1. 윈도우 함수 탐색 함수 연습 문제1-1) 유저의 다음 접속 월과 다다음 접속 월 구하기SELECT user_id , visit_month , LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS lead_visit_month , LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) AS lead2_visit_month FROM advanced.analytics_function_01 1-2) 이전 접속월 구하기SELECT user_id , visit_month , LAG(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS before_visit_month FROM advanced.analytics_function_01 1-3) 다음 접속 월까지의 간격 구하기SELECT * , after_visit_month - visit_month AS diff_month FROM ( SELECT user_id , visit_month , LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month FROM advanced.analytics_function_01 ) 1-4) 첫번째와 마지막 방문 월 구하기SELECT user_id , visit_month , FIRST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS first_visit_month , LAST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS last_visit_month FROM advanced.analytics_function_01 2. 윈도우 함수 Frame 연습 문제SELECT * , SUM(amount) OVER () AS amount_total , SUM(amount) OVER (ORDER BY order_id) AS cumulative_sum , SUM(amount) OVER (PARTITION BY user_id ORDER BY order_id) AS user_cumulative_sum , AVG(amount) OVER (ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_avg FROM advanced.orders 3. 윈도우 함수 연습 문제1-1) 사용자 별 쿼리를 실행한 총 횟수 구하기SELECT user , team , query_date , COUNT(*) OVER(PARTITION BY user) AS total_query_cnt FROM advanced.query_logs ORDER BY user, query_date. -- 검증 1-2) 주차 별로 팀 내에서 쿼리를 많이 실행한 수, 단 랭킹이 1등인 경우만 구하기SELECT week_number , team , user , query_cnt , RANK() OVER(PARTITION BY week_number, team ORDER BY query_cnt DESC) AS team_rank FROM ( SELECT EXTRACT(WEEK FROM query_date) AS week_number , team , user , COUNT(query_date) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) QUALIFY team_rank=1 ORDER BY week_number 1-3) 쿼리 실행 시점 기준 1주 전 쿼리 실행 수 구하기WITH query_cnt_by_team AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number , team , user , COUNT(query_date) AS query_cnt FROM advanced.query_logs GROUP BY EXTRACT(WEEK FROM query_date) , team , user ) SELECT * , LAG(query_cnt) OVER(PARTITION BY user ORDER BY week_number) AS prev_week_query_cnt FROM query_cnt_by_team ORDER BY user 1-4) 시간의 흐름에 따라 일자 별 유저가 실행한 누적 쿼리 수 구하기WITH query_cnt_by_user AS ( SELECT user , team , query_date , COUNT(query_date) AS query_count FROM advanced.query_logs GROUP BY ALL ) SELECT user , team , query_date , query_count , SUM(query_count) OVER(PARTITION BY user ORDER BY query_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_query_cnt -- , SUM(query_count) OVER(PARTITION BY user ORDER BY query_date) AS cumulative_query_cnt2 FROM query_cnt_by_user -- 검증 -- QUALIFY cumulative_query_cnt != cumulative_query_cnt2 ORDER BY user, query_date*집계 분석 함수를 사용하고 ORDER BY가 있는 경우 Frame의 디폴트 값은 UNBOUNDED PRECEDING ~ CURRENT ROW 이다. 1-5) 값이 null인 경우 직전 값으로 채우기null 값이 2번 연속 있는 경우가 있어 COALESCE를 사용했습니다.WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date -- , number_of_orders , COALESCE( number_of_orders , LAG(number_of_orders, 1) OVER(ORDER BY date) , LAG(number_of_orders, 2) OVER(ORDER BY date) ) AS number_of_orders FROM raw_data 성윤님의 풀이를 보고 LAST_VALUE + IGNORE NULLS를 사용하면 연속되는 null의 수에 상관없이 직전 값을 채울 수 있어 깔끔하다는 생각이 들었습니다!SELECT * , LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS last_value_orders FROM raw_data 1-6) 같은 데이터 셋에서 1일 전 부터 현재까지의 평균 구하기SELECT * , AVG(last_value_orders) OVER(ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM ( SELECT date , LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS last_value_orders FROM raw_data ) 1-7) 20초를 기준으로 세션 아이디 구하기WITH base AS ( SELECT event_date , user_pseudo_id , event_name , DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime FROM advanced.app_logs WHERE user_pseudo_id='1997494153.8491999091' AND event_date="2022-08-18" ORDER BY event_timestamp ) , base_second_dff AS ( SELECT * , DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) AS second_diff FROM ( SELECT * , LAG(event_datetime) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) before_event_datetime FROM base ) ) , result AS( SELECT * -- CASE 문 사용 가능 , IF(second_diff - before_second_diff > 20 OR second_diff IS NULL, 1, null) AS session_start FROM ( SELECT * , LAG(second_diff) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS before_second_diff FROM base_second_dff ) ) SELECT * , SUM(session_start) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_id FROM result ✔ 느낀 점프레임은 처음 보는 개념이라 초반에는 어색했지만 문제를 풀면서 익숙해질 수 있었다. 누적합을 구할 때 등 유용하게 사용할 것 같다."쿼리가 길어지는 것을 무서워하지말고 쿼리를 잘 수정할 수 있는 구조를 만들자" 일할 때에도 동료들이 수정하고 테스트하기 쉬운 구조인가?를 생각해보자.CTE의 이름을 짓는건 언제나 고민된다. 서브쿼리를 적절하게 사용하자! ✔ 새롭게 알게된 점ROW_NUMBER를 사용할 때 중복값이 있으면 순서가 보장되지 않을 수 있다. 이때는 OVER 안에서 id를 사용하자.FIRST_VALUE는 MIN과 같이 동작하지만, LAST_VALUE는 MAX가 아니다.전체 집계 값을 구하고 싶을 때에는 OVER 안에 아무것도 안 쓰면 된다.윈도우 함수에 조건을 걸고 싶을 땐 QUALIFY를 사용하자. 연속된 NULL값을 채우는 경우 LAST_VALUE와 IGNORE NULLS조합을 사용하자. 

강의 속도 조절 기능이 사라진거같은데 예전처럼 플레이어에 배속기능을 만들어 주세요

4번에 GROUPBY를 안써도 되는 이유가 무엇인가요??

1. 윈도우 함수 연습문제 풀이문제 1)사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리를 작성. 단, GROUP BY를 사용해서 집계하는 것이 아닌 query_logs의 데이터 우측에 새로운 컬럼을 만들어주세요SELECT user, team, query_date, COUNT(user) OVER (PARTITION BY user) AS total_query_cnt FROM advanced.query_logs ORDER BY query_date, team, user ;문제 2)주차별로 팀 내에서 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요.WITH weekly_team_query_cnt AS ( SELECT EXTRACT(week FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT week_number, team, user, query_cnt, ROW_NUMBER() OVER (PARTITION BY week_number, team ORDER BY query_cnt DESC) AS team_rank FROM weekly_team_query_cnt WHERE 1=1 QUALIFY team_rank = 1 ORDER BY week_number, team ;문제 3)(2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요WITH weekly_team_query_cnt AS ( SELECT EXTRACT(week FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT user, team, week_number, query_cnt, LAG(query_cnt, 1) OVER (PARTITION BY user ORDER BY week_number) AS prev_week_query_cnt FROM weekly_team_query_cnt ORDER BY user, team, week_number ;문제 4)시간의 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성WITH timeseries_user_query_count AS ( SELECT user, team, query_date, COUNT(user) AS query_count FROM advanced.query_logs GROUP BY ALL ) SELECT user, team, query_date, query_count, SUM(query_count) OVER(PARTITION BY user ORDER BY query_date) AS cumulative_query_count FROM timeseries_user_query_count문제 5)다음 데이터는 주문 횟수를 나타낸 데이터입니다. 만약 주문 횟수가 없으면 NULL로 기록됩니다. 이런 데이터에서 NULL 값이라고 되어있는 부분을 바로 이전 날짜의 값으로 채워주는 쿼리를 작성해주세요.WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, LAST_VALUE(number_of_orders IGNORE NULLS) OVER (ORDER BY date) AS number_of_orders FROM raw_data ;문제 6)5번 문제에서 NULL을 채운 후, 2일 전 ~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요(이동 평균)WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, number_of_orders, AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM ( SELECT * EXCEPT(number_of_orders), LAST_VALUE(number_of_orders IGNORE NULLS) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) ;문제 7)** 새로운 Session 부여하기본인 해결 쿼리) WITH create_new_session_app_logs_raw_data AS ( SELECT *, LAG(event_datetime) OVER (PARTITION BY user_pseudo_id ORDER BY event_timestamp) AS before_event_datetime FROM ( SELECT event_date, event_timestamp, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id, FROM advanced.app_logs WHERE event_date = '2022-08-18' AND user_pseudo_id = '1997494153.8491999091' ) ), create_new_session_app_logs AS ( SELECT *, IF(second_diff is null OR second_diff > 20, 1, null) AS session_start FROM ( SELECT *, EXTRACT(second FROM event_datetime - before_event_datetime) AS second_diff FROM create_new_session_app_logs_raw_data ) ) SELECT *, SUM(session_start) OVER (ORDER BY event_timestamp) AS session_id FROM create_new_session_app_logs ;카일스쿨님 해결 쿼리)본인 해결 쿼리와의 차이점 second_diff 새로운 컬럼 생성 시 DATETIME_DIFF(col1, col2, 시간단위)라는 함수 적용CASE WHEN문을 적용하여 직관적이고 수정하기 용이한 형태의 쿼리 사용 WITH base AS ( SELECT event_date, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id, FROM advanced.app_logs WHERE event_date = "2022-08-18" AND user_pseudo_id = "1997494153.8491999091" ORDER BY event_timestamp ), diff_data AS ( SELECT *, DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) AS second_diff # second_diff 기반으로 새로운 세션의 시작일지, 아닐지를 판단할 수 있음 FROM ( SELECT *, LAG(event_datetime, 1) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS prev_event_datetime, # event_datetime이랑 prev_event_datetime을 뺴서 20초가 넘으면 새로운 세션으로 정의 # 20초가 넘지 않으면 기존 세션 FROM base ) ) SELECT *, # 누적합을 사용해서 session_number를 만들었다! SUM(session_start) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_num # session을 구할 때 쿼리가 길어질 수 있음. 하루에 접속을 여러번 하는 서비스 => session기반이 좋을 수 있고, 아니라고 한다면 일자별 집계가 나을 수 있음 FROM ( SELECT *, CASE WHEN prev_event_datetime IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 # session을 나누는 기준 초. 데이터를 탐색하면서 결정 ELSE NULL END AS session_start # session이 시작됨을 알리는 session_start FROM diff_data ) ORDER BY event_datetime ; 

2. 윈도우 함수 주요 Point💡윈도우 함수 FRAME 개념은 익숙해지는 것이 중요qualify 개념 ⭐누적합 개념 사용해 세션 구하기 !! </aside> 2-1. 윈도우 함수 rows, range calculation연습문제 FRAME/* - 토픽: frame 연습문제 - 추가 컬럼 -- 1) amount_total -- 2) cumulative_sum -- 3) cumulative_sum_by_user -- 4) last_5_orders_avg_amount */ select * , sum(amount) over() as amount_total , sum(amount) over(order by order_id) as cumulative_sum , sum(amount) over(partition by user_id order by order_id) as cumulative_sum_by_user , sum(amount) over(order by order_id rows between 5 preceding and 1 preceding) as last_5_orders_avg_amount from advanced.orders ; 2-2. Qualify 개념윈도우 함수로 생성된 변수를 having 처리하는 것처럼 사용할 수 있음E.g. qualify amount_total ≥ 5002-3. 윈도우 함수 연습문제 모음윈도우 함수 연습문제/* - 연습문제 1 - window 활용 > partition by 활용 목적 */ select * , count(*) over(partition by user) as total_query_cnt from advanced.query_logs order by user, query_date ; /* - 연습문제 2 - window 활용 > partition by & qualify 활용 (훨씬 편하다!) - 해설 풀이 자체는 처음에 group by 쓰면 더 유용 (그럴 것 같음) */ with tmp as ( select week_number , team , user , count(*) over(partition by week_number, team, user) as query_cnt from ( select * , extract(week from query_date) as week_number from advanced.query_logs ) t ) select * , row_number() over(partition by week_number, team order by query_cnt desc) as team_rnk from tmp qualify team_rnk = 1 order by week_number, team ; /* - 연습문제 3 - window 활용 > lag 함수 목적 */ with tmp as ( select user , team , week_number , count(*) as query_count from ( select * , extract(week from query_date) as week_number from advanced.query_logs ) t group by user, team, week_number ) select * , lag(query_count) over(partition by user order by week_number) as prev_week_query_count from tmp t ; /* - 연습문제 4 - window 활용 > 누적 count 목적 - order by, partition by 개념 명확히 익히기에 좋음 - 해설 풀이와는 다름 (group by 후 > query_cnt를 sum하는 방식으로 풀이함) > 출제의도 frame 개념 파악 */ select distinct user , team , query_date , count(*) over(partition by user, query_date) as query_count ,count(*) over(partition by user order by query_date) as cumulative_query_count from advanced.query_logs order by 1, 2, 3 ; /* - 연습문제 6 - window 활용 > moving_avg 산출 문제 */ WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date , coalesce(number_of_orders, lag(number_of_orders) over(order by date)) as number_of_orders1 , Last_value(number_of_orders ignore nulls) over(order by date) as number_of_orders2 FROM raw_data ; /* - 연습문제 5 - window 활용 > coalesce & lag 활용목적 */ WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) select * , avg(number_of_orders) over(order by date rows between 2 preceding and current row) as moving_avg from ( SELECT date -- , coalesce(number_of_orders, lag(number_of_orders) over(order by date)) as number_of_orders1 , Last_value(number_of_orders ignore nulls) over(order by date) as number_of_orders FROM raw_data ) ; 2-4. 세션 구하기 문제세션 구하기 연습문제/* - 연습문제 7 - session 구하기 - diff & 누적합 개념으로 세션 구하는 것이 포인트 (누적합 개념 신선!) */ with base as ( select event_date , datetime(timestamp_micros(event_timestamp), 'Asia/Seoul') as event_datetime , event_name , user_id , user_pseudo_id from advanced.app_logs ), diff_date as ( select * , datetime_diff(event_datetime, prev_event_datetime, second) as second_diff from ( select * , lag(event_datetime) over(partition by user_pseudo_id order by event_datetime) as prev_event_datetime from base ) ) select * , sum(Session_Start) over(partition by user_pseudo_id order by event_datetime) as session_num from ( select * , case when prev_event_Datetime is null then 1 when second_diff >= 20 then 1 else 0 end as session_start from diff_date ) order by event_Datetime, user_pseudo_id ;

package com.example.stock.repository; import org.springframework.data.redis.core.RedisTemplate; import org.springframework.stereotype.Component; import java.time.Duration; @Component public class RedisLockRepository { private RedisTemplate<String, String> redisTemplate; public RedisLockRepository(RedisTemplate<String, String> redisTemplate) { this.redisTemplate = redisTemplate; } public Boolean lock(Long key) { return redisTemplate .opsForValue() .setIfAbsent(generateKey(key), "lock", Duration.ofMillis(3_000)); } public void unlock(Long key) { redisTemplate.delete(generateKey(key)); } private String generateKey(Long key) { return key.toString(); } } package com.example.stock.facade; import com.example.stock.repository.RedisLockRepository; import com.example.stock.service.StockService; import org.springframework.stereotype.Component; @Component public class LettuceLockStockFacade { private final RedisLockRepository redisLockRepository; private final StockService stockService; public LettuceLockStockFacade(RedisLockRepository redisLockRepository, StockService stockService) { this.redisLockRepository = redisLockRepository; this.stockService = stockService; } public void decrease(Long key, Long quantity) throws InterruptedException { while(!redisLockRepository.lock(key)){ Thread.sleep(100); } try{ stockService.decrease(key, quantity); }finally { redisLockRepository.unlock(key); } } } package com.example.stock.facade; import com.example.stock.domain.Stock; import com.example.stock.repository.StockRepository; import org.junit.jupiter.api.AfterEach; import org.junit.jupiter.api.BeforeEach; import org.junit.jupiter.api.Test; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.boot.test.context.SpringBootTest; import java.util.concurrent.CountDownLatch; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import static org.junit.jupiter.api.Assertions.*; @SpringBootTest class LettuceLockStockFacadeTest { @Autowired private LettuceLockStockFacade lettuceLockStockFacade; @Autowired private StockRepository stockRepository; @BeforeEach public void before(){ stockRepository.saveAndFlush(new Stock(1L, 100L)); } @AfterEach public void after(){ stockRepository.deleteAll(); } @Test public void 동시에_100개의_요청() throws InterruptedException { int threadCount = 100; ExecutorService executorService = Executors.newFixedThreadPool(32); CountDownLatch latch = new CountDownLatch(threadCount); for(int i = 0; i < threadCount; i++){ executorService.submit(() -> { try{ lettuceLockStockFacade.decrease(1L, 1L); } catch (InterruptedException e) { throw new RuntimeException(e); } finally { latch.countDown(); } }); } latch.await(); Stock stock = stockRepository.findById(1L).orElseThrow(); assertEquals(0, stock.getQuantity()); } }이렇게 코딩했고 dockre에 redis를 설치하고 window Shell에서 데이터가 남아 있을까봐 1번 key 값을 삭제해서 돌렸는데도 오류가 발생했습니다.또한 레디스가 실행이 되지 않은 채 했을까봐 ping을 입력했는데 pong이라는 응답을 받았습니다. Hibernate: drop table if exists stock 2024-11-03T22:08:40.414+09:00 DEBUG 17108 --- [ Test worker] org.hibernate.SQL : create table stock (id bigint not null auto_increment, product_id bigint, quantity bigint, version bigint, primary key (id)) engine=InnoDB Hibernate: create table stock (id bigint not null auto_increment, product_id bigint, quantity bigint, version bigint, primary key (id)) engine=InnoDB 2024-11-03T22:08:40.439+09:00 INFO 17108 --- [ Test worker] j.LocalContainerEntityManagerFactoryBean : Initialized JPA EntityManagerFactory for persistence unit 'default' 2024-11-03T22:08:41.136+09:00 INFO 17108 --- [ Test worker] o.s.d.j.r.query.QueryEnhancerFactory : Hibernate is in classpath; If applicable, HQL parser will be used. 2024-11-03T22:08:42.092+09:00 WARN 17108 --- [ Test worker] JpaBaseConfiguration$JpaWebConfiguration : spring.jpa.open-in-view is enabled by default. Therefore, database queries may be performed during view rendering. Explicitly configure spring.jpa.open-in-view to disable this warning 2024-11-03T22:08:42.990+09:00 INFO 17108 --- [ Test worker] c.e.s.facade.LettuceLockStockFacadeTest : Started LettuceLockStockFacadeTest in 6.21 seconds (process running for 7.345) 2024-11-03T22:08:43.504+09:00 DEBUG 17108 --- [ Test worker] org.hibernate.SQL : insert into stock (product_id,quantity,version) values (?,?,?) Hibernate: insert into stock (product_id,quantity,version) values (?,?,?) 2024-11-03T22:08:44.203+09:00 DEBUG 17108 --- [ Test worker] org.hibernate.SQL : select s1_0.id,s1_0.product_id,s1_0.quantity,s1_0.version from stock s1_0 where s1_0.id=? Hibernate: select s1_0.id,s1_0.product_id,s1_0.quantity,s1_0.version from stock s1_0 where s1_0.id=? 2024-11-03T22:08:44.245+09:00 DEBUG 17108 --- [ Test worker] org.hibernate.SQL : select s1_0.id,s1_0.product_id,s1_0.quantity,s1_0.version from stock s1_0 Hibernate: select s1_0.id,s1_0.product_id,s1_0.quantity,s1_0.version from stock s1_0 2024-11-03T22:08:44.252+09:00 DEBUG 17108 --- [ Test worker] org.hibernate.SQL : delete from stock where id=? and version=? Hibernate: delete from stock where id=? and version=? Expected :0 Actual :100 <Click to see difference> org.opentest4j.AssertionFailedError: expected: <0> but was: <100> at org.junit.jupiter.api.AssertionFailureBuilder.build(AssertionFailureBuilder.java:151) at org.junit.jupiter.api.AssertionFailureBuilder.buildAndThrow(AssertionFailureBuilder.java:132) at org.junit.jupiter.api.AssertEquals.failNotEqual(AssertEquals.java:197) at org.junit.jupiter.api.AssertEquals.assertEquals(AssertEquals.java:182) at org.junit.jupiter.api.AssertEquals.assertEquals(AssertEquals.java:177) at org.junit.jupiter.api.Assertions.assertEquals(Assertions.java:639) at com.example.stock.facade.LettuceLockStockFacadeTest.동시에_100개의_요청(LettuceLockStockFacadeTest.java:55) at java.base/java.lang.reflect.Method.invoke(Method.java:568) at java.base/java.util.ArrayList.forEach(ArrayList.java:1511) at java.base/java.util.ArrayList.forEach(ArrayList.java:1511) Java HotSpot(TM) 64-Bit Server VM warning: Sharing is only supported for boot loader classes because bootstrap classpath has been appended  어디를 더 수정해야 하는지 모르겠습니다..

 집계 함수Aggregation FunctionGROUP BY와 함께 사용한 COUNT나 AVG는 컬럼의 특정 Value를 기준으로 집계하면서 하나의 값을 반환합니다. 윈도우 함수다음과 같은 경우 유용한 함수A유저의 전 주문 수량은 얼마나 될까?B유저가 물건을 구입하기 전에 상세보기 페이지를 얼마나 방문했을까?C유저의 특정 구매 시점별 누적 구매 횟수는?특정 카테고리 내에서 제일 많이 방문한 제품은?이런 경우 모두 윈도우 함수(분석 함수)를 사용하면 편리함윈도움 함수를 사용하지 않으면 서브 쿼리와 JOIN 등을 사용해서 만들어야 함 집계 함수와 다르게 윈도우 함수는 각행마다 단일 값을 반환함종류 탐색 함수 : LEAD, LAG, FIRST_VALUE, LAST_VALUE번호 지정 함수 : RANK, DENSE_RANK, PERCENT_RANK, CUME_DIST, NTITLE집계 분석 함수: 집계 함수들, AVG, COUNT, SUM, MAX, MIN 문법함수 이름(컬럼, OFFSET) OVER (PARTITION BY 파티션_컬럼 ORDER BY 정렬_컬럼)OFFSET : 값을 가져올 행의 위치, 기본 값은 1이고 생략 가능 함수 이름 (컬럼) OVER (PARTITON BY 파티션_컬럼 ORDER BY 정렬_컬럼)필요에 따라 PARTITION BY는 생략 가능 #문제1 SELECT user_id, visit_month, lead(visit_month, 1) over(partition by user_id order by visit_month asc) as next_visit, lead(visit_month, 2) over(partition by user_id order by visit_month asc) as next_next_visit FROM workspace.analytics_function_01 ORDER BY user_id, visit_month ; #문제2 SELECT user_id, visit_month, lead(visit_month, 1) over(partition by user_id order by visit_month asc) as next_visit, lead(visit_month, 2) over(partition by user_id order by visit_month asc) as next_next_visit, lag(visit_month, 1) over(partition by user_id order by visit_month asc) as prev_visit FROM workspace.analytics_function_01 ORDER BY user_id, visit_month ; #문제3 SELECT user_id, visit_month, lead(visit_month, 1) over(partition by user_id order by visit_month asc) as next_visit_month, lead(visit_month, 1) over(partition by user_id order by visit_month asc) - visit_month as next_visit_month_diff FROM workspace.analytics_function_01 ORDER BY user_id, visit_month ; #추가문제 SELECT DISTINCT user_id, first_value(visit_month) over(partition by user_id order by visit_month asc rows between unbounded preceding and unbounded following) as first_visit_month, last_value(visit_month) over(partition by user_id order by visit_month asc rows between unbounded preceding and unbounded following) as last_visit_month FROM workspace.analytics_function_01 ORDER BY user_id ; #문제4 SELECT *, sum(amount) over() as total_amount, sum(amount) over(order by order_id asc rows between unbounded preceding and current row) as cumulative_sum, sum(amount) over(partition by user_id order by order_id asc rows between unbounded preceding and current row) as cumulative_sum_by_user, avg(amount) over(order by order_id asc rows between 5 preceding and 1 preceding) as last_five_orders_avg_amount FROM workspace.orders ORDER BY order_id ; #연습문제1 SELECT *, count(*) over(partition by user) as query_count_by_users FROM workspace.query_logs ; #연습문제2 SELECT query_weeknum, team, user, query_count, rank() over(partition by query_weeknum, team order by query_count desc) as query_rank FROM ( SELECT extract(week from query_date) as query_weeknum, team, user, count(1) as query_count FROM workspace.query_logs GROUP BY ALL ) QUALIFY query_rank = 1 ORDER BY query_weeknum ; #연습문제3 SELECT team, user, query_weeknum, query_count, lag(query_count, 1) over(partition by team, user order by query_weeknum asc) as prev_week_query_count FROM ( SELECT team, user, extract(week from query_date) as query_weeknum, count(1) as query_count FROM workspace.query_logs GROUP BY ALL ) #연습문제4 SELECT team, user, query_date, query_count, sum(query_count) over(partition by team, user order by query_date asc rows between unbounded preceding and current row) as cumulative_sum FROM ( SELECT team, user, query_date, count(1) as query_count FROM workspace.query_logs GROUP BY ALL ) ORDER BY team, user, query_date ; #연습문제5 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, ifnull(number_of_orders, real_prev_number_of_orders) as number_of_orders FROM ( SELECT date, number_of_orders, last_value(number_of_orders ignore nulls) over(order by date asc rows between unbounded preceding and 1 preceding) as real_prev_number_of_orders FROM raw_data ) ORDER BY date asc ; #연습문제6 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, number_of_orders, avg(number_of_orders) over(order by date asc rows between 2 preceding and current row) as moving_avg FROM ( SELECT date, ifnull(number_of_orders, real_prev_number_of_orders) as number_of_orders FROM ( SELECT date, number_of_orders, last_value(number_of_orders ignore nulls) over(order by date asc rows between unbounded preceding and 1 preceding) as real_prev_number_of_orders FROM raw_data ) ) ORDER BY date asc ; #연습문제7 WITH total_logs AS ( SELECT user_pseudo_id, event_name, timestamp_micros(event_timestamp) as event_datetime FROM workspace.app_logs ) SELECT user_pseudo_id, event_name, event_datetime, prev_event_datetime, second_diff, sum(session_change) over(partition by user_pseudo_id order by event_datetime asc) as session_id FROM ( SELECT *, case when event_datetime = first_event_datetime then 1 end as session_id, case when second_diff is null or second_diff >= 20 then 1 else 0 end as session_change FROM ( SELECT *, datetime_diff(event_datetime, prev_event_datetime, second) as second_diff FROM ( SELECT user_pseudo_id, event_name, event_datetime, lag(event_datetime, 1) over(partition by user_pseudo_id order by event_datetime asc) as prev_event_datetime, first_value(event_datetime) over(partition by user_pseudo_id order by event_datetime asc) as first_event_datetime FROM total_logs ) ) ) ORDER BY user_pseudo_id, event_datetime ; 

안녕하세요복습을 위해 수강연장 가능할지 문의드립니다.가능할 경우 연장해주시면 감사하겠습니다!

문제 1. SELECT *, lead(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_1month, lead(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_2month, FROM advanced.analytics_function_01 문제 2. SELECT *, lead(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_1month, lead(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_2month, lag(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS pre_1month FROM advanced.analytics_function_01 문제 3. SELECT *, SUM(amount) OVER (PARTITION BY user_id ORDER BY order_date, order_id ) AS cumulative_sum_by_user 문제 4. SELECT *, AVG(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_orders_avg_amout 문제 1. SELECT *, COUNT(query_date) OVER (PARTITION BY user ORDER BY user) AS total_query_cnt FROM advanced.query_logs 문제 2. WITH table AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT *, RANK() OVER(PARTITION BY week_number ORDER BY query_cnt) AS team_rank FROM table QUALIFY team_rank = 1 문제 3. WITH base AS ( SELECT EXTRACT(week FROM query_date) AS week_number, team, user, count(user) AS query_cnt FROM advanced.query_logs GROUP BY 1,2,3 ) SELECT *, lag(query_cnt) OVER(PARTITION BY team, user ORDER BY week_number asc) AS prev_week_query_cnt FROM base QUALIFY team_rank = 1 문제 4. WITH base AS ( SELECT user, team, query_date, COUNT(user) as query_count FROM advanced.query_logs GROUP BY 1,2,3 ) SELECT user, team, query_date, query_count, SUM(query_count) OVER(PARTITION BY team, user ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_query_count FROM base ORDER BY team, user, query_date; 문제 5. WITH raw_data AS( SELECT DATE'2024-05-01'AS date,15 AS number_of_orders UNION ALL SELECT DATE'2024-05-02',13 UNION ALL SELECT DATE'2024-05-03',NULL UNION ALL SELECT DATE'2024-05-04',16 UNION ALL SELECT DATE'2024-05-05',NULL UNION ALL SELECT DATE'2024-05-06',18 UNION ALL SELECT DATE'2024-05-07',20 UNION ALL SELECT DATE'2024-05-08',NULL UNION ALL SELECT DATE'2024-05-09',13 UNION ALL SELECT DATE'2024-05-10',14 UNION ALL SELECT DATE'2024-05-11',NULL UNION ALL SELECT DATE'2024-05-12',NULL ) SELECT date, IF(number_of_orders is null , last_value(number_of_orders IGNORE NULLS) OVER(ORDER BY date asc), number_of_orders) AS number_of_orders_not_null FROM raw_data; 문제 6. WITH raw_data AS( SELECT DATE'2024-05-01'AS date,15 AS number_of_orders UNION ALL SELECT DATE'2024-05-02', 13 UNION ALL SELECT DATE'2024-05-03', NULL UNION ALL SELECT DATE'2024-05-04', 16 UNION ALL SELECT DATE'2024-05-05', NULL UNION ALL SELECT DATE'2024-05-06', 18 UNION ALL SELECT DATE'2024-05-07', 20 UNION ALL SELECT DATE'2024-05-08', NULL UNION ALL SELECT DATE'2024-05-09', 13 UNION ALL SELECT DATE'2024-05-10', 14 UNION ALL SELECT DATE'2024-05-11', NULL UNION ALL SELECT DATE'2024-05-12', NULL ), filled_data AS ( SELECT * EXCEPT(number_of_orders), LAST_VALUE(number_of_orders IGNORE NULLS) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) SELECT *, AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM filled_data 문제 7. WITH base AS ( SELECT event_date, event_timestamp, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id FROM advanced.app_logs WHERE event_date = '2022-08-18' ), diff_date AS ( SELECT *, DATETIME_DIFF(event_datetime, pre_event_time, second) AS date_diff_sec FROM ( SELECT *, LAG(event_datetime, 1) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime asc) AS pre_event_time FROM base ) ), session_start AS ( SELECT *, CASE WHEN pre_event_time IS NULL THEN 1 WHEN date_diff_sec >= 20 THEN 1 END AS start_session FROM diff_date ) SELECT event_date, event_datetime, event_name, user_id, user_pseudo_id, date_diff_sec, SUM(start_session) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_id FROM session_start ORDER BY user_pseudo_id, event_datetime;

2-4. 윈도우 함수 탐색 함수 연습 문제문제 1) User들의 다음 접속 월과 다다음 접속 월을 구하는 쿼리를 작성해주세요 select *, lead(visit_month) over (partition by user_id order by visit_month) as next_visit_month, lead(visit_month, 2) over (partition by user_id order by visit_month) as next_visit_month2 from advanced.analytics_function_01 ; 결과문제 2) User들의 다음 접속 월과 다다음 접속 월, 이전 접속 월을 구하는 쿼리 작성select *, lead(visit_month, 1) over (partition by user_id order by visit_month) as next_visit_month, lead(visit_month, 2) over (partition by user_id order by visit_month) as next_visit_month2, lag(visit_month, 1) over (partition by user_id order by visit_month) as pre_visit_month from advanced.analytics_function_01 order by user_id, visit_month결과문제 3) user가 접속했을 때, 다음접속까지의 간격을 구하시오 select *, next_visit_month - visit_month as term From ( select *, lead(visit_month, 1) over (partition by user_id order by visit_month) as next_visit_month from advanced.analytics_function_01) order by user_id, visit_month ;결과추가문제 : 이 데이터셋을 User_id의 첫번째 방문 월, 마지막 방문월을 구하는 쿼리 작성 select *, first_value(visit_month) over (partition by user_id order by visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as first_visit_month, last_value(visit_month) over (partition by user_id order by visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as last_visit_month from advanced.analytics_function_01 order by user_id, visit_month ; 결과2-8. 윈도우 함수 Frame연습 문제Frame 연습문제select *, sum(amount) over() as amount_total, sum(amount) over(order by order_id) as cumulative_sum, -- 누적 sum(amount) over(partition by user_id order by order_id) as cumulative_sum_user, -- 누적 avg(amount) over(order by order_id rows between 5 preceding and 1 preceding) as last_5_avg from `advanced.orders` order by order_date, user_id ;결과 2-11. 윈도우 함수 연습 문제(1번)1) 사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리를 작성해주세요. 단, GROUP BY를 사용해서 집계하는 것이 아닌 query_logs의 데이터의 우측에 새로운 컬럼을 만들어주세요.select *, count(query_date) over() as total_cnt, # 전체 count(query_date) over(partition by user) as total_query_cnt from advanced.query_logs order by user, query_date ;결과2-11. 윈도우 함수 연습 문제(2번~6번)2) 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요with query_cnt_by_team as ( select extract(WEEK from query_date) as week_number, team, user, count(user) as query_cnt from advanced.query_logs group by all) select *, rank() over(partition by week_number, team order by query_cnt desc) as team_rank from query_cnt_by_team qualify team_rank = 1 order by week_number, team, query_cnt desc ;결과3) (2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요with query_cnt_by_team as ( select user, team, extract(WEEK from query_date) as week_number, count(user) as query_cnt from advanced.query_logs group by all) select *, LAG(query_cnt, 1) OVER (partition by user ORDER BY week_number) AS prev_week_query_count -- user 단위, 전주차 week_number from query_cnt_by_team -- order by user -- team, query_cnt desc ; 결과4) 시간의 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요select *, sum(query_cnt) over(partition by user order by query_date) as cumulative_sum, -- 왜되는거지? : frame의 default값 : unbounded preceding ~current row sum(query_cnt) over(partition by user order by query_date rows between unbounded preceding and current row) as cumulative_sum2 from (select user, team, query_date, count(*) as query_cnt from advanced.query_logs group by all) -- 검증 -- qualify cumulative_sum != cumulative_sum2 order by user, query_date ;결과5) 다음 데이터는 주문 횟수를 나타낸 데이터입니다. 만약 주문 횟수가 없으면 NULL로 기록됩니다. 이런 데이터에서 NULL 값이라고 되어있는 부분을 바로 이전 날짜의 값으로 채워주는 쿼리를 작성해주세요WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT *, LAG(number_of_orders) over(order by date) as number_of_orders2, ifnull(number_of_orders, LAG(number_of_orders) over(order by date)) as number_of_orders3, LAST_VALUE(number_of_orders) over(order by date) as last_value_orders, LAST_VALUE(number_of_orders ignore nulls) over(order by date) as last_value_orders2 FROM raw_data ;결과6) 5번 문제에서 NULL을 채운 후, 2일 전 ~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요(이동 평균)WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL) -- select * from raw_data; select * , avg(number_of_orders) over(order by date rows between 2 preceding and current row) as moving_avg from (SELECT * except(number_of_orders), LAST_VALUE(number_of_orders ignore nulls) over(order by date) as number_of_orders FROM raw_data) ;결과2-11. 윈도우 함수 연습 문제(7번)with base as ( select event_date, datetime(timestamp_micros(event_timestamp), 'Asia/Seoul') as event_datetime, event_name, user_id, user_pseudo_id from advanced.app_logs where 1=1 and event_date = "2022-08-18" and user_pseudo_id = '1997494153.8491999091'), diff_data as( select *, datetime_diff(event_datetime, prev_event_datetime, second) as second_diff -- second로 간격설정 # second_diff기반으로 새로운 세션의 시작일지 아닐지 판단할 수 있음 from(select *, lag(event_datetime, 1) over(partition by user_pseudo_id order by event_datetime) as prev_event_datetime # event_datetime이랑 prev_event_datetime을 빼서 20초가 넘으면 새로운 세션으로 정의, # 20초가 넘지 않으면 기존 세션 from base)) select *, sum(session_start) over(partition by user_pseudo_id order by event_datetime) as session_num from(select *, case when prev_event_datetime is null then 1 when second_diff >= 20 then 1 else 0 end as session_start # session이 시작됨을 알리는 session_start from diff_data) order by event_datetime limit 20;결과 느낀점마지막 문제같은 실무에 사용 할 법한 예제 문제들을 더 만들어서 풀어봐야 할 것 같다.session기준에 대한 리서치를 더 해봐야 할 것 같다. 

문제 1 SELECT *, LEAD(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month) AS lead_visit_month , LEAD(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS lead2_visit_month FROM advanced.analytics_function_01 ORDER BY user_id 문제 2 SELECT *, LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS lead_visit_month, LEAD(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS lead2_visit_month, -- 다다음달 LAG(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS lag_visit_month -- 이전 달 (LAG) FROM `advanced.analytics_function_01` ORDER BY user_id, visit_month 문제 3 SELECT *, LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS after_visit_month, LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) - visit_month AS diff # 윈도우 함수를 이렇게 쓰는게 좋을까? => 중복된 쿼리는 줄이는 것이 좋을 수 있음 FROM `advanced.analytics_function_01` - 총 정리 문제 문제 1 SELECT *, COUNT(query_date) OVER(PARTITION BY user ORDER BY query_date) AS total_query_cnt # 38 row FROM `advanced.query_logs` 문제 2 WITH base AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL) SELECT *, RANK() OVER(PARTITION BY team ORDER BY query_cnt) AS rk FROM base QUALIFY rk = 1 ORDER BY week_number, team, query_cnt DESC 문제 3 WITH base AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL) SELECT *, LAG(query_cnt, 1) OVER(PARTITION BY user ORDER BY week_number) AS prev_week_cnt FROM base 문제 4 SELECT *, SUM(query_cnt) OVER(PARTITION BY user ORDER BY query_date) AS cumulative_sum, SUM(query_cnt) OVER(PARTITION BY user ORDER BY query_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_sum2 FROM( SELECT query_date, team, user, COUNT(user) AS query_cnt FROM advanced.q 문제 5 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT *, LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS last_value_orders FROM raw_data 문제 6 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) , filled_date AS( SELECT * EXCEPT(number_of_orders), LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS number_of_orders FROM raw_data) SELECT *, AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM filled_date 문제 7 WITH base AS( SELECT EVENT_DATE, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id FROM advanced.app_logs WHERE event_date = '2022-08-18' AND user_pseudo_id = "1997494153.8491999091" ) , diff_date AS(SELECT *, DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) AS second_diff FROM ( SELECT *, LAG(EVENT_DATETIME, 1) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS prev_event_datetime FROM base ORDER BY base.event_datetime)) SELECT *, SUM(session_start) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_num FROM ( SELECT *, CASE WHEN prev_event_datetime IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 ELSE 0 END AS session_start FROM diff_date)    

탐색함수 연습 문제CREATE OR REPLACE TABLE advanced.analytics_function_01 AS ( SELECT 1004 AS user_id, 1 AS visit_month UNION ALL SELECT 1004, 3 UNION ALL SELECT 1004, 7 UNION ALL SELECT 1004, 8 UNION ALL SELECT 2112, 3 UNION ALL SELECT 2112, 6 UNION ALL SELECT 2112, 7 UNION ALL SELECT 3912, 4 ) SELECT * FROM advanced.analytics_function_01 --# 탐색 함수 1 SELECT user_id, visit_month, LEAD(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month, LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month FROM advanced.analytics_function_01 ORDER BY user_id --# 탐색 함수 2 SELECT user_id, visit_month, LEAD(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month, LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month, LAG(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS before_visit_month FROM advanced.analytics_function_01 ORDER BY user_id --# 탐색 함수 3 SELECT *, after_visit_month - visit_month AS diff FROM ( SELECT user_id, visit_month, LEAD(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month, FROM advanced.analytics_function_01 ORDER BY user_id ) --# 추가 문제 SELECT user_id, visit_month, FIRST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS fisrt_visit, LAST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) AS LAST_visit, FROM advanced.analytics_function_01 ORDER BY user_id프레임 연습 문제SELECT *, SUM(amount) OVER() AS amount_total, SUM(amount) OVER(ORDER BY order_id) AS cumulative_sum, SUM(amount) OVER(PARTITION BY user_id ORDER BY order_id) AS cumulative_sum_by_user, AVG(amount) OVER(ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_avg FROM advanced.orders ORDER BY order_id윈도우함수 연습 문제#1 SELECT *, COUNT(query_date) OVER (PARTITION BY user) as total_query_cnt FROM advanced.query_logs ORDER BY user, query_date #2 WITH TBL AS( SELECT EXTRACT(week from query_date) AS week_number ,team ,user ,COUNT(query_date) AS total_query_cnt FROM advanced.query_logs GROUP BY week_number, team, user ) SELECT week_number , team , user , total_query_cnt as query_count , RANK() OVER (PARTITION BY team ORDER BY total_query_cnt desc) AS team_rank FROM TBL QUALIFY team_rank = 1 ORDER BY week_number,team #3 WITH TBL AS( SELECT EXTRACT(week from query_date) AS week_number ,team ,user ,COUNT(query_date) AS total_query_cnt FROM advanced.query_logs GROUP BY week_number, team, user ) SELECT user ,team ,week_number ,total_query_cnt as query_count ,LAG(total_query_cnt) OVER (PARTITION BY user order by week_number) AS prev_week_query_count FROM TBL ORDER BY user, week_number; #4 SELECT user , team , query_date , query_count , SUM(query_count) OVER (PARTITION BY user ORDER BY query_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_query_count FROM (SELECT user ,team ,query_date ,COUNT(query_date) AS query_count FROM advanced.query_logs GROUP BY user,team,query_date ) ORDER BY user, query_date #5 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date , LAST_VALUE(number_of_orders ignore NULLS) OVER (ORDER BY date) AS number_of_orders FROM raw_data #6 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date , number_of_orders , AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) FROM ( SELECT date ,LAST_VALUE(number_of_orders ignore NULLS) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) #7 WITH base AS ( SELECT event_date , event_timestamp , DATETIME(TIMESTAMP_MICROS(event_timestamp),'Asia/Seoul') AS event_datetime , user_id ,user_pseudo_id FROM advanced.app_logs ), diff_sessions AS ( SELECT * , DATETIME_DIFF(event_datetime,prev_event_datetime, SECOND) AS second_diff FROM ( SELECT *, LAG(event_datetime) OVER (PARTITION BY user_pseudo_id ORDER BY base.event_datetime) AS prev_event_datetime FROM base ) ) SELECT * , SUM(session_start) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_start FROM ( SELECT * , CASE WHEN second_diff IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 ELSE 0 END as session_start FROM diff_sessions ) ORDER BY event_datetime

안녕하세요,직무 변경으로 여러 고민을 하던 중, 그로스마케터 직무에 관심이 생겨 강의 수강하고 있습니다.입문자가 들어도 이해가 쉽도록 깔끔하고 명확하게 설명해주셔서 도움이 많이 됩니다!수강평 등록했으며, 강의 교안 아래 메일로 보내주시면 감사하겠습니다!imlucky0824@gmail.com

탐색함수 연습문제문제 1) user들의 다음 접속 월과 다다음 접속 월을 구하는 쿼리를 작성해주세요. LEAD는 반드시 정렬이 먼저 되어야 함 -> ORDER BY 추가 SELECT *, LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month ASC) AS lead_visit_month, LEAD(visit_month,2) OVER (PARTITION BY user_id ORDER BY visit_month ASC) AS lead2_visit_month FROM advanced.analytics_function_01 문제 2) user들의 다음 접속 월과 다다음 접속 월, 이전 접속 월을 구하는 쿼리를 작성해주세요. SELECT *, LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month ASC) AS after_visit_month, LEAD(visit_month,2) OVER (PARTITION BY user_id ORDER BY visit_month ASC) AS after2_visit_month, LAG(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month ASC) AS before_visit_month ORDER BY user_id, visit_month FROM advanced.analytics_function_01 문제 3) 유저가 접속 했을 때, 다음 접속까지의 간격 SELECT *, after_visit_month - visit_month AS diff FROM ( SELECT *, LEAD(visit_month, 1) OVER (PARTITION BY user_id ORDER BY visit_month) AS after_visit_month FROM advanced.analytics_function_01 )   FRAME 연습문제-- amount_total: 전체 SUM -- cumulative_sum : row 시점에 누적 SUM -- cumulative_sum_by_user : row 시점에 유저별 누적 SUM -- last_5_orders_avg_amount : order_id 기준으로 정렬하고, 직전 5개 주문의 평균 amount -- 집계분석함수() OVER(PARTITION BY ~~~ ORDER BY ~~~ ROWS BETWEEN A AND B) SELECT *, SUM(amount) OVER() AS amount_total, -- OVER()에 아무것도 들어가지 않을 수 있음 SUM(amount) OVER(ORDER BY order_id) AS cumulative_sum, SUM(amount) OVER(PARTITION BY user_id ORDER BY order_id) AS cumulative_sum_by_user, AVG(amount) OVER(ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) FROM advanced.orders ORDER BY user_id, order_date    ***QUALIFY 연습문제QUALIFY(조건 설정) WHERE 대신 QUALIFY를 사용하면 윈도우 함수의 결과에 대해 필터링할 수 있음 WHERE과 같이 사용하는 경우엔 WHERE 아래에 작성하면 됨 SELECT order_id, order_date, user_id, amount, SUM(amount) OVER (PARTITION BY user_id) AS amount_total FROM advanced.orders WHERE 1=1 QUALIFY amount_total >= 500   윈도우 함수 연습문제 (1~7) 문제 1SELECT user, team, query_date, COUNT(query_date) OVER (PARTITION BY user) AS total_query_cnt FROM advanced.query_logs ORDER BY user, query_date 문제 2 WITH base AS ( SELECT EXTRACT(week FROM query_date) AS week_number, user, team, COUNT(query_date) AS total_query_cnt FROM advanced.query_logs GROUP BY week_number, user, team ) SELECT week_number, team, user, total_query_cnt, RANK() OVER (PARTITION BY team ORDER BY total_query_cnt DESC) AS ranking_in_team FROM base QUALIFY ranking_in_team = 1 ORDER BY week_number, team  문제 2WITH base AS ( SELECT EXTRACT(week FROM query_date) AS week_number, user, team, COUNT(query_date) AS total_query_cnt FROM advanced.query_logs GROUP BY week_number, user, team ) SELECT week_number, team, user, total_query_cnt, RANK() OVER (PARTITION BY team ORDER BY total_query_cnt DESC) AS ranking_in_team FROM base QUALIFY ranking_in_team = 1 ORDER BY week_number, team 문제 3WITH base AS ( SELECT EXTRACT(week FROM query_date) AS week_number, user, team, COUNT(query_date) AS query_cnt FROM advanced.query_logs GROUP BY week_number, user, team ) SELECT user, team, week_number, query_cnt, LAG(query_cnt) OVER (PARTITION BY user ORDER BY week_number) AS prev_week_query_cnt FROM base ORDER BY user, week_number 문제 4문제 4 SELECT user, team, query_date, query_count, SUM(query_count) OVER (PARTITION BY user ORDER BY query_date, ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_query_count FROM ( SELECT user, team, query_date, COUNT(query_date) AS query_count FROM advanced.query_logs GROUP BY 1,2,3 ) ORDER BY user, query_date 문제 5문제 5 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, LAST_VALUE(number_of_orders ignore nulls) OVER (ORDER BY date) AS number_of_orders FROM raw_data  문제 6WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT date, number_of_orders, AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM ( SELECT date, LAST_VALUE(number_of_orders ignore nulls) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) 문제 7문제 7 WITH base AS ( SELECT event_date, event_timestamp, DATETIME(TIMESTAMP_MICROS(event_timestamp),'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id, LAG(DATETIME(TIMESTAMP_MICROS(event_timestamp),'Asia/Seoul')) OVER (PARTITION BY user_pseudo_id ORDER BY event_timestamp) AS before_event_datetime FROM advanced.app_logs WHERE 1=1 AND event_date = '2022-08-18' ) SELECT *, DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) AS second_diff, CASE WHEN DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) IS NULL OR DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) >= 20 THEN 1 ELSE 0 END AS session_start, SUM(CASE WHEN DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) >= 20 THEN 1 ELSE 0 END) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) + 1 AS session_temp FROM base  

1. 윈도우 함수 연습문제(1) 사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리는 작성해주세요.단, GROUP BY를 사용해서 집계하는 것이 아닌 query_logs의 데이터의 우측에 새로운 컬럼을 만들어주세요.SELECT *, COUNT(query_date) OVER(PARTITION BY user) AS cnt_query FROM advanced.query_logs(2) 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요.WITH query_cnt_by_team AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT *, RANK() OVER(PARTITION BY week_number, team ORDER BY query_cnt) AS rk FROM query_cnt_by_team QUALIFY rk = 1 ORDER BY week_number, team, query_cnt DESC*주차별 데이터는 EXTRACT 사용(3) (2번 문제에서 사용한 수치별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요.WITH query_cnt_by_team AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT *, LAG(query_cnt) OVER(PARTITION BY user ORDER BY week_number) AS prev_week_query_cnt FROM query_cnt_by_team(4) 시간이 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요.-- 누적 쿼리 : frame. 과거의 시간(UNBOUNDED PRECEDING) 부터 CUREENT ROW 까지 -- frame의 default SELECT *, SUM(query_cnt) OVER(PARTITION BY user ORDER BY query_date) AS cumulative_sum FROM ( SELECT query_date, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) -- QUALIFY cumulative_sum != cumulative_sum2 -- WHERE, QUALIFY 조건 설정해서 2가지 조건이 같은지 비교 -> 같으면 != 연산 결과에 반환하는 값 없음 ORDER BY user, query_date(5) 다음 데이터는 주문 횟수를 나타낸 데이터입니다. 만약 주문 횟수가 없으면 NULL로 기록됩니다. 이런 데이터에서 NULL 값이라고 되어있는 부분을 바로 이전 날짜의 값으로 채워주는 쿼리를 작성해주세요.WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT *, LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS last_value_orders FROM raw_data*IGNORE NULLS 사용(6) 5번 문제에서 NULL을 채운 후, 2일전 ~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요. (이동 평균)WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) , filled_data AS ( SELECT * EXCEPT(number_of_orders), LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date) AS number_of_orders FROM raw_data ) SELECT *, AVG(number_of_orders) OVER(ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) as moving_avg FROM filled_data*WITH문 두번 사용 시, ',' 사용(7) app_logs 테이블에서 Custom Session을 만들어 주세요. 이전 이벤트 로그와 20초가 지나면 새로운 Session을 만들어 주세요. Session은 숫자로 (1, 2, 3, ...) 표시해도 됩니다.2022-08-18일의 user_pseudo_id(1997494153.8491999091)은 session_id가 4까지 나옵니다. WITH base AS( SELECT event_date, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_date, user_id, user_pseudo_id FROM advanced.app_logs WHERE event_date = "2022-08-18" AND user_pseudo_id = "1997494153.8491999091" ) , diff_data AS ( SELECT *, DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) AS second_diff FROM ( SELECT *, LAG(event_datetime, 1) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS prev_event_datetime FROM base ORDER BY event_datetime ) ) SELECT *, SUM(session_start) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_num -- session을 구할 때 쿼리가 길어질 수 있음. 하루에 접속을 여러번 하는 서비스 -> session 기반이 좋을 수 있고, 아니라고 하면 유저 집계가 나올 수 있음 FROM ( SELECT *, CASE WHEN prev_event_datetime IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 # 세션을 나누는 기준 초, 데이터를 탐색하면서 결정. ELSE NULL END AS session_start FROM diff_data ) *세션 기준을 직접 정의 가능  

1번 SELECT *, LEAD(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_visit_month, LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) AS nnext_visit_month FROM advanced.analytics_function_01 2번 SELECT *, LEAD(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_visit_month, LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) AS nnext_visit_month, LAG(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month) AS previous_visit_month FROM advanced.analytics_function_01 3번 SELECT *, SUM(amount) OVER (PARTITION BY user_id ORDER BY order_date, order_id ) AS cumulative_sum_by_user 4번 SELECT *, AVG(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_orders_avg_amout ==================================================================================== 총정리 문제 1번 SELECT *, COUNT(query_date) OVER (PARTITION BY user ORDER BY user) AS total_query_cnt FROM advanced.query_logs 2번 WITH table AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT *, RANK() OVER(PARTITION BY week_number ORDER BY query_cnt) AS team_rank FROM table QUALIFY team_rank = 1 3번 WITH table AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number, team, user, COUNT(user) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT *, LAG(query_cnt) OVER (PARTITION BY user ORDER BY week_number) AS prev_week_query_cnt FROM table 4번 WITH query_count_table AS ( SELECT *, COUNT(*) AS query_count FROM advanced.query_logs GROUP BY ALL ) SELECT *, SUM(query_count) OVER (PARTITION BY user ORDER BY query_date) AS cululative_query_count FROM query_count_table 5번 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07' , 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL UNION ALL ) SELECT *, LAST_VALUE(number_of_order IGNORE NULLS) OVER (ORDER BY date) AS last_value_orders FROM raw_data 6번 WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07' , 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ), filled_data AS ( SELECT * EXCEPT(number_of_orders), **LAST_VALUE**(number_of_orders **IGNORE NULLS**) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) SELECT *, AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM filled_data 7번 WITH base AS( SELECT event_date, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, event_name, user_id, user_pseudo_id FROM advanced.app_logs WHERE event_date = '2022-08-18' AND user_pseudo_id = "1997494153.8491999091" ), diff_data AS( SELECT *, DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) AS second_diff FROM ( SELECT *, LAG(event_datetime, 1) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS prev_event_datetime FROM base ) ) SELECT *, SUM(session_start) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_num FROM( SELECT *, CASE WHEN prev_event_datetime IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 ELSE NULL END AS session_start FROM diff_data ) ORDER BY event_datetime   

학습 관련 질문을 남겨주세요. 상세히 작성하면 더 좋아요!질문과 관련된 영상 위치를 알려주면 더 빠르게 답변할 수 있어요먼저 유사한 질문이 있었는지 검색해보세요6:42 부터 quantile 에서강의를 따라하면 에러가 나는데 어디서 수정을 해야할까요

def cnt_word1(filepath): with open(filepath,'r') as file: txt = file.read() txt = txt.replace(',',' ') print(txt) print(cnt_word1("../source/22-1.txt"))   이렇게 그대로 사용했는데 아무것도 안뜹니다오류메세지가 없는걸로 보아 오류는 아닌데 (그대로 입력) 파일 경로가 문제일까요?  

윈도우 함수(탐색 함수) 연습 문제문제 1. user들의 다음 접속 월과 다다음 접속 월을 구하는 쿼리를 작성해주세요.SELECT * , LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS next1_visit_month , LEAD(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS next2_visit_month FROM advanced.analytics_function_01 ORDER BY user_id, visit_month 문제 2. user들의 다음 접속 월과 다다음 접속 월, 이전 접속 월을 구하는 쿼리를 작성해주세요.SELECT * , LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS next1_visit_month , LEAD(visit_month, 2) OVER(PARTITION BY user_id ORDER BY visit_month) AS next2_visit_month , LAG(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month) AS before_visit_month FROM advanced.analytics_function_01 ORDER BY user_id, visit_month 문제 3. user가 접속했을 때 다음 접속까지의 간격을 구하시오.SELECT * , next_visit_month - visit_month AS diff_month FROM ( SELECT * , LEAD(visit_month, 1) OVER(PARTITION BY user_id ORDER BY visit_month) AS next_visit_month FROM advanced.analytics_function_01 ORDER BY user_id, visit_month ) AS tmp 추가 문제. 유저의 첫 번째 방문 월, 마지막 방문 월을 구하는 쿼리를 작성해주세요.SELECT * , FIRST_VALUE(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS first_visit_month , LAST_VALUE(visit_month) OVER(PARTITION BY user_id ORDER BY visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_visit_month FROM advanced.analytics_function_01 ORDER BY user_id, visit_month  윈도우 함수(프레임, QUALIFY) 연습 문제문제 1. -- amount_total: 전체 SUM -- cumulative_sum: row 시점에 누적 SUM -- cumulative_sum_user: row 시점에 유저별 누적 SUM -- last_5_avg: order_id 기준으로 정렬하고, 직전 5개의 주문의 평균 amount SELECT * , SUM(amount) OVER(ORDER BY order_id ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS amount_total , SUM(amount) OVER(ORDER BY order_id) AS cumulative_sum , SUM(amount) OVER(PARTITION BY user_id ORDER BY order_id) AS cumulative_sum_user , AVG(amount) OVER(ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_avg FROM `inflearn-bigquery-437203.advanced.orders` ORDER BY order_id 문제 2. QUALIFY 로 윈도우 함수 조건 걸기SELECT * , SUM(amount) OVER(ORDER BY order_id ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS amount_total_1 , SUM(amount) OVER() AS amout_total_2 -- OVER 안에 아무것도 쓰지 않는 경우도 있다. 그러면 amount_total와 같은 결과가 계산된다. , SUM(amount) OVER(ORDER BY order_id) AS cumulative_sum , SUM(amount) OVER(PARTITION BY user_id ORDER BY order_id) AS cumulative_sum_user , AVG(amount) OVER(ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_avg FROM `inflearn-bigquery-437203.advanced.orders` QUALIFY amount_total_1 >= 500  윈도우 함수 연습 문제 (1~7번)문제 1. 사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리를 작성해주세요. 단, GROUP BY를 사용해서 집계하는 것이 아닌 query_logs의 데이터의 우측에 새로운 컬럼을 만들어주세요.SELECT * , COUNT(*) OVER(PARTITION BY user) AS total_query_cnt FROM advanced.query_logs 문제 2. 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요.WITH query_cnt_by_team AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number , team , user , COUNT(*) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT * , RANK() OVER(PARTITION BY week_number, team ORDER BY query_cnt DESC) AS rk FROM query_cnt_by_team QUALIFY rk = 1 -- 윈도우 함수 결과에 대해 조건 걸 때 사용 ORDER BY week_number, team, query_cnt DESC 문제 3. (2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요.WITH query_cnt_by_team AS ( SELECT EXTRACT(WEEK FROM query_date) AS week_number , team , user , COUNT(*) AS query_cnt FROM advanced.query_logs GROUP BY ALL ) SELECT user , team , week_number , query_cnt , LAG(query_cnt, 1) OVER(PARTITION BY user ORDER BY week_number) as prev_week_query_cnt FROM query_cnt_by_team ORDER BY user, team, week_number 문제 4. 시간의 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요.SELECT * , SUM(query_cnt) OVER(PARTITION BY user ORDER BY query_date) AS cumulative_query_cnt FROM ( SELECT * , COUNT(*) AS query_cnt FROM advanced.query_logs GROUP BY ALL) AS tmp -- 윈도우 함수 집계 함수 프레임 Default -- : RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW 문제 5. 다음 데이터는 주문 횟수를 나타낸 데이터. 만약 주문 횟수가 없으면 NULL로 기록된다. 이런 데이터에서 NULL 값이라고 되어 있는 부분을 바로 이전 날짜의 값으로 채워주는 쿼리를 작성해주세요. (어려움!)WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) SELECT * , LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS last_value_orders FROM raw_data 문제 6. 5번 문제에서 NULL을 채운 후, 2일 전~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요. (이동평균)WITH raw_data AS ( SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL SELECT DATE '2024-05-03', NULL UNION ALL SELECT DATE '2024-05-04', 16 UNION ALL SELECT DATE '2024-05-05', NULL UNION ALL SELECT DATE '2024-05-06', 18 UNION ALL SELECT DATE '2024-05-07', 20 UNION ALL SELECT DATE '2024-05-08', NULL UNION ALL SELECT DATE '2024-05-09', 13 UNION ALL SELECT DATE '2024-05-10', 14 UNION ALL SELECT DATE '2024-05-11', NULL UNION ALL SELECT DATE '2024-05-12', NULL ) , filled_data AS ( SELECT * , LAST_VALUE(number_of_orders IGNORE NULLS) OVER(ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS last_value_orders FROM raw_data ) SELECT date , last_value_orders , AVG(last_value_orders) OVER(ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM filled_data 문제 7. 세션 id 만들기 WITH step1 AS ( SELECT event_date , event_timestamp , DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime , event_name , user_id , user_pseudo_id FROM advanced.app_logs WHERE event_date = "2022-08-18" AND user_pseudo_id = "1997494153.8491999091" ) -- 이전 이벤트가 발생한 시각을 나타내는 컬럼 추가 , step2 AS ( SELECT * , LAG(event_datetime, 1) OVER(PARTITION BY user_pseudo_id ORDER BY event_datetime) AS before_event_datetime FROM step1 ) -- 현재 이벤트와 이전 이벤트의 발생 시간의 차이를 나타내는 컬럼 추가 , step3 AS ( SELECT * , DATETIME_DIFF(event_datetime, before_event_datetime, SECOND) AS second_diff FROM step2 ) -- 세션 id 부여하는 컬럼 추가 SELECT * , SUM(session_start) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_id FROM ( SELECT * , IF(second_diff IS NULL OR second_diff > 20, 1, NULL) AS session_start FROM step3 ) AS tmp 

 탐색 함수 연습 문제 -- 문제 1) user들의 다음 접속 월과 다다음 접속 월을 구하는 쿼리를 작성해주세요. -- SELECT -- user_id, -- visit_month, -- LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) as next_visit_month, -- LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) as next_next_visit_month -- FROM advanced.analytics_function_01 -- 문제 2) user들의 다음 접속 월과 다다음 접속 월, 이전 접속 월을 구하는 쿼리를 작성해주세요. -- SELECT -- user_id, -- visit_month, -- LEAD(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) as next_visit_month, -- LEAD(visit_month, 2) OVER (PARTITION BY user_id ORDER BY visit_month) as next_next_visit_month, -- LAG(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month) as last_visit_month -- FROM advanced.analytics_function_01 -- 추가 문제 : 이 데이터셋을 기준으로 user_id의 첫 방문월, 마지막 방문월을 구하는 쿼리를 작성해주세요. SELECT DISTINCT user_id, FIRST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as first_visit_month, LAST_VALUE(visit_month) OVER (PARTITION BY user_id ORDER BY visit_month ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as last_visit_month FROM advanced.analytics_function_01  FramePIVOT 연습 문제에서 사용한 테이블 활용 SELECT *, SUM(amount) OVER () as amount_total, SUM(amount) OVER (ORDER BY order_id) as cumluative_sum, SUM(amount) OVER (PARTITION BY user_id ORDER BY order_id) as cumluative_sum_user, AVG(amount) OVER (ORDER BY order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) as last_5_avg FROM advanced.orders ORDER BY order_id  윈도우 함수 연습문제 -- 1) 사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리를 작성해주세요. 단, GROUP BY를 사용해서 집계하는 것이 아닌 query_logs의 데이터의 우측에 새로운 컬럼을 만들어주세요. -- SELECT -- *, -- COUNT(*) OVER (PARTITION BY user) as total_query_count -- FROM advanced.query_logs -- 2) 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요 -- SELECT -- DISTINCT *, -- RANK() OVER (PARTITION BY week_number, team ORDER BY query_cnt desc) as team_rank -- FROM -- ( -- SELECT -- week_number, -- team, -- user, -- COUNT(query_date) OVER (PARTITION BY user, week_number) as query_cnt -- FROM -- (SELECT -- *, -- EXTRACT(WEEK FROM query_date) as week_number -- FROM advanced.query_logs -- ) -- ) -- QUALIFY team_rank = 1 -- ORDER BY week_number, team -- 문제 의도 : 원본 데이터는 row 마다 데이터가 있고, 그걸 집계해서 활용. GROUP BY 사용 후에 윈도우 함수 -- 3) (2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요 -- SELECT -- *, -- LAG(query_cnt) OVER (PARTITION BY user ORDER BY week_number) as prev_week_query_count -- FROM -- (SELECT -- distinct week_number, -- team, -- user, -- COUNT(query_date) OVER (PARTITION BY user, week_number) as query_cnt -- FROM -- (SELECT -- *, -- EXTRACT(WEEK FROM query_date) as week_number -- FROM advanced.query_logs -- ) -- ) -- ORDER BY user, week_number -- 4) 시간의 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요 -- SELECT -- distinct *, -- COUNT(query_date) OVER (PARTITION BY user, query_date ORDER BY query_date) as query_cnt, -- COUNT(query_date) OVER (PARTITION BY user ORDER BY query_date) as cumulative_query_cnt -- FROM advanced.query_logs -- ORDER BY user, query_date -- 출제 의도 : Default Frame 이해 // 집계분석에서 Frame의 Default 값 = BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW -- 5) 다음 데이터는 주문 횟수를 나타낸 데이터입니다. 만약 주문 횟수가 없으면 NULL로 기록됩니다. 이런 데이터에서 NULL 값이라고 되어있는 부분을 바로 이전 날짜의 값으로 채워주는 쿼리를 작성해주세요 -- WITH raw_data AS ( -- SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL -- SELECT DATE '2024-05-03', NULL UNION ALL -- SELECT DATE '2024-05-04', 16 UNION ALL -- SELECT DATE '2024-05-05', NULL UNION ALL -- SELECT DATE '2024-05-06', 18 UNION ALL -- SELECT DATE '2024-05-07', 20 UNION ALL -- SELECT DATE '2024-05-08', NULL UNION ALL -- SELECT DATE '2024-05-09', 13 UNION ALL -- SELECT DATE '2024-05-10', 14 UNION ALL -- SELECT DATE '2024-05-11', NULL UNION ALL -- SELECT DATE '2024-05-12', NULL -- ) -- SELECT -- date, -- CASE WHEN number_of_orders is not null THEN number_of_orders -- ELSE LAG(number_of_orders) OVER (ORDER BY date) -- END AS number_of_orders -- FROM raw_data -- 수정 -- SELECT -- *, -- LAST_VALUE(number_of_orders IGNORE NULLS) OVER (ORDER BY date) as last_value_orders -- FROM raw_data -- 출제 의도 : first_value, last_value 사용할 때 null을 포함하고 싶으면 ignore nulls 활용 -- 6) 5번 문제에서 NULL을 채운 후, 2일 전 ~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요(이동 평균) -- WITH raw_data AS ( -- SELECT DATE '2024-05-01' AS date, 15 AS number_of_orders UNION ALL SELECT DATE '2024-05-02', 13 UNION ALL -- SELECT DATE '2024-05-03', NULL UNION ALL -- SELECT DATE '2024-05-04', 16 UNION ALL -- SELECT DATE '2024-05-05', NULL UNION ALL -- SELECT DATE '2024-05-06', 18 UNION ALL -- SELECT DATE '2024-05-07', 20 UNION ALL -- SELECT DATE '2024-05-08', NULL UNION ALL -- SELECT DATE '2024-05-09', 13 UNION ALL -- SELECT DATE '2024-05-10', 14 UNION ALL -- SELECT DATE '2024-05-11', NULL UNION ALL -- SELECT DATE '2024-05-12', NULL -- ) -- SELECT -- *, -- AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) as moving_avg -- FROM -- (SELECT -- date, -- CASE WHEN number_of_orders is not null THEN number_of_orders -- ELSE LAG(number_of_orders) OVER (ORDER BY date) -- END AS number_of_orders -- FROM raw_data -- ) -- 7) app_logs 테이블에서 Custom Session을 만들어 주세요. 이전 이벤트 로그와 20초가 지나면 새로운 Session을 만들어 주세요. Session은 숫자로 (1, 2, 3 ...) 표시해도 됩니다 -- 2022-08-18일의 user_pseudo_id(1997494153.8491999091)은 session_id가 4까지 나옵니다 WITH base AS ( SELECT event_date, DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') as event_datetime, user_id, user_pseudo_id FROM advanced.app_logs WHERE event_date = '2022-08-18' and user_pseudo_id = '1997494153.8491999091' ), diff_data AS ( SELECT *, DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) as second_diff FROM ( SELECT *, LAG(event_datetime) OVER (PARTITION BY user_pseudo_id ORDER BY base.event_datetime) as prev_event_datetime FROM base ) ) SELECT *, SUM(session_start) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_num -- session 을 구할 때 쿼리가 길어질 수 있음. 하루에 접속을 여러번 하는 서비스 -> session 기반이 좋을 수 있고, 아니라면 일자별 유저 집계가 나을 수 있음 FROM ( SELECT *, CASE WHEN second_diff IS NULL THEN 1 WHEN second_diff >= 20 THEN 1 #세션을 나누는 기준. 데이터 탐색 후 결정 / 보통 앱 로그는 30초 or 60초 ELSE 0 END as session_start FROM diff_data ) ORDER BY event_datetime -- 세션 정리 -- 이전 이벤트 로그와 현재 이벤트 로그의 diff => 초나 분을 구한다 -- 기준을 가지고 그 기준 보다 높으면 새로운 세션이라고 한다 -- 첫번째 값엔 null이 있을 수 있어, 이 부분도 챙겨야 한다 -- 새로운 세션, session_start 기반으로 누적합 => session_num이 된다