How do I get a random record when using :group?
#paintings = Painting.all(:group => "user_id", :order => "created_at DESC")
This gives me the latest painting for each user. Now I would like to select a random painting from each user instead of the latest. The order of the paintings should still be the same, so that the user that have been the most active will get his/her random painting displayed first.
painting150 (user1)
painting200 (user2)
painting231 (user3)
Is this possible?
Best regards.
Asbjørn Morell.
This answer is specific to Rails, but since you are using ActiveRecord, I am assuming it should be fine.
unique_paintings = []
#paintings.group_by(&:user_id).each do |user_id, paintings|
unique_paintings << paintings[rand(paintings.size-1)]
end
unique_paintings.sort_by(&:created_at)
The group_by most certainly messes up the created_at sort you did in the query, so I did a sort_by as the last step. You might want to get rid of it in the query since you'll have to do it anyway here.
#painting = #paintings[rand(#paintings.size-1)]
(or paintings.count, dont know the right method yet)
Assuming you have MySQL, you can try:
#paintings = Painting.all(:group => "user_id", :order => "RAND()")
you could do something like this but it will suffer as your number of records grow
#paintings = Painting.find(:all, :order => 'RAND()').map{ |i| i.user_id }.uniq
Related
After quite a bit of searching, I'm still a bit lost. There are a few other similar questions out there that deal with paginating multiple models, but they are either unanswered or they pagainate each model separately.
I need to paginate all records of an Account at once.
class Account
:has_many :emails
:has_many :tasks
:has_many :notes
end
So, I'd like to find the 30 most recent "things" no matter what they are. Is this even possible with the current pagination solutions out there?
Like using some combination of eager loading and Kaminari or will_paginate?
Or, should I first set up a polymorphic join of all these things, called Items. Then paginate the most recent 30 items, then do a lookup of the associated records of those items.
And if so, I'm not really sure what that code should look like. Any suggestions?
Which way is better? (or even possible)
Rails 3.1, Ruby 1.9.2, app not in production.
with will_paginate :
#records = #do your work and fetch array of records you want to paginate ( various types )
then do the following :
current_page = params[:page] || 1
per_page = 10
#records = WillPaginate::Collection.create(current_page, per_page, records.size) do |pager|
pager.replace(#records)
end
then in your view :
<%=will_paginate #records%>
Good question... I'm not sure of a "good" solution, but you could do a hacky one in ruby:
You'd need to first fetch out the 30 latest of each type of "thing", and put them into an array, indexed by created_at, then sort that array by created_at and take the top 30.
A totally non-refactored start might be something like:
emails = Account.emails.all(:limit => 30, :order => :created_at)
tasks = Account.tasks.all(:limit => 30, :order => :created_at)
notes = Account.notes.all(:limit => 30, :order => :created_at)
thing_array = (emails + tasks + notes).map {|thing| [thing.created_at, thing] }
# sort by the first item of each array (== the date)
thing_array_sorted = thing_array.sort_by {|a,b| a[0] <=> b[0] }
# then just grab the top thirty
things_to_show = thing_array_sorted.slice(0,30)
Note: not tested, could be full of bugs... ;)
emails = account.emails
tasks = account.tasks
notes = account.notes
#records = [emails + tasks + notes].flatten.sort_by(&:updated_at).reverse
#records = WillPaginate::Collection.create(params[:page] || 1, 30, #records.size) do |pager|
pager.replace(#records)
end
Thats it... :)
How do I select a single random record for each user, but order the Array by the latest record pr. user.
If Foo uploads a new painting, I would like to select a single random record from foo. This way a user that uploads 10 paintings won't monopolize all the space on the front page, but still get a slot on the top of the page.
This is how I did it with Rails 2.x running on MySQL.
#paintings = Painting.all.reverse
first_paintings = []
#paintings.group_by(&:user_id).each do |user_id, paintings|
first_paintings << paintings[rand(paintings.size-1)]
end
#paintings = (first_paintings + (Painting.all - first_paintings).reverse).paginate(:per_page => 9, :page => params[:page])
The example above generates a lot of SQL query's and is properly badly optimized. How would you pull this off with Rails 3.1 running on PostgreSQL? I have 7000 records..
#paintings = Painting.all.reverse = #paintings = Painting.order("id desc")
If you really want to reverse the order of the the paintings result set I would set up a scope then just use that
Something like
class Painting < ActiveRecord::Base
scope :reversed, order("id desc")
end
Then you can use Painting.reversed anywhere you need it
You have definitely set up a belongs_to association in your Painting model, so I would do:
# painting.rb
default_scope order('id DESC')
# paintings_controller.rb
first_paintings = User.includes(:paintings).collect do |user|
user.paintings.sample
end
#paintings = (first_paintings + Painting.where('id NOT IN (?)', first_paintings)).paginate(:per_page => 9, :page => params[:page])
I think this solution results in the fewest SQL queries, and is very readable. Not tested, but I hope you got the idea.
You could use the dynamic finders:
Painting.order("id desc").find_by_user_id!(user.id)
This is assuming your Paintings table contains a user_id column or some other way to associate users to paintings which it appears you have covered since you're calling user_id in your initial code. This isn't random but using find_all_by_user_id would allow you to call .reverse on the array if you still wanted and find a random painting.
I have a method that selects 5 random users who are following a certain user, and adds them to an array.
Relationship.find_all_by_followee_id( user.id ).shuffle[0,4].each do |follower|
follower = User.find(follower.user_id)
array.push follower
end
return array
I'm wondering, is this an efficient way of accomplishing this? My main concern is with the find_all_by_followee_id call. This returns a list of all the relationships where the specified user is being followed (this could be in the 100,000s). And then I shuffle that entire list, and then I trim it to the first 5. Is there a more efficient way to do this?
You can try this:
Relationship.find_all_by_followee_id( user.id, :order => 'rand()', :limit => 5 ) do |follower|
follower = User.find(follower.user_id)
array.push follower
end
return array
Btw, this will work with MySql. If you are using PostgreSQL or anything else you may need to change the rand() with any valid random function that your DB supports.
Some minor changes to make it a little more clean:
return Relationship.find_all_by_followee_id( user.id, :order => 'rand()', :limit => 5 ).collect {|follower| User.find(follower.user_id) }
You can also use a join in there in order to prevent the 5 selects but it won't make much difference.
Edit1:
As #mike.surowiec mentioned.
"Just for everyones benefit, translating this to the non-deprecated active record query syntax looks like this:"
Relationship.where(:followee_id => user.id).order( "random()" ).limit( 5 ).collect {|follower| User.find(follower.user_id) }
I have this rails find that i need to get the id as well but if i put the id in the :select wont it effect the query and is there another way to get the id
#past_requests = Request.find_all_by_artist(name, :conditions => ["showdate < ?", Time.now], :select => "distinct venue, showdate")
#past_requests = Request.find_all_by_artist(name, :group => "venue, showdate")
code long for view. i'm remove your condition. sorry about this. Hope it'll helpful for you. :)
To be fair in cases where distinct returns a single row out of maybe 5 duplicates who's to say which id out of those 5 should be displayed in your result? I'm afraid what you are asking for is not practical. Maybe you misunderstand what distinct is used for ? Give us more info pls.
Right now I have a table called Campaigns that has many Hits, if I call say:
Campaign.find(30).hits
Which takes 4 seconds, or 4213 ms.
If I call this instead:
campaign = Campaign.find(30)
campaign.hits.count
Does it still load all of the hits, then count? Or does it see I am counting and avoids loading all of the hits? (Which is currently 300,000+ rows).
I am trying to figure out a smart way to load/count my hits. I am thinking about adding a method to my Campaign.rb model, like:
def self.total_hits
find :first, :select => 'COUNT(id) as hits', :conditions => ["campaign_id = ?", self.id]
end
I know that query won't load from the hits table, but that is just an example of counting it from a self made query, apposed to Ruby on Rails doing this for me.
Would this memcache query be more effecient? (I have it running, but doesn't seem to be any better/faster/slower, just the same speed.)
def self.hits
Rails.cache.fetch("Campaign_Hits_#{self.campaign_id}", :expires_in => 40) {
find(:first, :select => 'COUNT(id) as hits', :conditions => ["campaign_id = ?", self.campaign_id]).hits
}
end
Any suggestions would be great!
How about:
Campaign.find(30).hits.count
You might also consider adding the following in hit.rb (assuming a one-to-many relationship between campaigns and hits).
belongs_to :campaign, :counter_cache => true
You then need a column in the campaigns table called hits_count. This will avoid hitting hits altogether if you're only getting the count.
You can check the API for the full rundown.
My ActiveRecord might be a little rusty, so forgive me if so, but IIRC Campaign.find(30).hits is at least two separate queries. How does Campaign.find(30, :include => [ :hits ]).hits do? That should perform a single query.