Я пытаюсь получить среднее число всех недель за каждый месяц. Как я могу сделать это в MySQL? Ниже приводится то, что у меня есть, а также то, что мне нужно:
SELECT
date_part(year, date_of_act) AS CALENDAR_YEAR,
date_part(week, date_of_act) AS CALENDAR_WEEK_OF_YEAR,
(num_active_divs / total_num_divs) AS "Weekly average for each year of week"
FROM
Infotable
GROUP BY calendar_year, calendar_week_of_year
ORDER BY calendar_year, calendar_week_of_year
What I have: What I have: What I have:
Calendar_Year Days of the year Calendar_Week_of_year Weekly average for each year of week
2018, Jan 01 - Jan 07, 1, 0.266855
2018, Jan 08 - Jan 14, 2, 0.297223
2018, Jan 15 - Jan 21, 3, 0.308583
2018, Jan 22 - Jan 28, 4, 0.309994
2018, Jan 29 - Feb 04, 5, 0.317419
2018, Feb 05 - Feb 11, 6, 0.316454
2018, Feb 12 - Feb 18, 7, 0.313929
2018, Feb 19 - Feb 25, 8, 0.315489
2018, Feb 26 - Mar 04, 9, 0.3218
2018, Mar 05 - Mar 11, 10, 0.308509
2018, Mar 12 - Mar 18, 11, 0.302866
2018, Mar 19 - Mar 25, 12, 0.31586
What I need: What I need: What I need: What I need:
Calendar_Year Calendar_Week_of_year Weekly average for each year of week Calendar_Month_of_year Week_of_month Average for all weeks in that month
2018, Jan 01 - Jan 07, 1, 0.266855, 1, 1
2018, Jan 08 - Jan 14, 2, 0.297223, 1, 2
2018, Jan 15 - Jan 21, 3, 0.308583, 1, 3
2018, Jan 22 - Jan 28, 4, 0.309994, 1, 4, (0.266855+0.297223+0.308583+0.309994)/4
2018, Jan 29 - Feb 04, 5, 0.317419, 2, 1
2018, Feb 05 - Feb 11, 6, 0.316454, 2, 2
2018, Feb 12 - Feb 18, 7, 0.313929, 2, 3
2018, Feb 19 - Feb 25, 8, 0.315489, 2, 4, (0.317419+0.316454+0.313929+0.315489)/4
2018, Feb 26 - Mar 04, 9, 0.32180, 3, 1
2018, Mar 05 - Mar 11, 10, 0.308509, 3, 2
2018, Mar 12 - Mar 18, 11, 0.302866, 3, 3
2018, Mar 19 - Mar 25, 12, 0.31586, 3, 4, (0.3218+0.308509+0.302866+0.31586)/4
Один из вариантов здесь - присоединиться к подзапросу, который находит среднемесячные значения:
SELECT
t1.*,
t2.avg_weekly_average
FROM yourTable t1
INNER JOIN
(
SELECT Calendar_Year, (Calendar_Week_of_year - 1) / 4 AS month,
AVG(weekly_average) AS avg_weekly_average
FROM yourTable
GROUP BYCalendar_Year, (Calendar_Week_of_year - 1) / 4
) t2
ON t1.Calendar_Year = t2.Calendar_Year AND
(t1.Calendar_Week_of_year - 1) / 4 = t2.month;
Если вы используете MySQL 8+, тогда вы можете использовать AVG
в качестве аналитической функции здесь. В этом случае запрос становится немного менее кратким:
SELECT *,
AVG(weekly_average) OVER (PARTITION BY Calendar_Year,
(Calendar_Week_of_year - 1) / 4) avg_weekly_average
FROM yourTable t1;
Calendar_Year Calendar_Week_of_year Weekly average for each year of week Average for all weeks in that month Expected value 2018 1 0.266637 0.204148429 0.29542225 2018 2 0.29698 0.23631892 0.29542225 2018 3 0.308331 0.238056842 0.29542225 2018 4 0.309741 0.246393239 0.29542225 2018 5 0.31716 0.259385717 0.315565 2018 6 0.316196 0.259694425 0.315565 2018 7 0.313673 0.261424683 0.315565 2018 8 0.315231 0.260448818 0.315565 2018 9 0.321537 0.262954257 0.31200375 2018 10 0.308257 0.263811621 0.31200375 2018 11 0.302619 0.25810876 0.31200375 2018 12 0.315602 0.264338671 0.312003