In my rails app, I have Users and Listings. The Listings belong to a User. Listing has user_id and its filled with users id who is creating the listing.
A user can be a premium user, gold user or silver user.
What I want is for each premium user, select one random listing to show in premium listings.
I can do it in O(n**2) time or n+1 query as follow:
users_id = User.where(:role => "premium").pluck[:id]
final_array = Array.new
users_id.each do |id|
final_array << Listing.where(:user_id => id).sample(1)
end
final_array
Is there a better way of doing this?
You could try this:
listings = Listing.select(
<<~SQL
DISTINCT ON (users.id) users.id,
listings.*,
row_number() OVER (PARTITION BY users.id ORDER BY random())
SQL
)
.joins(:user)
.includes(:user)
.where(users: { role: :premium })
It gives a random Listing for every premium user.
It produces the only request to db and also it won't make an extra request for getting listing's user, so you are free to do something like this:
listings.each do |listing|
p listing.user
end
random_user_listings = []
User.includes(:listings).where(role: "premium").find_each do |user|
random_user_listings << user.listings.sample(1)
end
random_user_listings
To avoid N+1 query you need to combine them, perform query one time like this:
list = Listing.includes(:user).where(:role => "premium").sample(1)
Feel free to deal with list instead of Listing. Because now you're dealing with variable, not Query.
ids = list.pluck(:user_id).uniq
Getting array of ids like above and doing further steps as you did (but with list, not Listing)
Need to be noticed that, when you deal with Model you're dealing with QUERY. Avoiding doing that in loop statement.
Related
Considering I have the following relationships:
class House(Model):
name = ...
class User(Model):
"""The standard auth model"""
pass
class Alert(Model):
user = ForeignKey(User)
house = ForeignKey(House)
somevalue = IntegerField()
Meta:
unique_together = (('user', 'property'),)
In one query, I would like to get the list of houses, and whether the current user has any alert for any of them.
In SQL I would do it like this:
SELECT *
FROM house h
LEFT JOIN alert a
ON h.id = a.house_id
WHERE a.user_id = ?
OR a.user_id IS NULL
And I've found that I could use prefetch_related to achieve something like this:
p = Prefetch('alert_set', queryset=Alert.objects.filter(user=self.request.user), to_attr='user_alert')
houses = House.objects.order_by('name').prefetch_related(p)
The above example works, but houses.user_alert is a list, not an Alert object. I only have one alert per user per house, so what is the best way for me to get this information?
select_related didn't seem to work. Oh, and surely I know I can manage this in multiple queries, but I'd really want to have it done in one, and the 'Django way'.
Thanks in advance!
The solution is clearer if you start with the multiple query approach, and then try to optimise it. To get the user_alerts for every house, you could do the following:
houses = House.objects.order_by('name')
for house in houses:
user_alerts = house.alert_set.filter(user=self.request.user)
The user_alerts queryset will cause an extra query for every house in the queryset. You can avoid this with prefetch_related.
alerts_queryset = Alert.objects.filter(user=self.request.user)
houses = House.objects.order_by('name').prefetch_related(
Prefetch('alert_set', queryset=alerts_queryset, to_attrs='user_alerts'),
)
for house in houses:
user_alerts = house.user_alerts
This will take two queries, one for houses and one for the alerts. I don't think you require select related here to fetch the user, since you already have access to the user with self.request.user. If you want you could add select_related to the alerts_queryset:
alerts_queryset = Alert.objects.filter(user=self.request.user).select_related('user')
In your case, user_alerts will be an empty list or a list with one item, because of your unique_together constraint. If you can't handle the list, you could loop through the queryset once, and set house.user_alert:
for house in houses:
house.user_alert = house.user_alerts[0] if house.user_alerts else None
I have two models
class User
has_many :subscriptions
end
and
class Subscription
belongs_to :user
end
one one of my pages I would like to display a list of all users ordered by the number of subscriptions each user has. I am not to good with sql queries but I think that
list = Users.all.joins(:subscriptions).group("user.id").order("count(subscriptions.id) DESC")
dose the job. Now to my problem, when I try to count the total number of objects in list, using list.count, I get a hash with user.id and subscription count, like this
{11 => 5,
8 => 7,
1 => 11,
...}
not the total number of users in list.. .count works fine if I have a list sorted by for example user name (which is in the user table). I would really like to use .count since it in a module for pagination thats in a gem but any ideas is great!
Thanks!
We can just use a single query to finish this:
User.joins("LEFT OUTER JOIN ( SELECT user_id, COUNT(*) as num_subscriptions
FROM subscriptions
GROUP BY user_id
) AS temp
ON temp.user_id = users.id")
.order("temp.num_subscriptions DESC")
Basically, my idea is to try to query the number of subscription for each user_id in the subquery, then join with User. I used LEFT OUTER JOIN, because there will be several users which don't have any subscriptions
Improve option: You can define a scope inside User, it would be more beautiful for later usage:
user.rb
class User < ActiveRecord::Base
has_many :subscriptions
scope :sorted_by_num_subscriptions, -> {
joins("LEFT OUTER JOIN ( SELECT user_id, COUNT(*) as num_subscriptions
FROM subscriptions
GROUP BY user_id
) AS temp
ON temp.user_id = users.id")
.order("temp.num_subscriptions DESC")
}
end
Then just use it:
User.sorted_by_num_subscriptions
When grouping, the count method changes it's behavior and indeed, instead of returning the total count of records, it returns a hash of the counts for each group (see the docs for more info). So what you get with list.count is simply a hash of the subscription counts for each user.
So, your query is correct and all you need is to sum up the individual counts in the groups. This can be done simply by:
total_count = list.count.values.sum
If it is the pagination code that calls just a bare count that makes the issue, usually the pagination code is able to accept a parameter with total count. For example, will_paginate accepts the total_entries parameter, so you should be able to pass it the total count like this:
list.paginate(page: 2, total_entries: list.count.values.sum)
I'm trying to order a list of locations based on the number of times a user has viewed them. Am using the impressionist gem for the sake of it.
The problem I'm having is that my query completely excludes those locations the user's never viewed. I need to display these at the bottom of the results and order by the created_at timestamp.
I can do this to get a list of location_ids:
#location_ids = #user.impressions.
select('count(id) as counter, impressionable_id').
group(:impressionable_id).
order('counter DESC').
#location_ids.map(&:impressionable_id)
Which gives [3,5,8,44,99] and so on..
However, that doesn't get me far so I tried this:
#user.locations.
joins(:impressions).
select("count(impressions.id) as counter, impressionable_id, locations.location_name, locations.id").
group(:impressionable_id).
order("counter desc")
Which is better but it omits those locations with zero views.
How should I do this to get all the locations?
By default, Rails uses an inner join when you use .joins. That's why you don't see the locations with no associated impressions. You need to tell it to use a left join instead, probably like so:
#user.locations.
joins("left join impressions on impressions.impressionable_id = locations.id and impressions.impressionable_type = 'Location'").
select("count(impressions.id) as counter, impressionable_id, locations.location_name, locations.id").
group('locations.id').
order("counter desc")
I have 2 models - Restaurant and Feature. They are connected via has_and_belongs_to_many relationship. The gist of it is that you have restaurants with many features like delivery, pizza, sandwiches, salad bar, vegetarian option,… So now when the user wants to filter the restaurants and lets say he checks pizza and delivery, I want to display all the restaurants that have both features; pizza, delivery and maybe some more, but it HAS TO HAVE pizza AND delivery.
If I do a simple .where('features IN (?)', params[:features]) I (of course) get the restaurants that have either - so or pizza or delivery or both - which is not at all what I want.
My SQL/Rails knowledge is kinda limited since I'm new to this but I asked a friend and now I have this huuuge SQL that gets the job done:
Restaurant.find_by_sql(['SELECT restaurant_id FROM (
SELECT features_restaurants.*, ROW_NUMBER() OVER(PARTITION BY restaurants.id ORDER BY features.id) AS rn FROM restaurants
JOIN features_restaurants ON restaurants.id = features_restaurants.restaurant_id
JOIN features ON features_restaurants.feature_id = features.id
WHERE features.id in (?)
) t
WHERE rn = ?', params[:features], params[:features].count])
So my question is: is there a better - more Rails even - way of doing this? How would you do it?
Oh BTW I'm using Rails 4 on Heroku so it's a Postgres DB.
This is an example of a set-iwthin-sets query. I advocate solving these with group by and having, because this provides a general framework.
Here is how this works in your case:
select fr.restaurant_id
from features_restaurants fr join
features f
on fr.feature_id = f.feature_id
group by fr.restaurant_id
having sum(case when f.feature_name = 'pizza' then 1 else 0 end) > 0 and
sum(case when f.feature_name = 'delivery' then 1 else 0 end) > 0
Each condition in the having clause is counting for the presence of one of the features -- "pizza" and "delivery". If both features are present, then you get the restaurant_id.
How much data is in your features table? Is it just a table of ids and names?
If so, and you're willing to do a little denormalization, you can do this much more easily by encoding the features as a text array on restaurant.
With this scheme your queries boil down to
select * from restaurants where restaurants.features #> ARRAY['pizza', 'delivery']
If you want to maintain your features table because it contains useful data, you can store the array of feature ids on the restaurant and do a query like this:
select * from restaurants where restaurants.feature_ids #> ARRAY[5, 17]
If you don't know the ids up front, and want it all in one query, you should be able to do something along these lines:
select * from restaurants where restaurants.feature_ids #> (
select id from features where name in ('pizza', 'delivery')
) as matched_features
That last query might need some more consideration...
Anyways, I've actually got a pretty detailed article written up about Tagging in Postgres and ActiveRecord if you want some more details.
This is not "copy and paste" solution but if you consider following steps you will have fast working query.
index feature_name column (I'm assuming that column feature_id is indexed on both tables)
place each feature_name param in exists():
select fr.restaurant_id
from
features_restaurants fr
where
exists(select true from features f where fr.feature_id = f.feature_id and f.feature_name = 'pizza')
and
exists(select true from features f where fr.feature_id = f.feature_id and f.feature_name = 'delivery')
group by
fr.restaurant_id
Maybe you're looking at it backwards?
Maybe try merging the restaurants returned by each feature.
Simplified:
pizza_restaurants = Feature.find_by_name('pizza').restaurants
delivery_restaurants = Feature.find_by_name('delivery').restaurants
pizza_delivery_restaurants = pizza_restaurants & delivery_restaurants
Obviously, this is a single instance solution. But it illustrates the idea.
UPDATE
Here's a dynamic method to pull in all filters without writing SQL (i.e. the "Railsy" way)
def get_restaurants_by_feature_names(features)
# accepts an array of feature names
restaurants = Restaurant.all
features.each do |f|
feature_restaurants = Feature.find_by_name(f).restaurants
restaurants = feature_restaurants & restaurants
end
return restaurants
end
Since its an AND condition (the OR conditions get dicey with AREL). I reread your stated problem and ignoring the SQL. I think this is what you want.
# in Restaurant
has_many :features
# in Feature
has_many :restaurants
# this is a contrived example. you may be doing something like
# where(name: 'pizza'). I'm just making this condition up. You
# could also make this more DRY by just passing in the name if
# that's what you're doing.
def self.pizza
where(pizza: true)
end
def self.delivery
where(delivery: true)
end
# query
Restaurant.features.pizza.delivery
Basically you call the association with ".features" and then you use the self methods defined on features. Hopefully I didn't misunderstand the original problem.
Cheers!
Restaurant
.joins(:features)
.where(features: {name: ['pizza','delivery']})
.group(:id)
.having('count(features.name) = ?', 2)
This seems to work for me. I tried it with SQLite though.
I'm really struggling on this one.
I need to be able to sort my user by the number of positive vote received on their comment.
I have a table userprofile, a table comment and a table likeComment.
The table comment has a foreign key to its user creator and the table likeComment has a foreign key to the comment liked.
To get the number of positive vote a user received I do :
LikeComment.objects.filter(Q(type = 1), Q(comment__user=user)).count()
Now I want to be able to get all the users sorted by the ones that have the most positive votes. How do I do that ? I tried to use extra and JOIN but this didn't go anywhere.
Thank you
It sounds like you want to perform a filter on an annotation:
class User(models.Model):
pass
class Comment(models.Model):
user = models.ForeignKey(User, related_name="comments")
class Like(models.Model):
comment = models.ForeignKey(Comment, related_name="likes")
type = models.IntegerField()
users = User \
.objects \
.all()
.extra(select = {
"positive_likes" : """
SELECT COUNT(*) FROM app_like
JOIN app_comment on app_like.comment_id = app_comment.id
WHERE app_comment.user_id = app_user.id AND app_like.type = 1 """})
.order_by("positive_likes")
models.py
class UserProfile(models.Model):
.........
def like_count(self):
LikeComment.objects.filter(comment__user=self.user, type=1).count()
views.py
def getRanking( anObject ):
return anObject.like_count()
def myview(request):
users = list(UserProfile.objects.filter())
users.sort(key=getRanking, reverse=True)
return render(request,'page.html',{'users': users})
Timmy's suggestion to use a subquery is probably the simplest way to solve this kind of problem, but subqueries almost never perform as well as joins, so if you have a lot of users you may find that you need better performance.
So, re-using Timmy's models:
class User(models.Model):
pass
class Comment(models.Model):
user = models.ForeignKey(User, related_name="comments")
class Like(models.Model):
comment = models.ForeignKey(Comment, related_name="likes")
type = models.IntegerField()
the query you want looks like this in SQL:
SELECT app_user.id, COUNT(app_like.id) AS total_likes
FROM app_user
LEFT OUTER JOIN app_comment
ON app_user.id = app_comment.user_id
LEFT OUTER JOIN app_like
ON app_comment.id = app_like.comment_id AND app_like.type = 1
GROUP BY app_user.id
ORDER BY total_likes DESCENDING
(If your actual User model has more fields than just id, then you'll need to include them all in the SELECT and GROUP BY clauses.)
Django's object-relational mapping system doesn't provide a way to express this query. (As far as I know—and I'd be very happy to be told otherwise!—it only supports aggregation across one join, not across two joins as here.) But when the ORM isn't quite up to the job, you can always run a raw SQL query, like this:
sql = '''
SELECT app_user.id, COUNT(app_like.id) AS total_likes
# etc (as above)
'''
for user in User.objects.raw(sql):
print user.id, user.total_likes
I believe this can be achieved with Django's queryset:
User.objects.filter(comments__likes__type=1)\
.annotate(lks=Count('comments__likes'))\
.order_by('-lks')
The only problem here is that this query will miss users with 0 likes. Code from #gareth-rees, #timmy-omahony and #Catherine will include also 0-ranked users.