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()
Related
I have a little problem with getting latest foreign key value in my django app. Here are my two models:
class Stock(models.Model):
...
class Dividend(models.Model):
date = models.DateField('pay date')
stock = models.ForeignKey(Stock, related_name="dividends")
class Meta:
ordering = ["date"]
I would like to get latest dividend from stock object. So basically this - stock.dividends.latest('date'). However, everytime I call stock.dividends.latest('date'), it fires up sql query to get latest dividend. I have latest() method in for cycle for every stock I have. I would like to avoid these sql queries. May I somehow define new method in class Stock that would get latest dividend within sql query for stock object?
I cannot change default ordering from "date" to "-date".
Using select_related('dividends') loads dividends objects with stock, but latest probably uses order_by and it requires sql query anyway. :(
EDIT1: To make more clear what I want, here is an example. Let's say I have 100 symbols in shares.keys():
for stock in Stock.objects.filter(symbol__in=shares.keys()): # 1 sql query
latest_dividend = stock.dividends.latest('date') # 100 sql queries
... #do something with latest dividend
Well and in some cases I might have 500 symbols in shares.keys(). That is why I need to avoid making sql queries on getting latest dividend for stock.
I have the same problem with you, so I tested many Django queries. Finally, I found out that we can use this:
Stock.objects.all().annotate(latest_date=Max('dividends__date')).filter(dividends__date=F('latest_date')).values('dividends')
I'm not sure my solution is the best, but here it is (works only with PostgreSQL):
stocks = list(Stock.objects.filter(**something))
dividends = Dividend.objects.filter(
stock__in=stocks,
).order_by(
'stock_id',
'-date'
).distinct(
'stock_id',
)
dividends_dict = {d.stock_id: d for d in dividends}
for stock in stocks:
stock.latest_dividend = dividends_dict.get(stock.id)
I'm a little confused by your question, I'm assuming you are trying to access the dividends from your stock object in order to limit your queries to the database. I believe that is the least number queries of possible.
stock_options = stock.objects.get(pk=your_query)
order_options = stock.dividend_set.order_by('-date')[:5]
likeon: Thanks for your answer. But I think I can avoid initializing that large dictionary (I have 5000 stocks and 280 000 dividends). But your list gave me an idea. Your code requires 2 sql queries. Here is my example (EDIT1).
for stock in Stock.objects.filter(symbol__in=shares.keys())\
.prefetch_related('dividends'): # 2 sql queries
latest_dividend = list(stock.dividends.all())[-1] # 0 sql queries
... #do something with latest_dividend
My code also requires 2 sql queries, but I do not have to reorder it and create list from stocks and all 280 000 dividends (I only create dict from current stock dividends every cycle). May be creating one dict is quicker than creating len(shares.keys()) dicts, not sure.
I thought there would be easier solution (avoid creating list/dictionary from dividends), but this is good enough for now. Thanks for answers!
As long as I understood you can do it this way:
stock.dividends.last()
as implementation in Django is like this:
def first(self):
"""Return the first object of a query or None if no match is found."""
for obj in (self if self.ordered else self.order_by('pk'))[:1]:
return obj
Also, you can use .latest(*fields, field_name=None) too.
I'd like to update a table with Django - something like this in raw SQL:
update tbl_name set name = 'foo' where name = 'bar'
My first result is something like this - but that's nasty, isn't it?
list = ModelClass.objects.filter(name = 'bar')
for obj in list:
obj.name = 'foo'
obj.save()
Is there a more elegant way?
Update:
Django 2.2 version now has a bulk_update.
Old answer:
Refer to the following django documentation section
Updating multiple objects at once
In short you should be able to use:
ModelClass.objects.filter(name='bar').update(name="foo")
You can also use F objects to do things like incrementing rows:
from django.db.models import F
Entry.objects.all().update(n_pingbacks=F('n_pingbacks') + 1)
See the documentation.
However, note that:
This won't use ModelClass.save method (so if you have some logic inside it won't be triggered).
No django signals will be emitted.
You can't perform an .update() on a sliced QuerySet, it must be on an original QuerySet so you'll need to lean on the .filter() and .exclude() methods.
Consider using django-bulk-update found here on GitHub.
Install: pip install django-bulk-update
Implement: (code taken directly from projects ReadMe file)
from bulk_update.helper import bulk_update
random_names = ['Walter', 'The Dude', 'Donny', 'Jesus']
people = Person.objects.all()
for person in people:
r = random.randrange(4)
person.name = random_names[r]
bulk_update(people) # updates all columns using the default db
Update: As Marc points out in the comments this is not suitable for updating thousands of rows at once. Though it is suitable for smaller batches 10's to 100's. The size of the batch that is right for you depends on your CPU and query complexity. This tool is more like a wheel barrow than a dump truck.
Django 2.2 version now has a bulk_update method (release notes).
https://docs.djangoproject.com/en/stable/ref/models/querysets/#bulk-update
Example:
# get a pk: record dictionary of existing records
updates = YourModel.objects.filter(...).in_bulk()
....
# do something with the updates dict
....
if hasattr(YourModel.objects, 'bulk_update') and updates:
# Use the new method
YourModel.objects.bulk_update(updates.values(), [list the fields to update], batch_size=100)
else:
# The old & slow way
with transaction.atomic():
for obj in updates.values():
obj.save(update_fields=[list the fields to update])
If you want to set the same value on a collection of rows, you can use the update() method combined with any query term to update all rows in one query:
some_list = ModelClass.objects.filter(some condition).values('id')
ModelClass.objects.filter(pk__in=some_list).update(foo=bar)
If you want to update a collection of rows with different values depending on some condition, you can in best case batch the updates according to values. Let's say you have 1000 rows where you want to set a column to one of X values, then you could prepare the batches beforehand and then only run X update-queries (each essentially having the form of the first example above) + the initial SELECT-query.
If every row requires a unique value there is no way to avoid one query per update. Perhaps look into other architectures like CQRS/Event sourcing if you need performance in this latter case.
Here is a useful content which i found in internet regarding the above question
https://www.sankalpjonna.com/learn-django/running-a-bulk-update-with-django
The inefficient way
model_qs= ModelClass.objects.filter(name = 'bar')
for obj in model_qs:
obj.name = 'foo'
obj.save()
The efficient way
ModelClass.objects.filter(name = 'bar').update(name="foo") # for single value 'foo' or add loop
Using bulk_update
update_list = []
model_qs= ModelClass.objects.filter(name = 'bar')
for model_obj in model_qs:
model_obj.name = "foo" # Or what ever the value is for simplicty im providing foo only
update_list.append(model_obj)
ModelClass.objects.bulk_update(update_list,['name'])
Using an atomic transaction
from django.db import transaction
with transaction.atomic():
model_qs = ModelClass.objects.filter(name = 'bar')
for obj in model_qs:
ModelClass.objects.filter(name = 'bar').update(name="foo")
Any Up Votes ? Thanks in advance : Thank you for keep an attention ;)
To update with same value we can simply use this
ModelClass.objects.filter(name = 'bar').update(name='foo')
To update with different values
ob_list = ModelClass.objects.filter(name = 'bar')
obj_to_be_update = []
for obj in obj_list:
obj.name = "Dear "+obj.name
obj_to_be_update.append(obj)
ModelClass.objects.bulk_update(obj_to_be_update, ['name'], batch_size=1000)
It won't trigger save signal every time instead we keep all the objects to be updated on the list and trigger update signal at once.
IT returns number of objects are updated in table.
update_counts = ModelClass.objects.filter(name='bar').update(name="foo")
You can refer this link to get more information on bulk update and create.
Bulk update and Create
Which one would be better for performance?
We take a slice of products. which make us impossible to bulk update.
products = Product.objects.filter(featured=True).order_by("-modified_on")[3:]
for product in products:
product.featured = False
product.save()
or (invalid)
for product in products.iterator():
product.update(featured=False)
I have tried QuerySet's in statement too as following.
Product.objects.filter(pk__in=products).update(featured=False)
This line works fine on SQLite. But, it rises following exception on MySQL. So, I couldn't use that.
DatabaseError: (1235, "This version of MySQL doesn't yet support
'LIMIT & IN/ALL/ANY/SOME subquery'")
Edit: Also iterator() method causes re-evaluate the query. So, it is bad for performance.
As #Chris Pratt pointed out in comments, the second example is invalid because the objects don't have update methods. Your first example will require queries equal to results+1 since it has to update each object. That might really be costly if you have 1000 products. Ideally you do want to reduce this to a more fixed expense if possible.
This is a similar situation to another question:
Django: Cannot update a query once a slice has been taken
That being said, you would have to do it in at least 2 queries, but you have to be a bit sneaky on how to construct the LIMIT...
Using Q objects for complex queries:
# get the IDs we want to exclude
products = Product.objects.filter(featured=True).order_by("-modified_on")[:3]
# flatten them into just a list of ids
ids = products.values_list('id', flat=True)
# Now use the Q object to construct a complex query
from django.db.models import Q
# This builds a list of "AND id NOT EQUAL TO i"
limits = [~Q(id=i) for i in ids]
Product.objects.filter(featured=True, *limits).update(featured=False)
In some cases it's acceptable to cache QuerySet in array
products = list(products)
Product.objects.filter(pk__in=products).update(featured=False)
Small optimization with values_list
products_id = list(products.values_list('id', flat=True)
Product.objects.filter(pk__in=products_id).update(featured=False)
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)
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.