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

문제 1번SELECT 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 2번WITH 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 3번WITH 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 4번SELECT 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 5번WITH 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 6번WITH 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 7번WITH 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

@RequestParam 개념을 공부중인데 제가 이해한게 맞는지 여쭤보고 싶어서요 메서드의 파라미터 자리에 작성하는 애노테이션이다.파라미터의 이름과 같읕 메서드를 찾아 값을 바인딩 한다.void String event(@RequestParam (value=“abc”) String abc){ ~ }이라는 코드가 있을 때 abc의 값을 파라미터 abc의 값으로 저장해준다.이름이 같을 경우에는 void String event(@RequestParam String abc){ ~ } 으로 생략 가능하다. 이게 맞을까요?

 <FRAME 문제> 1번문제: 총 수량(amount_total), 수량의 누적 합(cumulativ_sum), 유저별 수량의 누적 합(cumulative_sum(user)), 최근 5개 수량의 평균(last_5_avg) 출력쿼리를 작성하는 목표, 확인할 지표: 수량의 총합 또는 누적 합 구하기쿼리 계산 방법: 윈도우 함수 - AVG, SUM데이터의 기간: X사용할 테이블: advanced.ordersJOIN KEY: X데이터 특징: XSELECT order_id, -- 각 주문의 고유 식별자 order_date, -- 주문 날짜 user_id, -- 주문한 사용자 ID amount, -- 주문 금액 SUM(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS amount_total, -- 주문 데이터 전체의 총 금액을 계산합니다. -- 'ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING'은 모든 행을 참조한다는 의미로, 테이블의 전체 합계를 계산합니다. SUM(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_sum, -- 날짜와 주문 ID 순서로 누적 합계를 계산합니다. -- 'ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW'는 현재 행까지의 모든 이전 행을 포함하여 누적 합계를 계산합니다. SUM(amount) OVER (PARTITION BY user_id ORDER BY order_date, order_id ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS cumulative_sum_by_user, -- 사용자별로 누적 합계를 계산합니다. -- 'PARTITION BY user_id'는 각 사용자별로 데이터를 분리해 누적 합계를 계산하도록 합니다. -- 즉, 각 사용자가 주문한 금액의 누적 합계를 구합니다. AVG(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_orders_avg_amount -- 현재 행을 기준으로 바로 이전 5개의 행의 주문 금액 평균을 계산합니다. -- 'ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING'은 현재 행 이전의 5개 행에서 1개 행까지 포함하여 평균을 구합니다. -- 이를 통해 최근 5건의 주문에 대한 평균 금액을 알 수 있습니다. FROM advanced.orders -- advanced 데이터셋의 orders 테이블에서 데이터를 가져옵니다. --QUALIFY last_5_orders_avg_amount >= 150 -- 마지막 5건의 주문 평균 금액이 150 이상인 주문만을 결과로 필터링하기 위한 조건입니다. 현재는 주석 처리되어 있어 실행되지 않습니다. ORDER BY order_id; -- 주문 ID 기준으로 결과를 정렬하여 각 주문에 대해 계산된 값을 순서대로 확인합니다.   <총 정리 문제> 1번문제:사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리를 작성해주세요. 우측에 새로운 컬럼을 만들어주세요. SELECT * -- 모든 열을 선택합니다. advanced.query_logs 테이블의 모든 컬럼을 그대로 포함합니다. , COUNT(user) OVER (PARTITION BY user) AS total_query_cnt -- 'user'별로 쿼리 실행 횟수를 계산합니다. -- 'PARTITION BY user'를 사용하여 각 사용자별로 나누고, 그 안에서 해당 사용자가 몇 번의 쿼리를 실행했는지 계산합니다. -- 이 값은 각 사용자가 테이블에 몇 번 등장했는지를 의미합니다. FROM advanced.query_logs -- advanced 데이터셋의 query_logs 테이블에서 데이터를 가져옵니다. ORDER BY query_date, user; -- 'query_date'와 'user'를 기준으로 데이터를 정렬합니다. 'query_date'를 우선적으로, 같은 날짜 내에서는 'user'를 기준으로 정렬합니다.   2번문제: 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후, 실행한 수를 활용해 랭킹을 구해주세요. 단, 랭킹이 1등인 사람만 결곽가 보이도록 해주세요.SELECT * -- 모든 열을 선택합니다. 서브쿼리에서 생성된 모든 컬럼을 포함합니다. , RANK() OVER (PARTITION BY week_number, team ORDER BY query_cnt DESC) AS team_rank -- RANK() 함수를 사용하여 각 팀 내에서 쿼리 실행 횟수(query_cnt)에 따라 랭킹을 부여합니다. -- 'PARTITION BY week_number, team'은 팀과 주차별로 데이터를 나누어 그 안에서 랭킹을 계산합니다. -- 'ORDER BY query_cnt DESC'는 쿼리 실행 횟수에 따라 랭킹을 내림차순으로 정렬합니다. -- 이 결과로 각 팀 내에서 쿼리 실행 횟수에 따라 랭킹이 부여된 'team_rank'라는 새로운 열이 추가됩니다. FROM (SELECT EXTRACT(WEEK FROM query_date) AS week_number -- query_date에서 주차를 추출하여 'week_number' 열을 만듭니다. , team -- 팀을 나타내는 컬럼을 선택합니다. , user -- 사용자를 나타내는 컬럼을 선택합니다. , COUNT(user) AS query_cnt -- 사용자가 쿼리를 실행한 횟수를 집계하여 'query_cnt'라는 열로 만듭니다. FROM advanced.query_logs -- advanced 데이터셋의 query_logs 테이블에서 데이터를 가져옵니다. GROUP BY ALL) -- 모든 선택된 컬럼을 그룹화합니다. 여기서는 'week_number', 'team', 'user'로 데이터를 그룹화합니다. QUALIFY team_rank = 1 -- QUALIFY 절은 필터링 조건을 적용하는 역할을 하며, 여기서는 RANK가 1인 경우만 선택합니다. -- 즉, 각 팀 내에서 쿼리 실행 횟수가 가장 많은 사용자(1등)만 필터링됩니다. ORDER BY week_number, team; -- 주차별로, 그리고 각 주차 내에서 팀별로 데이터를 정렬합니다.    3번문제: (2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 기준 1주 전에 쿼리 실행 수를 별도의 컬럼으로 확인할 수 있는 쿼리를 작성해주세요.SELECT * -- 모든 열을 선택합니다. 서브쿼리에서 생성된 모든 컬럼을 포함합니다. , LAG(query_cnt) OVER (PARTITION BY user ORDER BY week_number) AS prev_week_query_count -- LAG() 함수를 사용하여 이전 주차(week)의 쿼리 실행 수를 가져옵니다. -- 'PARTITION BY user'를 통해 각 사용자별로 데이터를 나누고, 각 사용자에 대해 'week_number' 순서로 정렬하여 이전 주차의 값을 가져옵니다. -- 결과적으로, 이전 주차의 쿼리 실행 횟수를 'prev_week_query_count'라는 새로운 열로 추가합니다. FROM (SELECT user -- 쿼리를 실행한 사용자를 나타내는 열입니다. , team -- 사용자가 속한 팀을 나타내는 열입니다. , EXTRACT(WEEK FROM query_date) AS week_number -- query_date에서 주차(week)를 추출하여 'week_number'라는 열로 만듭니다. , COUNT(user) AS query_cnt -- 사용자가 실행한 쿼리 횟수를 집계하여 'query_cnt' 열로 만듭니다. FROM advanced.query_logs -- advanced 데이터셋의 query_logs 테이블에서 데이터를 가져옵니다. GROUP BY ALL) -- 선택된 모든 열을 그룹화합니다. ORDER BY user, week_number; -- 최종 결과를 사용자별로 정렬하고, 그다음 각 사용자 내에서 주차별로 정렬합니다.  4번문제:시간의 흐름에 따라, 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요.SELECT * -- 모든 열을 선택합니다. 서브쿼리에서 생성된 모든 컬럼을 포함합니다. , SUM(query_cnt) OVER (PARTITION BY user ORDER BY query_date) AS cumulative_query_count -- SUM 윈도우 함수를 사용하여 누적 쿼리 실행 수를 계산합니다. -- 'PARTITION BY user'를 통해 각 사용자별로 데이터를 나누고, 각 사용자에 대해 'query_date' 순서로 정렬하여 누적 값을 계산합니다. -- 결과적으로, 각 사용자에 대한 누적 쿼리 실행 횟수가 'cumulative_query_count'라는 새로운 열로 추가됩니다. FROM (SELECT user -- 쿼리를 실행한 사용자를 나타내는 열입니다. , team -- 사용자가 속한 팀을 나타내는 열입니다. , query_date -- 쿼리가 실행된 날짜를 나타내는 열입니다. , COUNT(user) AS query_cnt -- 사용자가 쿼리를 실행한 횟수를 집계하여 'query_cnt' 열로 만듭니다. FROM advanced.query_logs -- advanced 데이터셋의 query_logs 테이블에서 데이터를 가져옵니다. GROUP BY ALL) -- 선택된 모든 열을 그룹화하여 중복되지 않게 집계합니다. ORDER BY user, query_date; -- 최종 결과를 사용자별로 정렬하고, 그다음 각 사용자의 쿼리 실행 날짜별로 정렬합니다.  5번문제: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 ( -- 날짜와 해당 날짜의 주문 수를 포함한 임시 테이블을 생성합니다. 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 ) -- 임시 데이터를 활용하여 쿼리의 기능을 확인하기 위한 서브쿼리입니다. SELECT * -- 모든 열을 선택합니다 (날짜와 주문 수). , LAST_VALUE(number_of_orders IGNORE NULLS) OVER (ORDER BY date) AS number_of_orders -- NULL 값을 제외하고 마지막으로 발견된 'number_of_orders' 값을 가져옵니다. -- 'ORDER BY date'를 사용하여 날짜 순서대로 정렬하며, -- 각 행에서 가장 최근의 유효한 (NULL이 아닌) 'number_of_orders' 값을 반환합니다. FROM raw_data -- 데이터가 저장된 'raw_data' 임시 테이블에서 데이터를 조회합니다. ORDER BY date; -- 최종 결과를 날짜 순서대로 정렬하여 출력합니다.   6번문제: 5번 문제에서 NULL을 채운 후, 2일 전 ~ 현재 데이터의 평균을 구하는 쿼리를 작성해주세요.(이동 평균)WITH raw_data AS ( -- 임시 데이터를 정의합니다. 날짜(date)와 해당 날짜의 주문 수(number_of_orders)를 포함한 테이블을 생성합니다. 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 ) -- 누락된 데이터를 보완하고, 이동 평균을 계산하는 쿼리를 작성합니다. SELECT * -- 모든 열을 선택한 뒤, 이동 평균(moving_avg)을 추가로 계산합니다. , ROUND(AVG(number_of_orders) OVER (ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW), 1) AS moving_avg -- AVG 함수를 사용하여 현재 행을 포함해 이전 두 행의 값에 대한 이동 평균을 계산합니다. -- 계산된 결과를 소수점 첫 번째 자리까지 반올림합니다. FROM ( SELECT date, -- IFNULL 함수를 사용하여 NULL 값을 보완합니다. -- 이전 날짜의 주문 수를 가져와 현재 행의 주문 수가 NULL인 경우 보완합니다. IFNULL(number_of_orders, LAG(number_of_orders) OVER (ORDER BY date)) AS number_of_orders FROM raw_data ) -- 최종 결과를 날짜 순서로 정렬하여 출력합니다. ORDER BY date;   7번문제: 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 ( -- Step 1. 기초 데이터 추출 및 이전 이벤트 시간 계산 SELECT event_date, -- 이벤트가 발생한 날짜 event_timestamp, -- 마이크로초 단위의 이벤트 발생 시각 DATETIME(TIMESTAMP_MICROS(event_timestamp), 'Asia/Seoul') AS event_datetime, -- 이벤트 발생 시각을 'Asia/Seoul' 시간대로 변환하여 가독성을 높임 event_name, -- 이벤트의 이름 user_id, -- 사용자 ID user_pseudo_id, -- 사용자 고유의 익명화된 ID DATETIME(TIMESTAMP_MICROS(LAG(event_timestamp) OVER (PARTITION BY user_pseudo_id ORDER BY event_timestamp)), 'Asia/Seoul') AS before_event_datetime -- 이전 이벤트의 발생 시각을 'Asia/Seoul' 시간대로 변환 -- LAG() 함수를 사용해 각 사용자의 이전 이벤트 발생 시각을 가져옴 FROM advanced.app_logs WHERE event_date = '2022-08-18' -- 특정 날짜의 이벤트만 필터링 AND user_pseudo_id = '1997494153.8491999091' -- 특정 사용자에 대한 로그만 조회 ), -- Step 2. 세션 유지 시간 및 신규 세션 여부 계산 session_info AS ( SELECT *, TIMESTAMP_DIFF(event_datetime, before_event_datetime, SECOND) AS second_diff, -- 이전 이벤트와 현재 이벤트 사이의 시간 차이를 초 단위로 계산 CASE WHEN TIMESTAMP_DIFF(event_datetime, before_event_datetime, SECOND) >= 20 OR TIMESTAMP_DIFF(event_datetime, before_event_datetime, SECOND) IS NULL THEN 1 -- 두 이벤트 사이의 시간 차이가 20초 이상이거나 이전 이벤트가 없는 경우 새로운 세션 시작으로 간주 ELSE NULL END AS session_start -- 새로운 세션이 시작되면 1, 그렇지 않으면 NULL FROM base ) -- Step 3. 신규 세션 ID 세팅 SELECT *, SUM(session_start) OVER (PARTITION BY user_pseudo_id ORDER BY event_datetime) AS session_id -- 각 사용자의 이벤트 시퀀스에서 새로운 세션이 시작될 때마다 session_start 값을 누적하여 세션 ID를 부여 FROM session_info ORDER BY event_date, event_timestamp; -- 이벤트 발생 시각을 기준으로 정렬하여 결과를 출력    

문제 1번-- 1. 사용자별 쿼리를 실행한 횟수의 총합 -- GROUP BY 사용 X. 우측에 새로운 컬럼 생성하기. SELECT 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 2번-- 2. 주차별로 팀내에서 쿼리를 많이 실행한 수 → 랭킹 구하기 -- 팀별로 랭킹이 1위인 사람만 출력하기 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 3번-- 3. (2번 문제에서 사용한 주차별 쿼리 사용) 쿼리를 실행한 시점 -- 1주 전에 쿼리를 실행한 횟수를 별도의 컬럼으로 출력 WITH 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 4번-- 4. 일자별 유저가 쿼리한 횟수 누적합 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 5번-- 주문 횟수 테이블: 데이터 없으면 NULL로 표기됨. -- 5. NULL값을 바로 전달 데이터로 채우기 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 6번-- 6. (5번 완료 후) 2일 전 ~ 현재 데이터의 평균 구하기 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 date , LAST_VALUE(number_of_orders ignore nulls) OVER (ORDER BY date) AS number_of_orders FROM raw_data ) 문제 7번-- 7. app_logs 테이블에서 커스텀 세션 추가 -- 이전 이벤트 로그와 20초 이상 차이가 나면 새로운 세션으로 정의 -- 세션은 숫자로(1,2,3...) 표시 가능 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

안녕하세요, 강사님.다름이 아니라 스크롤라 반복이 되지 않아서 질문 드립니다.알려주신대로 펑션으로 스크롤라를 호출하고 인덱스 파일에 애니메이트도 선언했습니다만, 처음에만 애니메이션이 작동하고 스크롤을 내렸다 올릴 땐 작동하지 않습니다.스크롤을 초기화 해야 되는걸까요?챕터마다 넣어주신 캡쳐 파일에선 저랑 다른 부분을 찾기가 어려워 질문 남깁니다!    

문제 1.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_01문제2.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_01문제3.SELECT *, SUM(amount) OVER (PARTITION BY user_id ORDER BY order_date, order_id ) AS cumulative_sum_by_user문제 4.SELECT *, AVG(amount) OVER (ORDER BY order_date, order_id ROWS BETWEEN 5 PRECEDING AND 1 PRECEDING) AS last_5_orders_avg_amout 총정리 연습문제문제 1.SELECT *, COUNT(query_date) OVER (PARTITION BY user ORDER BY user) AS total_query_cnt FROM advanced.query_logs문제2.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 = 1문제3.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 table문제4.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_table문제5.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_data문제6.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_data문제 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_data AS( SELECT *, #이전 이벤트 시간과 현재 이벤트시간의 간격을 SECOND 초단위로 구하기 / second_diff를 기반으로 새로운 세션의 시작일지 아닐지 판단 DATETIME_DIFF(event_datetime, prev_event_datetime, SECOND) AS second_diff FROM ( SELECT *, # 직전 이벤트 시간을 prev_event_datetime으로 만들기 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

2주차윈도우 함수 - 탐색 함수프로덕트 측면에서 윈도움 탐색 함수를 활용해서 할 수 있는 것?어떤 user가 앱/웹에 접속한 후, 어떤 화면으로 이동했는지를 알 수 있음. 즉 시간의 흐름에 따른 행동을 볼 수 있음(다음 ROW의 Page를 확인 가능)엡 로그 상에서 직전 이벤트와 현재 이벤트가 동일한 것들을 필터링할 수 있음(LEAD한 값과 기존 컬럼과의 비교). 같은 Page를 연속으로 접근한 경우 하나로 처리해서 ‘퍼널’을 구할 때 활용할 수 있음.리텐션 쿼리를 작성할때 기준점을 만들 수 있음(유저의 첫 접속일, FIRST_VALUE 이용)2. 윈도우 함수의 번호 지정 함수코드 작성(윈도우 함수 Frame)# 윈도우 함수 Frame 연습 문제 SELECT *, SUM(amount) OVER() AS amount_total, # over안에 아무것도 들어가지 않는 경우도 있음!, 전체의 SUM. 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_orders_avg_amount FROM advanced.orders ORDER BY user_id 3. 윈도우 함수 연습 문제코드 작성# 문제 1) 사용자별 쿼리를 실행한 총 횟수를 구하는 쿼리 작성. # 단 , GROUP BY를 사용해서 집계하는 것이 아닌 데이터 우측에 새로운 컬럼 생성. SELECT *, COUNT(query_date) OVER (PARTITION BY user) AS total_query_cnt FROM advanced.query_logs order by user, query_date # 문제 2) 주차별로 팀 내에서 쿼리를 많이 실행한 수를 구한 후(먼저 집계 한 후) # 실행한 수를 활용해 랭킹을 구해주세요.(윈도우 함수 활용) # 단, 랭킹이 1등인 사람만 결과가 보이도록 해주세요. **문제의 의도 : 원본 데이터는 1 row 마다 데이터가 있고, 그걸 집계해서 사용(GROUP BY) => 그 다음에 윈도우 함수 사용.** with query_cnt_by_team AS ( SELECT EXTRACT(week from query_date) AS week_num, team, user, COUNT(query_date) AS query_cnt -- RANK() OVER(ORDER BY) FROM advanced.query_logs GROUP BY ALL ) SELECT *, RANK() OVER(PARTITION BY week_num, team ORDER BY query_cnt DESC) AS rk FROM query_cnt_by_team QUALIFY rk = 1 ORDER BY week_num, 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 작성. with query_cnt_by_team AS ( SELECT EXTRACT(week from query_date) AS week_num, team, user, COUNT(query_date) AS query_cnt -- RANK() OVER(ORDER BY) FROM advanced.query_logs GROUP BY ALL ) SELECT *, LAG(query_cnt,1) OVER(PARTITION BY user ORDER BY week_num) AS last_week_query_cnt FROM query_cnt_by_team # 문제 4) 시간의 흐름(query_date)에 따라 일자별로 유저가 실행한 누적 쿼리 수를 작성해주세요. Fram 관련 문제(누적합) 누적 쿼리 : 과거의 시간부터(UNBOUNDED PRECEDING)부터 현재까지(CURRENT ROWS) Frame의 Default 값 : UNBOUNDED PRECEDING ~ CURRENT ROWS => 출제 의도 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.query_logs GROUP BY ALL ) QUALIFY cumulative_sum != cumulative_sum2 ORDER BY user,query_date # 문제 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 작성. 출제 의도 : null을 제외하고 연산하고 싶을때는 IGONORE NULLS 활용!! 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) 5번 문제에서 NULL을 채운 후, 2일 전~ 현재 데이터의 평균을 구하는 쿼리를 작성(이동평균) 출제 의도 : Frame을 지정할 수 있는가? 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 *, LAST_VALUE(number_of_orders IGNORE nulls) OVER(ORDER BY date) AS number_of_order FROM raw_data ) SELECT *, AVG(number_of_order) OVER(ORDER BY date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS moving_avg FROM filled_date # 문제 7) app_logs 테이블에서 Custom Session을 만들어 주세요. # 이전 이벤트 로그와 20호가 지나면 새로운 Session을 만들어 주세요. # Session은 숫자로 (1,2,3,---) 표시해도 됩니다 # 2022-08-18일의 user_pseudo_id(1997494153.8491999091)은 sessio_id가 4까지 나옵니다. session을 구할 때 쿼리가 길어질 수 있음. 하루에 접속을 여러번 하는 서비스 => session 기반이 좋을 수 있고, 아니라고 하면 일자별 유저 집계가 나을 수 있음. - 세션 정리 - 이전 이벤트 로그와 현재 이벤트 로그의 diff => 초나 분을 구한다 - 그 기준을 가지고 기준보다 높으면 새로운 세션이라고 생각한다 - 첫번째 값엔 null이 있을 수 있어서, 이 부분도 챙겨야 한다.=> 1로 바꿔준다(Case 문 활용) - 새로운 세션, session_start 값을 기반으로 누적합 => session_num이 된다! 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_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 # event_datetime이랑 prev_datetime을 빼서 20초가 넘으면 새로운 세션으로 정의. # 20초가 넘지 않으면 기존 세션 / DATETIME_DIFF() => 차이를 구할 수 있음. 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 # session을 나누는 기준 초. 데이터 탐색하면서 결정. 보통 앱 록느에서는 30초, 60초 등으로 정함. ELSE 0 END AS session_start # 세션이 시작되됨을 알리는 session_start FROM diff_date )

이전 질문의 연장선 입니다. 말씀 주신대로 win XP랑 apmsetup을 주어진 실습 파일로 설치해도 connection timedout이 발생하여 질문 드립니다. 현재 환경은 VMware 17 Pro Kali 6.8.1 amd 64 / 10.10.10.7 WinXP Professional 입니다. /10.10.10.8  ping 명령어로 양방향 통신이 가능한걸 확인했습니다.  이전에설명대로 password: apmsetup, rhost: winXP ip 주소, username: root로 설정했지만 같은 에러가 발생하는데 따로 고려할 수 있는 부분이 있을까요?