顧客グループの好きな俳優を特定する

SQLで進めるデータドリブンな意思決定

Irene Ortner

Data Scientist at Applied Statistics

1つのクエリで SQL 文を組み合わせる

LEFT JOIN
WHERE
GROUP BY
HAVING
ORDER BY

レンタル履歴から顧客・俳優情報へ

問い: 特定の顧客グループにとって人気の俳優は誰か？

renting を以下と結合:

customers
actsin
actors

SELECT *
FROM renting as r
LEFT JOIN customers AS c
ON r.customer_id = c.customer_id
LEFT JOIN actsin as ai
ON r.movie_id = ai.movie_id
LEFT JOIN actors as a
ON ai.actor_id = a.actor_id;

男性顧客

男性顧客が視聴した作品で最も多く出演する俳優。

SELECT a.name, 
       COUNT(*)
FROM renting as r
LEFT JOIN customers AS c
ON r.customer_id = c.customer_id
LEFT JOIN actsin as ai
ON r.movie_id = ai.movie_id
LEFT JOIN actors as a
ON ai.actor_id = a.actor_id

WHERE c.gender = 'male'
GROUP BY a.name;

お気に入りの俳優は誰か？

最も多く視聴された俳優
視聴時の平均評価が最も高い俳優

SELECT a.name, 
       COUNT(*) AS number_views, 
       AVG(r.rating) AS avg_rating
FROM renting as r
LEFT JOIN customers AS c
ON r.customer_id = c.customer_id
LEFT JOIN actsin as ai
ON r.movie_id = ai.movie_id
LEFT JOIN actors as a
ON ai.actor_id = a.actor_id

WHERE c.gender = 'male'
GROUP BY a.name;

HAVING と ORDER BY を追加

SELECT a.name, 
       COUNT(*) AS number_views, 
       AVG(r.rating) AS avg_rating
FROM renting as r
LEFT JOIN customers AS c
ON r.customer_id = c.customer_id
LEFT JOIN actsin as ai
ON r.movie_id = ai.movie_id
LEFT JOIN actors as a
ON ai.actor_id = a.actor_id

WHERE c.gender = 'male'
GROUP BY a.name
HAVING AVG(r.rating) IS NOT NULL
ORDER BY avg_rating DESC, number_views DESC;

HAVING と ORDER BY を追加

| name               | number_views | avg_rating |
|--------------------|--------------|------------|
| Ray Romano         | 3            | 10.00      |
| Sean Bean          | 2            | 10.00      |
| Leonardo DiCaprio  | 3            | 9.33       |
| Christoph Waltz    | 3            | 9.33       |

Passons à la pratique !

SQLで進めるデータドリブンな意思決定