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@yuyasat

ActiveRecordにおけるGROUP BYの使い方

Last updated at Posted at 2017-12-31

0. はじめに

0-0. はじめに

以前、「Rails における内部結合、外部結合まとめ」という記事を書いたのですが、GROUP BYに関しては触れなかったなと思い、実務で割と使用していて毎回少しずつ悩むので、自分の備忘録の意味も込めて記載します。

0-1. RubyとRailsとPostgreSQLのバージョン

$ ruby -v
ruby 2.4.1p111 (2017-03-22 revision 58053) [x86_64-darwin16]
$ rails -v
Rails 5.1.2

=> SELECT version();
                                                   version
--------------------------------------------------------------------------------------------------------------
 PostgreSQL 9.5.3 on x86_64-apple-darwin15.4.0, compiled by Apple LLVM version 7.3.0 (clang-703.0.31), 64-bit

0-2. 今回扱うERD

ERD 説明
女優テーブルがある。
女優テーブルは映画テーブルを持つ。
映画テーブルは、映画に関連するキーワードを入れたタグテーブルを持つ。

0-3. モデル

Actress
class Actress < ApplicationRecord
  has_many :movies
  has_many :tags, through: :movies
end
Movie
class Movie < ApplicationRecord
  has_many :tags
  belongs_to :actress
end
Tag
class Tag < ApplicationRecord
  belongs_to :movie
end

0-4. 今回扱うデータ

actress
SELECT id, name FROM actresses;
 id |    name
----+------------
  1 | 多部未華子
  2 | 佐津川愛美
  3 | 新垣結衣
  4 | 堀北真希
  5 | 吉高由里子
  6 | 悠城早矢
movies
SELECT id, actress_id, title, year FROM movies;
 id | actress_id |        title        | year
----+------------+---------------------+------
  1 |          2 | 蝉しぐれ            | 2005
  2 |          1 | 夜のピクニック      | 2006
  3 |          4 | ALWAYS 三丁目の夕日 | 2005
  4 |          2 | 忍道-SHINOBIDO-     | 2012
  5 |          2 | 貞子vs伽椰子        | 2016
  6 |          4 | 県庁おもてなし課    | 2013
  7 |          5 | 真夏の方程式        | 2013
tags
SELECT id, movie_id, key FROM tags;
 id | movie_id |    key
----+----------+------------
  1 |        1 | 時代劇
  2 |        1 | 子役
  3 |        3 | 昭和
  4 |        5 | ホラー
  5 |        7 | ミステリー
  6 |        7 | 夏
  7 |        6 | 公務員
  8 |        6 | 地方活性
  9 |        1 | 夏
 10 |        4 | 時代劇
# 前回のレコードにid: 10のレコードが追加されています。

1. テーブルが1つの場合

1-1. 各女優の映画数を知りたい（idで表示）

Movie.group(:actress_id).count
=> {4=>2, 1=>1, 5=>1, 2=>3}

SELECT
  COUNT(*) AS count_all,
  "movies"."actress_id" AS movies_actress_id
FROM
  "movies"
GROUP BY
  "movies"."actress_id"

これだとacctress_idで出てくるので、分かりにくい。そこで、acctressesテーブルを結合して表示させる。

2. テーブルが2つに跨ぐ場合

2-1. 各女優の映画数を知りたい（名前で表示）（JOINする方法）

Movie.joins(:actress).group("actresses.name").count
=> {"多部未華子"=>1, "佐津川愛美"=>3, "堀北真希"=>2, "吉高由里子"=>1}

SELECT
  COUNT(*) AS count_all,
  actresses.name AS actresses_name
FROM
  "movies"
  INNER JOIN
    "actresses"
  ON
    "actresses"."id" = "movies"."actress_id"
GROUP BY
  actresses.name

2-2. 各女優の映画数を知りたい（名前で表示）（JOINしない方法）

  • 名前の表示の必要がなければ、moviesテーブルのactress_idでGROUP BYすればいいので、前述のようにMovie.group(:actress_id).countとすればよい。ここでactress_idがとれるので、それを事前読み込みするという方法もある。
  • n+1を考慮しない場合
Movie.group(:actress_id).select("actress_id, count(movies.id) as movie_count").map { |m| [m.actress.name, m.movie_count] }.to_h
# =>
  Movie Load (0.5ms)  SELECT actress_id, count(movies.id) as movie_count FROM "movies" GROUP BY "movies"."actress_id"
  Actress Load (0.4ms)  SELECT  "actresses".* FROM "actresses" WHERE "actresses"."id" = $1 LIMIT $2  [["id", 5], ["LIMIT", 1]]
  Actress Load (0.4ms)  SELECT  "actresses".* FROM "actresses" WHERE "actresses"."id" = $1 LIMIT $2  [["id", 4], ["LIMIT", 1]]
  Actress Load (0.4ms)  SELECT  "actresses".* FROM "actresses" WHERE "actresses"."id" = $1 LIMIT $2  [["id", 2], ["LIMIT", 1]]
  Actress Load (0.4ms)  SELECT  "actresses".* FROM "actresses" WHERE "actresses"."id" = $1 LIMIT $2  [["id", 1], ["LIMIT", 1]]
=> {"吉高由里子"=>1, "堀北真希"=>2, "佐津川愛美"=>3, "多部未華子"=>1}
  • n+1を考慮する場合
    • preloadやincludesを利用する。
Movie.group(:actress_id).select("actress_id, count(movies.id) as movie_count").preload(:actress).map { |m| [m.actress.name, m.movie_count] }.to_h
=> {"吉高由里子"=>1, "堀北真希"=>2, "佐津川愛美"=>3, "多部未華子"=>1}
Movie.group(:actress_id).select("actress_id, count(movies.id) as movie_count").includes(:actress).map { |m| [m.actress.name, m.movie_count] }.to_h
# =>
  Movie Load (0.5ms)  SELECT actress_id, count(movies.id) as movie_count FROM "movies" GROUP BY "movies"."actress_id"
  Actress Load (0.4ms)  SELECT "actresses".* FROM "actresses" WHERE "actresses"."id" IN (5, 4, 2, 1)
=> {"吉高由里子"=>1, "堀北真希"=>2, "佐津川愛美"=>3, "多部未華子"=>1}
  • ここで、preloadではなくeager_loadをつかうと外部結合をしてしまい、エラーになる。（PostgreSQLで検証。他は不明）
    • 挙動をいちいち覚えておけないので、GROUP BYしてダメならpreloadやincludesを用いると覚えておけば良さそう。
Movie.group(:actress_id).select("actress_id, count(movies.id) as movie_count").eager_load(:actress).map { |m| [m.actress.name, m.movie_count] }.to_h
# =>
  ActiveRecord::StatementInvalid: PG::GroupingError: ERROR:  column "movies.id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: ...LECT actress_id, count(movies.id) as movie_count, "movies"."...

2-3. 数が多い順に並び替える

2-3-1. 返り値がHashで良い場合

  • 発行されたSQLを見るとわかるが、RailsがCOUNT(*)にcount_allという名前をつけているのでこれでORDER BYする。
Movie.joins(:actress).group("actresses.name").order("count_all DESC").count
=> {"佐津川愛美"=>3, "堀北真希"=>2, "多部未華子"=>1, "吉高由里子"=>1}

SELECT
  COUNT(*) AS count_all,
  actresses.name AS actresses_name
FROM
  "movies"
  INNER JOIN
    "actresses"
  ON
    "actresses"."id" = "movies"."actress_id"
GROUP BY
  actresses.name
ORDER BY
  count_all DESC

Movie.group(:actress_id).select("actress_id, count(movies.id) as movie_count").order("movie_count desc").preload(:actress).map { |m| [m.actress.name, m.movie_count] }.to_h
# =>
  Movie Load (0.5ms)  SELECT actress_id, count(movies.id) as movie_count FROM "movies" GROUP BY "movies"."actress_id" ORDER BY movie_count desc
  Actress Load (0.4ms)  SELECT "actresses".* FROM "actresses" WHERE "actresses"."id" IN (2, 4, 5, 1)
=> {"佐津川愛美"=>3, "堀北真希"=>2, "吉高由里子"=>1, "多部未華子"=>1}

2-3-2. 返り値をActiveRecord_Relationにしたい場合

  • moviesでjoinして、count(actresses.id) descなどをorderに指定する。
  • Rails6.0からは、orderの中にSQLを直書きしてはいけないので、Arel.sqlで囲む。
Actress.joins(:movies).group('actresses.id').order(Arel.sql('count(actresses.id) desc'))
#=> [#<Actress:0x007f89619a7878
  id: 2,
  name: "佐津川愛美",
  created_at: Fri, 23 Mar 2018 01:03:30 JST +09:00,
  updated_at: Fri, 23 Mar 2018 01:03:30 JST +09:00>,
 #<Actress:0x007f89619a7738
  id: 4,
  name: "堀北真希",
  created_at: Fri, 23 Mar 2018 01:03:30 JST +09:00,
  updated_at: Fri, 23 Mar 2018 01:03:30 JST +09:00>,
 #<Actress:0x007f89619a75a8
  id: 5,
  name: "吉高由里子",
  created_at: Fri, 23 Mar 2018 01:03:30 JST +09:00,
  updated_at: Fri, 23 Mar 2018 01:03:30 JST +09:00>,
 #<Actress:0x007f89619a7468
  id: 1,
  name: "多部未華子",
  created_at: Fri, 23 Mar 2018 01:03:30 JST +09:00,
  updated_at: Fri, 23 Mar 2018 01:03:30 JST +09:00>]
SQL
SELECT
  "actresses".*
FROM
  "actresses"
INNER JOIN
  "movies"
ON
  "movies"."actress_id" = "actresses"."id"
GROUP BY
  actresses.id
ORDER BY
  count(actresses.id) desc

3. テーブルが３つに跨ぐ場合

3-1. 各女優のタグ数を知りたい

  • 二つ先のテーブルなので、moviesを経由して結合する。
Actress.joins(movies: :tags).group("actresses.name").count
=> {"佐津川愛美"=>5, "堀北真希"=>3, "吉高由里子"=>2}
SQL
SELECT
  COUNT(*) AS count_all,
  actresses.name AS actresses_name
FROM
  "actresses"
  INNER JOIN
    "movies"
  ON
    "movies"."actress_id" = "actresses"."id"
  INNER JOIN
    "tags"
  ON
    "tags"."movie_id" = "movies"."id"
GROUP BY
  actresses.name
  • 今回の場合、以下のようにActressモデルにthroughオプションをつけているのでmoviesを明示しなくても動く。
actress.rb
has_many :tags, through: :movies

 Actress.joins(:tags).group("actresses.name").count
=> {"佐津川愛美"=>5, "堀北真希"=>3, "吉高由里子"=>2}

3-2. 各女優の各タグごとの数を知りたい

  • 例えば、佐津川愛美には「ホラー」「夏」「子役」「時代劇」というタグがつけられているが、それぞれ何個あるかを知りたいという場合。
  • actresses.nameでGROUP BYしてからtags.keyでGROUP BYを行えば良い。（軸をactresses.name）
Actress.joins(:tags).group("actresses.name").group("tags.key").count
=> {["堀北真希", "地方活性"]=>1,
 ["佐津川愛美", "子役"]=>1,
 ["佐津川愛美", "夏"]=>1,
 ["佐津川愛美", "ホラー"]=>1,
 ["吉高由里子", "ミステリー"]=>1,
 ["堀北真希", "昭和"]=>1,
 ["堀北真希", "公務員"]=>1,
 ["佐津川愛美", "時代劇"]=>2,
 ["吉高由里子", "夏"]=>1}

SELECT
  COUNT(*) AS count_all,
  actresses.name AS actresses_name,
  tags.key AS tags_key
FROM
  "actresses"
  INNER JOIN
    "movies"
  ON
    "movies"."actress_id" = "actresses"."id"
  INNER JOIN
    "tags"
  ON
   "tags"."movie_id" = "movies"."id"
GROUP BY
   actresses.name, tags.key
  • 返り値を見ると、同じ女優でもまとまっていなくて見にくいのでORDER BY句をつけておくと良い。
Actress.joins(:tags).group("actresses.name").group("tags.key").order("actresses.name").count
=> {["堀北真希", "昭和"]=>1,
 ["堀北真希", "公務員"]=>1,
 ["堀北真希", "地方活性"]=>1,
 ["佐津川愛美", "夏"]=>1,
 ["佐津川愛美", "子役"]=>1,
 ["佐津川愛美", "ホラー"]=>1,
 ["佐津川愛美", "時代劇"]=>2,
 ["吉高由里子", "夏"]=>1,
 ["吉高由里子", "ミステリー"]=>1}

SELECT
  COUNT(*) AS count_all,
  actresses.name AS actresses_name,
  tags.key AS tags_key
FROM
  "actresses"
  INNER JOIN
    "movies"
  ON
    "movies"."actress_id" = "actresses"."id"
  INNER JOIN
    "tags"
  ON
    "tags"."movie_id" = "movies"."id"
GROUP BY
   actresses.name, tags.key
ORDER BY
   actresses.name
  • 各女優の中で、タグの数が多い順で並び替えたい場合
Actress.joins(:tags).group("actresses.name").group("tags.key").order("actresses.name, count_all DESC").count
=> {["堀北真希", "地方活性"]=>1,
 ["堀北真希", "昭和"]=>1,
 ["堀北真希", "公務員"]=>1,
 ["佐津川愛美", "時代劇"]=>2,
 ["佐津川愛美", "ホラー"]=>1,
 ["佐津川愛美", "夏"]=>1,
 ["佐津川愛美", "子役"]=>1,
 ["吉高由里子", "夏"]=>1,
 ["吉高由里子", "ミステリー"]=>1}

タグ数の多い女優から順に表示させたい場合

  • 佐津川愛美のタグ数が5、堀北真希が3、吉高由里子が2なので、この順に表示させたい。
    • 何かいい方法がないかと考えたが、思いつかなかったので、あらかじめタグ数でORDER BYしてから行うという方法をとってみた。
    • SQLで一発で書く方法教えてください。
# Lambdaにしたのは、遅延評価したかったため。実用的には変数に格納してしまえば良い。
ordered_actress_ids = -> {
  Actress.joins(:tags).group("actresses.id").order("count_all DESC").count.keys.map.with_index { |id, i| "WHEN #{id} THEN #{i} " }.join.strip
}
Actress.joins(:tags).group("actresses.id, actresses.name").group("tags.key").order("CASE actresses.id #{ordered_actress_ids.call} END").order("actresses.name, count_all DESC").count
=> {["佐津川愛美", "時代劇"]=>2,
 ["佐津川愛美", "ホラー"]=>1,
 ["佐津川愛美", "夏"]=>1,
 ["佐津川愛美", "子役"]=>1,
 ["堀北真希", "公務員"]=>1,
 ["堀北真希", "地方活性"]=>1,
 ["堀北真希", "昭和"]=>1,
 ["吉高由里子", "夏"]=>1,
 ["吉高由里子", "ミステリー"]=>1}

SELECT
  COUNT(*) AS count_all,
  actresses.id AS actresses_id
FROM
  "actresses"
  INNER JOIN
    "movies"
  ON
    "movies"."actress_id" = "actresses"."id"
  INNER JOIN
    "tags"
  ON
    "tags"."movie_id" = "movies"."id"
  GROUP BY
    actresses.id
  ORDER BY
    count_all DESC

SELECT
  COUNT(*) AS count_all,
  actresses.id,
  actresses.name AS actresses_id_actresses_name,
  tags.key AS tags_key
FROM
  "actresses"
  INNER JOIN
    "movies"
  ON
    "movies"."actress_id" = "actresses"."id"
  INNER JOIN
    "tags"
  ON
    "tags"."movie_id" = "movies"."id"
  GROUP BY
    actresses.id, actresses.name, tags.key
  ORDER BY CASE
    actresses.id WHEN 2 THEN 0 WHEN 4 THEN 1 WHEN 5 THEN 2 END
    , actresses.name, count_all DESC

その他（GROUP BYとは関係ないものも含む）

サブクエリ関係は、ActiveRecordでサブクエリ（副問い合わせ）と内部結合に移動。

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