인프런 커뮤니티 질문&답변

jihyun lee

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[빠짝스터디 1주차 과제] ARRAY, STRUCT 연습 문제/ PIVOT 연습문제/ 퍼널 분석 연습 문제

작성

1. 자료형: Array, Struct & 퍼널 분석

주요 학습 Point

기존 SQL 작성할 때 없던 개념: Array와 Struct 및 Unnest(Flatten) 개념파악
Pivot 등 퍼널 분석 기본 쿼리 작성 ✍🏼 및 목적에 맞는 퍼널분석 방법론 비교

1-1. 자료형: Array & Struct

연습문제 1-4

/* 연습문제 1
- 테이블: array_exercises
- Row 기준: 각 영화 Title 별
- Column 기준: 장르 Unnest
*/

select title
  , genre
from advanced.array_exercises
cross join unnest(genres) as genre -- genre가 array 형태임
;

  /* 연습문제 2
  - 테이블: array_exercises
  - Row 기준: 각 영화 Title 별
  - Column 기준: actor, character 
  */

  select title
      , actor
  from advanced.array_exercises
  cross join unnest(actors) as actor

  ;

  /* 연습문제 3
  - 테이블: array_exercises
  - Column 기준: title, actor, character, genre
  */

  select title
    , actor
    , genre
  from advanced.array_exercises
  cross join unnest(actors) as actor
  cross join unnest(genres) as genre
  limit 100
  ;

  /* 연습문제 4
  - 문제: app_logs 배열 풀기
  - 주어진 배열: event_params(>key, string_value or int_value)
  */

  select event_date
      , event_timestamp
      , event_name
      , param.key
      , param.value.string_value
      , param.value.int_value
      , user_id
      , user_pseudo_id
      , platform
  from advanced.app_logs
  cross join unnest(event_params) as param
  limit 10
  ;

1-2. Pivot 쿼리 작성

Pivot 연습문제

/* 연습문제 1
- 테이블: orders
- Row 기준: order_date
- Column 기준: user_id 별
- value 기준: sum(amount)
*/

select order_date
    , sum(if(user_id = 1, amount, 0)) as user_1
    , sum(if(user_id = 2, amount, 0)) as user_2
    , sum(if(user_id = 3, amount, 0)) as user_3        
from advanced.orders
group by order_date
order by order_date
;

/* 연습문제 2
- 테이블: orders
- Row 기준: user_id
- Column 기준: order_date 날짜 별
- value 기준: sum(amount)
*/

select user_id
  , sum(if(order_date = '2023-05-01', amount, 0)) as `2023-05-01`
  , sum(if(order_date = '2023-05-02', amount, 0)) as `2023-05-02`
  , sum(if(order_date = '2023-05-03', amount, 0)) as `2023-05-03`
  , sum(if(order_date = '2023-05-04', amount, 0)) as `2024-05-04`
  , sum(if(order_date = '2023-05-05', amount, 0)) as `2023-05-05`        
from advanced.orders
group by user_id
;

/* 연습문제 3
- 테이블: orders
- Row 기준: user_id
- Column 기준: order_date 날짜 별
- value 기준: 주문 있으면 1 없으면 0 (Indicator)
*/

select user_id
  , max(if(order_date = '2023-05-01', 1, 0)) as `2023-05-01`
  , max(if(order_date = '2023-05-02', 1, 0)) as `2023-05-02`
  , max(if(order_date = '2023-05-03', 1, 0)) as `2023-05-03`
  , max(if(order_date = '2023-05-04', 1, 0)) as `2024-05-04`
  , max(if(order_date = '2023-05-05', 1, 0)) as `2023-05-05`        
from advanced.orders
group by user_id
;

/* 연습문제4
- 테이블: app_logs
- Pivot
- where 기준: user_id = 32888, event_name: click_cart
*/

with tmp as (
  select user_id
      , event_date
      , event_name
      , user_pseudo_id
      , param.key
      , param.value.string_value
      , param.value.int_value
  from advanced.app_logs
  cross join unnest(event_params) as param
  where 1=1
  and user_id = 32888
  and event_name = 'click_cart' 
  and key = 'food_id'
)
select int_value
from tmp
  limit 10
	  ;

1-3. 퍼널 분석

퍼널 연습문제

/* 문제1 
- 테이블: app_logs
- 문제: 각 퍼널 유저수 집계 
- where 기준: 2022-08-01 ~ 2022-08-18
*/

WITH tmp AS (
SELECT  
  event_date,
  event_timestamp,
  event_name,
  user_id,
  user_pseudo_id,
  MAX(IF(param.key = 'firebase_screen', param.value.string_value, null)) AS screen_name,
  CONCAT(event_name, '-', MAX(IF(param.key = 'firebase_screen', param.value.string_value, null))) AS event_name_with_screen
FROM advanced.app_logs
cross join  UNNEST(event_params) AS param
WHERE event_date BETWEEN '2022-08-01' AND '2022-08-18'
GROUP BY 1,2,3,4,5
)
SELECT
  event_name_with_screen,
  CASE WHEN event_name_with_screen = 'screen_view-welcome' THEN 1
       WHEN event_name_with_screen = 'screen_view-home' THEN 2
       WHEN event_name_with_screen = 'screen_view-food_category' THEN 3
       WHEN event_name_with_screen = 'screen_view-restaurant' THEN 4
       WHEN event_name_with_screen = 'screen_view-cart' THEN 5
       WHEN event_name_with_screen = 'click_payment-cart' THEN 6 END AS step_number,
  COUNT(DISTINCT user_pseudo_id) AS cnt
FROM tmp
WHERE event_name IN ('screen_view', 'click_payment')
  AND screen_name IN ('welcome', 'home', 'food_category', 'restaurant', 'cart')
GROUP BY 1,2
ORDER by 2
;

/* 문제2
- 테이블: app_logs
- 문제: 일자별 퍼널 유저수 집계 
- where 기준: 2022-08-01 ~ 2022-08-18
*/

WITH tmp AS (
SELECT  
  event_date,
  event_timestamp,
  event_name,
  user_id,
  user_pseudo_id,
  MAX(IF(param.key = 'firebase_screen', param.value.string_value, null)) AS screen_name,
  CONCAT(event_name, '-', MAX(IF(param.key = 'firebase_screen', param.value.string_value, null))) AS event_name_with_screen
FROM advanced.app_logs
cross join  UNNEST(event_params) AS param
WHERE event_date BETWEEN '2022-08-01' AND '2022-08-18'
GROUP BY 1,2,3,4,5
)
SELECT event_date
     , event_name_with_screen
     , CASE WHEN event_name_with_screen = 'screen_view-welcome' THEN 1
       WHEN event_name_with_screen = 'screen_view-home' THEN 2
       WHEN event_name_with_screen = 'screen_view-food_category' THEN 3
       WHEN event_name_with_screen = 'screen_view-restaurant' THEN 4
       WHEN event_name_with_screen = 'screen_view-cart' THEN 5
       WHEN event_name_with_screen = 'click_payment-cart' THEN 6 END AS step_number
     , COUNT(DISTINCT user_pseudo_id) AS cnt
FROM tmp
WHERE event_name IN ('screen_view', 'click_payment')
  AND screen_name IN ('welcome', 'home', 'food_category', 'restaurant', 'cart')
GROUP BY 1,2,3
ORDER by 1, 3
;

/* 문제3
- 테이블: app_logs
- 문제: 집계한 데이터 pivot
- where 기준: 2022-08-01 ~ 2022-08-18
*/

WITH tmp AS (
SELECT  
  event_date,
  event_timestamp,
  event_name,
  user_id,
  user_pseudo_id,
  MAX(IF(param.key = 'firebase_screen', param.value.string_value, null)) AS screen_name,
  CONCAT(event_name, '-', MAX(IF(param.key = 'firebase_screen', param.value.string_value, null))) AS event_name_with_screen
FROM advanced.app_logs
cross join  UNNEST(event_params) AS param
WHERE event_date BETWEEN '2022-08-01' AND '2022-08-18'
GROUP BY 1,2,3,4,5
), tmp2 as (
SELECT event_date
     , event_name_with_screen
     , CASE WHEN event_name_with_screen = 'screen_view-welcome' THEN 1
       WHEN event_name_with_screen = 'screen_view-home' THEN 2
       WHEN event_name_with_screen = 'screen_view-food_category' THEN 3
       WHEN event_name_with_screen = 'screen_view-restaurant' THEN 4
       WHEN event_name_with_screen = 'screen_view-cart' THEN 5
       WHEN event_name_with_screen = 'click_payment-cart' THEN 6 END AS step_number
     , COUNT(DISTINCT user_pseudo_id) AS cnt
FROM tmp
WHERE event_name IN ('screen_view', 'click_payment')
  AND screen_name IN ('welcome', 'home', 'food_category', 'restaurant', 'cart')
GROUP BY 1,2,3
ORDER by 1, 3
)
select event_date
    , sum(if(step_number = 1, cnt, 0)) as `screen_view-welcome`
    , sum(if(step_number = 2, cnt, 0)) as `screen_view-home`
    , sum(if(step_number = 3, cnt, 0)) as `screen_view-food_category`
    , sum(if(step_number = 4, cnt, 0)) as `screen_view-restaurant`
    , sum(if(step_number = 5, cnt, 0)) as `screen_view-cart`
    , sum(if(step_number = 6, cnt, 0)) as `click_payment-cart`                
from tmp2
group by event_date
order by 1
;

Q1. 질문사항

pivot 쿼리 연습문제 2번에서, 모든 날짜를 기입하는 방식으로만 Pivot화 할 수 있는지 궁금합니다. 예를 들어, 날짜가 20개 있는 경우라면 일일이 날짜 기입하지 않고 일반화해서 pivot할 수 있는 방법이 있을지 여쭙습니다.

답변 1

인프런 AI 인턴

안녕하세요, 인프런 AI 인턴입니다.
jihyun lee님, SQL에서 다수의 날짜를 간편하게 Pivot 처리를 하려면 CASE문을 일일이 사용하는 대신 동적인 피벗을 사용할 수 있습니다. 일반적으로 RDBMS에 따라 접근 방식이 조금씩 다를 수 있지만, 예를 들어 MySQL이나 PostgreSQL에서는 동적 SQL을 사용하거나, CROSS JOIN을 통해 다중 날짜를 포함하도록 설계할 수 있습니다.

홉을 알기 위해 다음과 같은 접근 방식을 고려해 보세요:

동적 SQL 사용(주로 프로시저에서 사용 가능): SQL을 문자열로 구성한 뒤 해당 문자열을 실행하여 동적으로 열을 생성합니다.
CROSS JOIN을 활용한 Grouping: SQL을 조금 변형하여 각 날짜에 대해 그룹별로 데이터를 집계할 수 있습니다.

그러나, 주어진 데이터베이스의 기능에 따라 구체적인 방법이 달라질 수 있습니다. 따라서 각 데이터베이스의 문서를 참고하거나 Stack Overflow와 같은 커뮤니티에서 질문하여 더 많은 도움을 받을 수 있습니다.

Stack Overflow 링크: Spring Data on Stack Overflow

추가로 도움이 필요하시면 언제든지 문의해 주세요.

jihyun lee

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