django values not working - django

When I try to call values with more than 3 fields it seems to 'break' (ie. it doesn't group duplicate entries together)
My model is a through model with three fields, 2 ForeignKey and one DateTimeField
ProjectView(models.Model):
user = models.ForeignKey(User)
project = models.ForeignKey(Project)
datetime_created = models.DateTimeField()
I want to do:
ProjectView.objects.filter(datetime_created__gt=yesterday).values('project__id', 'project__title', 'project__thumbnail', 'project__creator_username')
If i get rid of any one of the values fields it groups them by same projects without duplicates, if there are 4 values it seems to do no grouping. Am i doing something wrong?

If you take a look at the docs for values, you'll see no guarantee of grouping or distinct. If you want that functionality, you'll have to call .order_by() and/or .distinct() when making you call to the ORM.
That it works at all is probably just a side effect of the SQL generated. If you want to see the SQL, take a look at Django-debug-toolbar

Related

django annotate queryset with field comparison result

I have a queryset like this:
predicts = Prediction.objects.select_related('match').filter(match_id=pk)
I need to annotate this with a new field is_correct. I need to compare two string fields and the result should be annotated in this new field. the fields that I want to compare are:
predict from Prediction table
result from Match table (that has been joined through select_related)
I need to know what expression should I put inside my annotate function; below I have my current code which throughs a TypeError exception:
predicts = predicts.annotate(is_correct=(F('predict') == F('result')))
all help will be greatly appreciated.
UPDATE:
I found an alternative solution that does the job for me (filtering the Prediction based on Match result using filter and exclude), but I still like to know how to address this specific case where the new annotated field is the result of the comparison between two other fields of the queryset. For those who may need it, in Django 2.2 and later the Nullif database function does a comparison between two fields.
You can use the extra function, a hook for injecting specific clauses into the SQL.
First of all, we must know the names of the apps and the models, or the name of the tables in the database.
Assuming that in your case, the two tables are called "app_prediction" and "app_match".
The sentence would be as follows:
Prediction.objects.select_related('match').extra(
select={'is_correct': "app_prediction.predict = app_match.result"}
)
This will add a field called is_correct in your result,
in the database, the fields and tables must be called in the same way.
It would be best to see the models.

Migrate Django model to unique_together constraint

I have a model with three fields
class MyModel(models.Model):
a = models.ForeignKey(A)
b = models.ForeignKey(B)
c = models.ForeignKey(C)
I want to enforce a unique constraint between these fields, and found django's unique_together, which seems to be the solution. However, I already have an existing database, and there are many duplicates. I know that since unique_together works at the database level, I need to unique-ify the rows, and then try a migration.
Is there a good way to go about removing duplicates (where a duplicate has the same (A,B,C)) so that I can run migration to get the unique_together contstraint?
If you are happy to choose one of the duplicates arbitrarily, I think the following might do the trick. Perhaps not the most efficient but simple enough and I guess you only need to run this once. Please verify this all works yourself on some test data in case I've done something silly, since you are about to delete a bunch of data.
First we find groups of objects which form duplicates. For each group, (arbitrarily) pick a "master" that we are going to keep. Our chosen method is to pick the one with lowest pk
from django.db.models import Min, Count
master_pks = MyModel.objects.values('A', 'B', 'C'
).annotate(Min('pk'), count=Count('pk')
).filter(count__gt=1
).values_list('pk__min', flat=True)
we then loop over each master, and delete all its duplicates
masters = MyModel.objects.in_bulk( list(master_pks) )
for master in masters.values():
MyModel.objects.filter(a=master.a, b=master.b, c=master.c
).exclude(pk=master.pk).del_ACCIDENT_PREVENTION_ete()
I want to add a slightly improved answer that will delete everything in a single query, instead of looping and deleting for each duplicate group. This will be much faster if you have a lot of records.
non_dupe_pks = list(
Model.objects.values('A', 'B', 'C')
.annotate(Min('pk'), count=Count('pk'))
.order_by()
.values_list('pk__min', flat=True)
)
dupes = Model.objects.exclude(pk__in=non_dupe_pks)
dupes.delete()
It's important to add order_by() in the first query otherwise the default ordering in the model might mess up with the aggregation.
You can comment out the last line and use dupes.count() to check if the query is working as expected.

Annotating a Django queryset with a left outer join?

Say I have a model:
class Foo(models.Model):
...
and another model that basically gives per-user information about Foo:
class UserFoo(models.Model):
user = models.ForeignKey(User)
foo = models.ForeignKey(Foo)
...
class Meta:
unique_together = ("user", "foo")
I'd like to generate a queryset of Foos but annotated with the (optional) related UserFoo based on user=request.user.
So it's effectively a LEFT OUTER JOIN on (foo.id = userfoo.foo_id AND userfoo.user_id = ...)
A solution with raw might look like
foos = Foo.objects.raw("SELECT foo.* FROM foo LEFT OUTER JOIN userfoo ON (foo.id = userfoo.foo_id AND foo.user_id = %s)", [request.user.id])
You'll need to modify the SELECT to include extra fields from userfoo which will be annotated to the resulting Foo instances in the queryset.
This answer might not be exactly what you are looking for but since its the first result in google when searching for "django annotate outer join" so I will post it here.
Note: tested on Djang 1.7
Suppose you have the following models
class User(models.Model):
name = models.CharField()
class EarnedPoints(models.Model):
points = models.PositiveIntegerField()
user = models.ForeignKey(User)
To get total user points you might do something like that
User.objects.annotate(points=Sum("earned_points__points"))
this will work but it will not return users who have no points, here we need outer join without any direct hacks or raw sql
You can achieve that by doing this
users_with_points = User.objects.annotate(points=Sum("earned_points__points"))
result = users_with_points | User.objects.exclude(pk__in=users_with_points)
This will be translated into OUTER LEFT JOIN and all users will be returned. users who has no points will have None value in their points attribute.
Hope that helps
Notice: This method does not work in Django 1.6+. As explained in tcarobruce's comment below, the promote argument was removed as part of ticket #19849: ORM Cleanup.
Django doesn't provide an entirely built-in way to do this, but it's not neccessary to construct an entirely raw query. (This method doesn't work for selecting * from UserFoo, so I'm using .comment as an example field to include from UserFoo.)
The QuerySet.extra() method allows us to add terms to the SELECT and WHERE clauses of our query. We use this to include the fields from UserFoo table in our results, and limit our UserFoo matches to the current user.
results = Foo.objects.extra(
select={"user_comment": "UserFoo.comment"},
where=["(UserFoo.user_id IS NULL OR UserFoo.user_id = %s)"],
params=[request.user.id]
)
This query still needs the UserFoo table. It would be possible to use .extras(tables=...) to get an implicit INNER JOIN, but for an OUTER JOIN we need to modify the internal query object ourself.
connection = (
UserFoo._meta.db_table, User._meta.db_table, # JOIN these tables
"user_id", "id", # on these fields
)
results.query.join( # modify the query
connection, # with this table connection
promote=True, # as LEFT OUTER JOIN
)
We can now evaluate the results. Each instance will have a .user_comment property containing the value from UserFoo, or None if it doesn't exist.
print results[0].user_comment
(Credit to this blog post by Colin Copeland for showing me how to do OUTER JOINs.)
I stumbled upon this problem I was unable to solve without resorting to raw SQL, but I did not want to rewrite the entire query.
Following is a description on how you can augment a queryset with an external raw sql, without having to care about the actual query that generates the queryset.
Here's a typical scenario: You have a reddit like site with a LinkPost model and a UserPostVote mode, like this:
class LinkPost(models.Model):
some fields....
class UserPostVote(models.Model):
user = models.ForeignKey(User,related_name="post_votes")
post = models.ForeignKey(LinkPost,related_name="user_votes")
value = models.IntegerField(null=False, default=0)
where the userpostvote table collect's the votes of users on posts.
Now you're trying to display the front page for a user with a pagination app, but you want the arrows to be red for posts the user has voted on.
First you get the posts for the page:
post_list = LinkPost.objects.all()
paginator = Paginator(post_list,25)
posts_page = paginator.page(request.GET.get('page'))
so now you have a QuerySet posts_page generated by the django paginator that selects the posts to display. How do we now add the annotation of the user's vote on each post before rendering it in a template?
Here's where it get's tricky and I was unable to find a clean ORM solution. select_related won't allow you to only get votes corresponding to the logged in user and looping over the posts would do bunch queries instead of one and doing it all raw mean's we can't use the queryset from the pagination app.
So here's how I do it:
q1 = posts_page.object_list.query # The query object of the queryset
q1_alias = q1.get_initial_alias() # This forces the query object to generate it's sql
(q1str, q1param) = q1.sql_with_params() #This gets the sql for the query along with
#parameters, which are none in this example
we now have the query for the queryset, and just wrap it, alias and left outer join to it:
q2_augment = "SELECT B.value as uservote, A.*
from ("+q1str+") A LEFT OUTER JOIN reddit_userpostvote B
ON A.id = B.post_id AND B.user_id = %s"
q2param = (request.user.id,)
posts_augmented = LinkPost.objects.raw(q2_augment,q1param+q2param)
voila! Now we can access post.uservote for a post in the augmented queryset.
And we just hit the database with a single query.
The two queries you suggest are as good as you're going to get (without using raw()), this type of query isn't representable in the ORM at present time.
You could do this using simonw's django-queryset-transform to avoid hard-coding a raw SQL query - the code would look something like this:
def userfoo_retriever(qs):
userfoos = dict((i.pk, i) for i in UserFoo.objects.filter(foo__in=qs))
for i in qs:
i.userfoo = userfoos.get(i.pk, None)
for foo in Foo.objects.filter(…).tranform(userfoo_retriever):
print foo.userfoo
This approach has been quite successful for this need and to efficiently retrieve M2M values; your query count won't be quite as low but on certain databases (cough MySQL cough) doing two simpler queries can often be faster than one with complex JOINs and many of the cases where I've most needed it had additional complexity which would have been even harder to hack into an ORM expression.
As for outerjoins:
Once you have a queryset qs from foo that includes a reference to columns from userfoo, you can promote the inner join to an outer join with
qs.query.promote_joins(["userfoo"])
You shouldn't have to resort to extra or raw for this.
The following should work.
Foo.objects.filter(
Q(userfoo_set__user=request.user) |
Q(userfoo_set=None) # This forces the use of LOUTER JOIN.
).annotate(
comment=F('userfoo_set__comment'),
# ... annotate all the fields you'd like to see added here.
)
The only way I see to do this without using raw etc. is something like this:
Foo.objects.filter(
Q(userfoo_set__isnull=True)|Q(userfoo_set__isnull=False)
).annotate(bar=Case(
When(userfoo_set__user_id=request.user, then='userfoo_set__bar')
))
The double Q trick ensures that you get your left outer join.
Unfortunately you can't set your request.user condition in the filter() since it may filter out successful joins on UserFoo instances with the wrong user, hence filtering out rows of Foo that you wanted to keep (which is why you ideally want the condition in the ON join clause instead of in the WHERE clause).
Because you can't filter out the rows that have an unwanted user value, you have to select rows from UserFoo with a CASE.
Note also that one Foo may join to many UserFoo records, so you may want to consider some way to retrieve distinct Foos from the output.
maparent's comment put me on the right way:
from django.db.models.sql.datastructures import Join
for alias in qs.query.alias_map.values():
if isinstance(alias, Join):
alias.nullable = True
qs.query.promote_joins(qs.query.tables)

Select DISTINCT individual columns in django?

I'm curious if there's any way to do a query in Django that's not a "SELECT * FROM..." underneath. I'm trying to do a "SELECT DISTINCT columnName FROM ..." instead.
Specifically I have a model that looks like:
class ProductOrder(models.Model):
Product = models.CharField(max_length=20, promary_key=True)
Category = models.CharField(max_length=30)
Rank = models.IntegerField()
where the Rank is a rank within a Category. I'd like to be able to iterate over all the Categories doing some operation on each rank within that category.
I'd like to first get a list of all the categories in the system and then query for all products in that category and repeat until every category is processed.
I'd rather avoid raw SQL, but if I have to go there, that'd be fine. Though I've never coded raw SQL in Django/Python before.
One way to get the list of distinct column names from the database is to use distinct() in conjunction with values().
In your case you can do the following to get the names of distinct categories:
q = ProductOrder.objects.values('Category').distinct()
print q.query # See for yourself.
# The query would look something like
# SELECT DISTINCT "app_productorder"."category" FROM "app_productorder"
There are a couple of things to remember here. First, this will return a ValuesQuerySet which behaves differently from a QuerySet. When you access say, the first element of q (above) you'll get a dictionary, NOT an instance of ProductOrder.
Second, it would be a good idea to read the warning note in the docs about using distinct(). The above example will work but all combinations of distinct() and values() may not.
PS: it is a good idea to use lower case names for fields in a model. In your case this would mean rewriting your model as shown below:
class ProductOrder(models.Model):
product = models.CharField(max_length=20, primary_key=True)
category = models.CharField(max_length=30)
rank = models.IntegerField()
It's quite simple actually if you're using PostgreSQL, just use distinct(columns) (documentation).
Productorder.objects.all().distinct('category')
Note that this feature has been included in Django since 1.4
User order by with that field, and then do distinct.
ProductOrder.objects.order_by('category').values_list('category', flat=True).distinct()
The other answers are fine, but this is a little cleaner, in that it only gives the values like you would get from a DISTINCT query, without any cruft from Django.
>>> set(ProductOrder.objects.values_list('category', flat=True))
{u'category1', u'category2', u'category3', u'category4'}
or
>>> list(set(ProductOrder.objects.values_list('category', flat=True)))
[u'category1', u'category2', u'category3', u'category4']
And, it works without PostgreSQL.
This is less efficient than using a .distinct(), presuming that DISTINCT in your database is faster than a python set, but it's great for noodling around the shell.
Update:
This is answer is great for making queries in the Django shell during development. DO NOT use this solution in production unless you are absolutely certain that you will always have a trivially small number of results before set is applied. Otherwise, it's a terrible idea from a performance standpoint.

Django DB, finding Categories whose Items are all in a subset

I have a two models:
class Category(models.Model):
pass
class Item(models.Model):
cat = models.ForeignKey(Category)
I am trying to return all Categories for which all of that category's items belong to a given subset of item ids (fixed thanks). For example, all categories for which all of the items associated with that category have ids in the set [1,3,5].
How could this be done using Django's query syntax (as of 1.1 beta)? Ideally, all the work should be done in the database.
Category.objects.filter(item__id__in=[1, 3, 5])
Django creates the reverse relation ship on the model without the foreign key. You can filter on it by using its related name (usually just the model name lowercase but it can be manually overwritten), two underscores, and the field name you want to query on.
lets say you require all items to be in the following set:
allowable_items = set([1,3,4])
one bruteforce solution would be to check the item_set for every category as so:
categories_with_allowable_items = [
category for category in
Category.objects.all() if
set([item.id for item in category.item_set.all()]) <= allowable_items
]
but we don't really have to check all categories, as categories_with_allowable_items is always going to be a subset of the categories related to all items with ids in allowable_items... so that's all we have to check (and this should be faster):
categories_with_allowable_items = set([
item.category for item in
Item.objects.select_related('category').filter(pk__in=allowable_items) if
set([siblingitem.id for siblingitem in item.category.item_set.all()]) <= allowable_items
])
if performance isn't really an issue, then the latter of these two (if not the former) should be fine. if these are very large tables, you might have to come up with a more sophisticated solution. also if you're using a particularly old version of python remember that you'll have to import the sets module
I've played around with this a bit. If QuerySet.extra() accepted a "having" parameter I think it would be possible to do it in the ORM with a bit of raw SQL in the HAVING clause. But it doesn't, so I think you'd have to write the whole query in raw SQL if you want the database doing the work.
EDIT:
This is the query that gets you part way there:
from django.db.models import Count
Category.objects.annotate(num_items=Count('item')).filter(num_items=...)
The problem is that for the query to work, "..." needs to be a correlated subquery that looks up, for each category, the number of its items in allowed_items. If .extra had a "having" argument, you'd do it like this:
Category.objects.annotate(num_items=Count('item')).extra(having="num_items=(SELECT COUNT(*) FROM app_item WHERE app_item.id in % AND app_item.cat_id = app_category.id)", having_params=[allowed_item_ids])