item for item in row ]) for row in cursor]
...or should I just quit()
from django.db.models import Count,Q
myfilter = [Q(aspect__startswith="es")|Q(brand__contains="pj"),tag__in=['in','out']]
# condition 1 OR condition2 AND condition3
qry = Log.objects.filter(*myfilter).values('position').annotate({"count":Count('position')})
print(qry.values())
I think maybe gets you the same answer using django ... close anyway I think
whether it is better or not i suppose is in the eye of the beholder
myfilter = Q(aspect__startswith="es") | Q(brand__contains="pj"), Q(tag__in=['in', 'out'])
qry = Log.objects.filter(*myfilter).values('position').annotate(Count('position')).order_by('-position__count')
had to add Q before tag__inbrackets on myfilter did not make a difference
annotate({"count":Count('position')}) gave an error QuerySet.annotate() received non-expression(s)
and to sort in descending order had to add
.order_by('-position__count')
output is a dictionary, but that's perfect for displaying on the website
So You were close. Now the data is matching mysql output and I understand it better how to do the QuerySets and filters. Thanks
Related
Let's say I have following models:
class Invoice(models.Model):
...
class Note(models.Model):
invoice = models.ForeignKey(Invoice, related_name='notes', on_delete=models.CASCADE)
text = models.TextField()
and I want to select Invoices that have some notes. I would write it using annotate/Exists like this:
Invoice.objects.annotate(
has_notes=Exists(Note.objects.filter(invoice_id=OuterRef('pk')))
).filter(has_notes=True)
This works well enough, filters only Invoices with notes. However, this method results in the field being present in the query result, which I don't need and means worse performance (SQL has to execute the subquery 2 times).
I realize I could write this using extra(where=) like this:
Invoice.objects.extra(where=['EXISTS(SELECT 1 FROM note WHERE invoice_id=invoice.id)'])
which would result in the ideal SQL, but in general it is discouraged to use extra / raw SQL.
Is there a better way to do this?
You can remove annotations from the SELECT clause using .values() query set method. The trouble with .values() is that you have to enumerate all names you want to keep instead of names you want to skip, and .values() returns dictionaries instead of model instances.
Django internaly keeps the track of removed annotations in
QuerySet.query.annotation_select_mask. So you can use it to tell Django, which annotations to skip even wihout .values():
class YourQuerySet(QuerySet):
def mask_annotations(self, *names):
if self.query.annotation_select_mask is None:
self.query.set_annotation_mask(set(self.query.annotations.keys()) - set(names))
else:
self.query.set_annotation_mask(self.query.annotation_select_mask - set(names))
return self
Then you can write:
invoices = (Invoice.objects
.annotate(has_notes=Exists(Note.objects.filter(invoice_id=OuterRef('pk'))))
.filter(has_notes=True)
.mask_annotations('has_notes')
)
to skip has_notes from the SELECT clause and still geting filtered invoice instances. The resulting SQL query will be something like:
SELECT invoice.id, invoice.foo FROM invoice
WHERE EXISTS(SELECT note.id, note.bar FROM notes WHERE note.invoice_id = invoice.id) = True
Just note that annotation_select_mask is internal Django API that can change in future versions without a warning.
Ok, I've just noticed in Django 3.0 docs, that they've updated how Exists works and can be used directly in filter:
Invoice.objects.filter(Exists(Note.objects.filter(invoice_id=OuterRef('pk'))))
This will ensure that the subquery will not be added to the SELECT columns, which may result in a better performance.
Changed in Django 3.0:
In previous versions of Django, it was necessary to first annotate and then filter against the annotation. This resulted in the annotated value always being present in the query result, and often resulted in a query that took more time to execute.
Still, if someone knows a better way for Django 1.11, I would appreciate it. We really need to upgrade :(
We can filter for Invoices that have, when we perform a LEFT OUTER JOIN, no NULL as Note, and make the query distinct (to avoid returning the same Invoice twice).
Invoice.objects.filter(notes__isnull=False).distinct()
This is best optimize code if you want to get data from another table which primary key reference stored in another table
Invoice.objects.filter(note__invoice_id=OuterRef('pk'),)
We should be able to clear the annotated field using the below method.
Invoice.objects.annotate(
has_notes=Exists(Note.objects.filter(invoice_id=OuterRef('pk')))
).filter(has_notes=True).query.annotations.clear()
I'm having loads of trouble translating some SQL into Django.
Imagine we have some cars, each with a unique VIN, and we record the dates that they are in the shop with some other data. (Please ignore the reason one might structure the data this way. It's specifically for this question. :-) )
class ShopVisit(models.Model):
vin = models.CharField(...)
date_in_shop = models.DateField(...)
mileage = models.DecimalField(...)
boolfield = models.BooleanField(...)
We want a single query to return a Queryset with the most recent record for each vin and update it!
special_vins = [...]
# Doesn't work
ShopVisit.objects.filter(vin__in=special_vins).annotate(max_date=Max('date_in_shop').filter(date_in_shop=F('max_date')).update(boolfield=True)
# Distinct doesn't work with update
ShopVisit.objects.filter(vin__in=special_vins).order_by('vin', '-date_in_shop).distinct('vin').update(boolfield=True)
Yes, I could iterate over a queryset. But that's not very efficient and it takes a long time when I'm dealing with around 2M records. The SQL that could do this is below (I think!):
SELECT *
FROM cars
INNER JOIN (
SELECT MAX(dateInShop) as maxtime, vin
FROM cars
GROUP BY vin
) AS latest_record ON (cars.dateInShop= maxtime)
AND (latest_record.vin = cars.vin)
So how can I make this happen with Django?
This is somewhat untested, and relies on Django 1.11 for Subqueries, but perhaps something like:
latest_visits = Subquery(ShopVisit.objects.filter(id=OuterRef('id')).order_by('-date_in_shop').values('id')[:1])
ShopVisit.objects.filter(id__in=latest_visits)
I had a similar model, so went to test it but got an error of:
"This version of MySQL doesn't yet support 'LIMIT & IN/ALL/ANY/SOME subquery"
The SQL it generated looked reasonably like what you want, so I think the idea is sound. If you use PostGres, perhaps it has support for that type of subquery.
Here's the SQL it produced (trimmed up a bit and replaced actual names with fake ones):
SELECT `mymodel_activity`.* FROM `mymodel_activity` WHERE `mymodel_activity`.`id` IN (SELECT U0.`id` FROM `mymodel_activity` U0 WHERE U0.`id` = (`mymodel_activity`.`id`) ORDER BY U0.`date_in_shop` DESC LIMIT 1)
I wonder if you found the solution yourself.
I could come up with only raw query string. Django Raw SQL query Manual
UPDATE "yourapplabel_shopvisit"
SET boolfield = True WHERE date_in_shop
IN (SELECT MAX(date_in_shop) FROM "yourapplabel_shopvisit" GROUP BY vin);
I have a general ledger table in my DB with the columns: member_id, is_credit and amount. I want to get the current balance of the member.
Ideally that can be got by two queries where the first query has is_credit == True and the second query is_credit == False something close to:
credit_amount = session.query(func.sum(Funds.amount).label('Debit_Amount')).filter(Funds.member_id==member_id, Funds.is_credit==True)
debit_amount = session.query(func.sum(Funds.amount).label('Debit_Amount')).filter(Funds.member_id==member_id, Funds.is_credit==False)
balance = credit_amount - debit_amount
and then subtract the result. Is there a way to have the above run in one query to give the balance?
From the comments you state that hybrids are too advanced right now, so I will propose an easier but not as efficient solution (still its okay):
(session.query(Funds.is_credit, func.sum(Funds.amount).label('Debit_Amount')).
filter(Funds.member_d==member_id).group_by(Funds.is_credit))
What will this do? You will recieve a two-row result, one has the credit, the other the debit, depending on the is_credit property of the result. The second part (Debit_Amount) will be the value. You then evaluate them to get the result: Only one query that fetches both values.
If you are unsure what group_by does, I recommend you read up on SQL before doing it in SQLAlchemy. SQLAlchemy offers very easy usage of SQL but it requires that you understand SQL as well. Thus, I recommend: First build a query in SQL and see that it does what you want - then translate it to SQLAlchemy and see that it does the same. Otherwise SQLAlchemy will often generate highly inefficient queries, because you asked for the wrong thing.
I'm trying to have my popular_query subquery remove dupe Place.id, but it doesn't remove it. This is the code below. I tried using distinct but it does not respect the order_by rule.
SimilarPost = aliased(Post)
SimilarPostOption = aliased(PostOption)
popular_query = (db.session.query(Post, func.count(SimilarPost.id)).
join(Place, Place.id == Post.place_id).
join(PostOption, PostOption.post_id == Post.id).
outerjoin(SimilarPostOption, PostOption.val == SimilarPostOption.val).
join(SimilarPost,SimilarPost.id == SimilarPostOption.post_id).
filter(Place.id == Post.place_id).
filter(self.radius_cond()).
group_by(Post.id).
group_by(Place.id).
order_by(desc(func.count(SimilarPost.id))).
order_by(desc(Post.timestamp))
).subquery().select()
all_posts = db.session.query(Post).select_from(filter.pick()).all()
I did a test printout with
print [x.place.name for x in all_posts]
[u'placeB', u'placeB', u'placeB', u'placeC', u'placeC', u'placeA']
How can I fix this?
Thanks!
This should get you what you want:
SimilarPost = aliased(Post)
SimilarPostOption = aliased(PostOption)
post_popularity = (db.session.query(func.count(SimilarPost.id))
.select_from(PostOption)
.filter(PostOption.post_id == Post.id)
.correlate(Post)
.outerjoin(SimilarPostOption, PostOption.val == SimilarPostOption.val)
.join(SimilarPost, sql.and_(
SimilarPost.id == SimilarPostOption.post_id,
SimilarPost.place_id == Post.place_id)
)
.as_scalar())
popular_post_id = (db.session.query(Post.id)
.filter(Post.place_id == Place.id)
.correlate(Place)
.order_by(post_popularity.desc())
.limit(1)
.as_scalar())
deduped_posts = (db.session.query(Post, post_popularity)
.join(Place)
.filter(Post.id == popular_post_id)
.order_by(post_popularity.desc(), Post.timestamp.desc())
.all())
I can't speak to the runtime performance with large data sets, and there may be a better solution, but that's what I managed to synthesize from quite a few sources (MySQL JOIN with LIMIT 1 on joined table, SQLAlchemy - subquery in a WHERE clause, SQLAlchemy Query documentation). The biggest complicating factor is that you apparently need to use as_scalar to nest the subqueries in the right places, and therefore can't return both the Post id and the count from the same subquery.
FWIW, this is kind of a behemoth and I concur with user1675804 that SQLAlchemy code this deep is hard to grok and not very maintainable. You should take a hard look at any more low-tech solutions available like adding columns to the db or doing more of the work in python code.
I don't want to sound like the bad guy here but... in my opinion your approach to the issue seems far less than optimal... if you're using postgresql you could simplify the whole thing using WITH ... but a better approach factoring in my assumption that these posts will be read much more often than updated would be to add some columns to your tables that are updated by triggers on insert/update to other tables, at least if performance is likely to ever become an issue this is the solution I'd go with
Not very familiar with sqlalchemy, so can't write it in clear code for you, but the only other solution I can come up with uses at least a subquery to select the things from order by for each of the columns in group by, and that will add significantly to your already slow query
Using SQLAlchemy, I have a one to many relation with two tables - users and scores. I am trying to query the top 10 users sorted by their aggregate score over the past X amount of days.
users:
id
user_name
score
scores:
user
score_amount
created
My current query is:
top_users = DBSession.query(User).options(eagerload('scores')).filter_by(User.scores.created > somedate).order_by(func.sum(User.scores).desc()).all()
I know this is clearly not correct, it's just my best guess. However, after looking at the documentation and googling I cannot find an answer.
EDIT:
Perhaps it would help if I sketched what the MySQL query would look like:
SELECT user.*, SUM(scores.amount) as score_increase
FROM user LEFT JOIN scores ON scores.user_id = user.user_id
WITH scores.created_at > someday
ORDER BY score_increase DESC
The single-joined-row way, with a group_by added in for all user columns although MySQL will let you group on just the "id" column if you choose:
sess.query(User, func.sum(Score.amount).label('score_increase')).\
join(User.scores).\
filter(Score.created_at > someday).\
group_by(User).\
order_by("score increase desc")
Or if you just want the users in the result:
sess.query(User).\
join(User.scores).\
filter(Score.created_at > someday).\
group_by(User).\
order_by(func.sum(Score.amount))
The above two have an inefficiency in that you're grouping on all columns of "user" (or you're using MySQL's "group on only a few columns" thing, which is MySQL only). To minimize that, the subquery approach:
subq = sess.query(Score.user_id, func.sum(Score.amount).label('score_increase')).\
filter(Score.created_at > someday).\
group_by(Score.user_id).subquery()
sess.query(User).join((subq, subq.c.user_id==User.user_id)).order_by(subq.c.score_increase)
An example of the identical scenario is in the ORM tutorial at: http://docs.sqlalchemy.org/en/latest/orm/tutorial.html#selecting-entities-from-subqueries
You will need to use a subquery in order to compute the aggregate score for each user. Subqueries are described here: http://www.sqlalchemy.org/docs/05/ormtutorial.html?highlight=subquery#using-subqueries
I am assuming the column (not the relation) you're using for the join is called Score.user_id, so change it if this is not the case.
You will need to do something like this:
DBSession.query(Score.user_id, func.sum(Score.score_amount).label('total_score')).group_by(Score.user_id).filter(Score.created > somedate).order_by('total_score DESC')[:10]
However this will result in tuples of (user_id, total_score). I'm not sure if the computed score is actually important to you, but if it is, you will probably want to do something like this:
users_scores = []
q = DBSession.query(Score.user_id, func.sum(Score.score_amount).label('total_score')).group_by(Score.user_id).filter(Score.created > somedate).order_by('total_score DESC')[:10]
for user_id, total_score in q:
user = DBSession.query(User)
users_scores.append((user, total_score))
This will result in 11 queries being executed, however. It is possible to do it all in a single query, but due to various limitations in SQLAlchemy, it will likely create a very ugly multi-join query or subquery (dependent on engine) and it won't be very performant.
If you plan on doing something like this often and you have a large amount of scores, consider denormalizing the current score onto the user table. It's more work to upkeep, but will result in a single non-join query like:
DBSession.query(User).order_by(User.computed_score.desc())
Hope that helps.