Question

每首诗作为两个投票，一个作为poem_id，other_poem_id，胜出＆amp;第二个记录是第一个记录的倒数。也许有更好的方法，但我试图找到一段时间内获胜率最高的诗歌。由于每次比较的双重记录，这令人困惑。我应该添加另一个表，结果，它有两个投票记录的comparison_id吗？

Here is a sample
poem_id:1 other_poem_id:2 wins:3
poem_id:2 other_poem_id:1 wins:3
so it is 50% rather than a running tally

   scope :recent, lambda {
    { :joins => "JOIN votes ON votes.poem_id = poems.id",
      :conditions => ["poems.created_at > ?", 8.days.ago],
      :order => "votes.wins DESC",
      :limit => 10 
    } 
   }

的ActiveRecord :: StatementInvalid：   SQLite3 :: SQLException：含糊不清   列名：created_at：SELECT
  “诗歌”。*来自“诗歌”加入   投票票数.poem_id = poems.id   WHERE（created_at＆gt;'2010-02-12   15：12：35.764252'）ORDER BY赢得DESC   限制10

编辑：我改变了架构，这是我现在正在使用的...
以下是一个跟踪诗歌排名的模型。我昨天刚刚写了第一稿。它看起来有点笨重，但我还不知道如何改进它。 DailyRanking.tabulate将每晚由cron调用。（在模型之后是比较的模式。）

# == Schema Information
# Schema version: 20100221120442
#
# Table name: daily_rankings
#
#  id          :integer         not null, primary key
#  poem_id     :integer
#  rank        :integer
#  percentile  :integer
#  wins        :integer
#  losses      :integer
#  draws       :integer
#  comparisons :integer
#  created_at  :datetime
#  updated_at  :datetime
#

class DailyRanking < ActiveRecord::Base
  belongs_to :poem
  class << self
    def tabulate
      # 1. get all comparisons over the past 24 hours
      comparisons = Comparison.day.all
      # 2. collect poem id for each time it wins

      # TODO make hash of "poem_id" => {:wins => a, :losses => b, :draws => c}

      a, results = 0, []

      while a < comparisons.size
        c = comparisons[a]
        if c.poem1_id == c.winner_id
          results << c.poem1_id
        elsif c.poem2_id == c.winner_id
          results << c.poem2_id
        end
        a += 1
      end
      # 3. presort by poem count

      a, unsorted_wins = 0, []

      until results.empty?
        unsorted_wins << [results.first, results.count(results.first)]
        results.delete(results.first)
      end
      # 4. sort by win count

      sorted_wins = unsorted_wins.sort { |a, b| b[1] <=> a[1] }

      # 5. repeat for losses
      a, results = 0, []

      while a < comparisons.size
        c = comparisons[a]
        if c.poem1_id == c.loser_id
          results << c.poem1_id
        elsif c.poem2_id == c.loser_id
          results << c.poem2_id
        end
        a += 1
      end

      unsorted_losses = []

      until results.empty?
        unsorted_losses << [results.first, results.count(results.first)]
        results.delete(results.first)
      end

      sorted_losses = unsorted_losses.sort { |a, b| b[1] <=> a[1] } 

      # 6. sort wins v losses

        # a. sort wins[poem] v losses[poem]
        # b. get poem and pct wins for wins[poem]
        # c. delete wins[poem] and losses[poem]
        # repeat

      unsorted_results, a = [], 0

      while a < sorted_wins.size
        poem_id = sorted_wins[a][0]
        wins = sorted_wins[a][1]
        losses = sorted_losses.select do |item| 
            item.second if item.first == poem_id
          end.compact.first.second

        unsorted_results << [ poem_id, wins / (wins + losses).to_f ]

        a += 1
      end

      # 7. sort by pct

      sorted_results = unsorted_results.sort { |a, b| b[1] <=> a[1] }

      # 8. persist rankings

      sorted_results.each_with_index do |result, index|
        ranking = find_or_create_by_rank(index + 1)
        ranking.poem_id = result.first
        ranking.save!
      end     
    end
  end
end

# == Schema Information
# Schema version: 20100221120442
#
# Table name: comparisons
#
#  id         :integer         not null, primary key
#  poem1_id   :integer
#  poem2_id   :integer
#  response   :string(4)       default("none"), not null
#  winner_id  :integer
#  loser_id   :integer
#  user_id    :integer
#  session_id :integer
#  ip         :string(15)
#  created_at :datetime
#  updated_at :datetime
#

class Comparison < ActiveRecord::Base

  scope :day,   lambda { { :conditions => ["created_at > ?", 1.day.ago] } }

end

Answer 1

我认为适用于您的SQL查询类似于SELECT poems.*, percentage as ((SELECT wins FROM votes WHERE poem_id = poem.id WHERE created_at > 8.days.ago) / (SELECT wins FROM votes WHERE other_poem_id = poem.id WHERE created_at > 8.days.ago)) ORDER BY percentage DESC LIMIT 10。至于如何优化并将其转换为Rails范围，我不确定。

但是，您遇到的错误是由["poems.created_at > ?", 8.days.ago]条件转换为SQL的方式引起的。 SQLite不知道你是在寻找poems.created_at还是votes.created_at（顺便说一句，从你的描述来看，我认为你想要votes.created_at）。

优化每日排名表

1 个答案: