I am running django with postgres and I need to query some record from a table, sorting them by rank, and get unique entry in respect of a foreign key.
Basically my model is something like this:
class BookingCatalog(models.Model):
.......
boat = models.ForeignKey(Boat, verbose_name=u"Boat", related_name="booking_catalog")
is_skippered = models.BooleanField(u'Is Skippered',choices=SKIPPER_CHOICE, default=False)
rank = models.IntegerField(u"Rank", default=0, db_index=True)
.......
The idea is to run something like this
BookingCatalog.objects.filter (...).order_by ('-rank', 'boat', 'is_skippered').distinct ('boat')
Unfortunately, this is not working since I am using postgres which raises this exception:
SELECT DISTINCT ON expressions must match initial ORDER BY expressions
What should I do instead?
The distinct argument has to match the first order argument. Try using this:
BookingCatalog.objects.filter(...) \
.order_by('boat', '-rank', 'is_skippered') \
.distinct('boat')
The way that I do this is to select the distinct objects first, then use those results to filter another queryset.
# Initial filtering
result = BookingCatalog.objects.filter(...)
# Make the results distinct
result = result.order_by('boat').distinct('boat')
# Extract the pks from the result
result_pks = result.values_list("pk", flat=True)
# Use those result pks to create a new queryset
restult_2 = BookingCatalog.objects.filter(pk__in=result_pks)
# Order that queryset
result_2 = result_2.order_by('-rank', 'is_skippered')
print(result_2)
I believe that this results in a single query being executed, which contains a subquery. I would love for someone who knows more about Django to confirm this though.
..ordering by -rank will give you the lowest rank of each duplicate, but your overall query results will be ordered by boat field
BookingCatalog.objects.filter (...).order_by('boat','-rank','is_skippered').distinct('boat')
For more info on, refer to Django documentation
including for Postgres
Related
Following on from this question: Django Postgresql ArrayField aggregation
I have an ArrayField of Categories and I would like to retrieve all unique values it has - however the results should be filtered so that only values starting with the supplied string are returned.
What's the "most Django" way of doing this?
Given an Animal model that looks like this:
class Animal(models.Model):
# ...
categories = ArrayField(
models.CharField(max_length=255, blank=True),
default=list,
)
# ...
Then, as per the other question's answer, this works for finding all categories, unfiltered.
all_categories = (
Animal.objects
.annotate(categories_element=Func(F('categories'), function='unnest'))
.values_list('categories_element', flat=True)
.distinct()
)
However, now, when I attempt to filter the result I get failure, not just with __startswith but all types of filter:
all_categories.filter(categories_element__startswith('ga'))
all_categories.filter(categories_element='dog')
Bottom of stacktrace is:
DataError: malformed array literal: "dog"
...
DETAIL: Array value must start with "{" or dimension information.
... and it appears that it's because Django tries to do a second UNNEST - this is the SQL it generates:
...) WHERE unnest("animal"."categories") = dog::text[]
If I write the query in PSQL then it appears to require a subquery as a result of the UNNEST:
SELECT categories_element
FROM (
SELECT UNNEST(animal.categories) as categories_element
) ul
WHERE ul.categories_element like 'Ga%';
Is there a way to get Django ORM to make a working query? Or should I just give up on the ORM and use raw SQL?
You probably have the wrong database design.
Tip: Arrays are not sets; searching for specific array elements can be
a sign of database misdesign. Consider using a separate table with a
row for each item that would be an array element. This will be easier
to search, and is likely to scale better for a large number of
elements.
http://www.postgresql.org/docs/9.1/static/arrays.html
I have two tables like so:
class Collection(models.Model):
name = models.CharField()
class Image(models.Model):
name = models.CharField()
image = models.ImageField()
collection = models.ForeignKey(Collection)
I'd like to retrieve the first image out of every collection. I have attempted:
image_list = Image.objects.order_by('collection.id').distinct('collection.id')
but it didn't work out the way I expected :(
Any ideas?
Thanks.
Don't use dots to separate fields that span relations in Django; the double-underscore convention is used instead -- it means "follow this relation to get to this field"
this is more correct:
image_list = Image.objects.order_by('collection__id').distinct('collection__id')
However, it probably doesn't do what you want.
The concept of "first" doesn't always apply in relational databases the way you seem to be using it. For all of the records in the image table with the same collection id, there is no record which is 'first' or 'last' -- they're all just records. You could put another field on that table to define a specific order, or you could order by id, or alphabetically by name, but none of those will happen by default.
What will probably work best for you is to get the list of collections with one query, and then get a single item per collection, in separate queries:
collection_ids = Image.objects.values_list('collection', flat=True).distinct()
image_list = [
Image.objects.filter(collection__id=c)[0] for c in collection_ids
]
If you want to apply an order to the Images, to define which is 'first', then modify it like this:
collection_ids = Image.objects.values_list('collection', flat=True).distinct()
image_list = [
Image.objects.filter(collection__id=c).order_by('-id')[0] for c in collection_ids
]
You could also write raw SQL -- MySQL aggregation has the interesting property that fields which are not aggregated over can still appear in the final output, and essentially take a random value from the set of matching records. Something like this might work:
Image.objects.raw("SELECT image.* FROM app_image GROUP BY collection_id")
This query should get you one image from each collection, but you will have no control over which one is returned.
As written in my comment, you cannot use specific fields with distinct under MySQL. However, you can achieve the same result with the following:
from itertools import groupby
all_images = Image.objects.order_by('collection__id')
images_by_collection = groupby(all_images, lambda image: image.collection_id)
image_list = sum([group for key, group in images_by_collection], [])
Unfortunately, this results in a "bigger" query to the DB (all images are retrieved).
dict([(c.id, c.image_set.all()[0]) for c in Collection.objects.all()])
That will create a dictionary of the first image (by default ordering) in each collection, keyed by the collection's id. Be aware, though, that this will generate 1+N queries, where N is the total number of collection objects.
To get around that, you'll either need to wait for Django 1.4 and prefetch_related or use something like django-batch-select.
First get the distinct result, then do your filters.
I think you should try this one.
image_list = Image.objects.distinct()
image_list = image_list.order_by('collection__id')
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)
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.
I've got this model:
class Visit(models.Model):
timestamp = models.DateTimeField(editable=False)
ip_address = models.IPAddressField(editable=False)
If a user visits multiple times in one day, how can I filter for unique rows based on the ip field? (I want the unique visits for today)
today = datetime.datetime.today()
yesterday = datetime.datetime.today() - datetime.timedelta(days=1)
visits = Visit.objects.filter(timestamp__range=(yesterday, today)) #.something?
EDIT:
I see that I can use:
Visit.objects.filter(timestamp__range=(yesterday, today)).values('ip_address')
to get a ValuesQuerySet of just the ip fields. Now my QuerySet looks like this:
[{'ip_address': u'127.0.0.1'}, {'ip_address': u'127.0.0.1'}, {'ip_address':
u'127.0.0.1'}, {'ip_address': u'127.0.0.1'}, {'ip_address': u'127.0.0.1'}]
How do I filter this for uniqueness without evaluating the QuerySet and taking the db hit?
# Hope it's something like this...
values.distinct().count()
What you want is:
Visit.objects.filter(stuff).values("ip_address").annotate(n=models.Count("pk"))
What this does is get all ip_addresses and then it gets the count of primary keys (aka number of rows) for each ip address.
With Alex Answer I also have the n:1 for each item. Even with a distinct() clause.
It's weird because this is returning the good numbers of items :
Visit.objects.filter(stuff).values("ip_address").distinct().count()
But when I iterate over "Visit.objects.filter(stuff).values("ip_address").distinct()" I got much more items and some duplicates...
EDIT :
The filter clause was causing me troubles. I was filtering with another table field and a SQL JOIN was made that was breaking the distinct stuff.
I used this hint to see the query that was really used :
q=Visit.objects.filter(myothertable__field=x).values("ip_address").distinct().count()
print q.query
I then reverted the class on witch I was making the query and the filter to have a join that doesn't rely on any "Visit" id.
hope this helps
The question is different from what the title suggests. If you want set-like behavior from the database, you need something like this.
x = Visit.objects.all().values_list('ip_address', flat=True).distinct()
It should give you something like this for x.
[1.2.3.4, 2.3.4.5, ...]
Where
len(x) == len(set(x))
Returns True