如何计算每小时的比率？

Question

在执行查询时，我对自己的进一步操作有点理解。

我有两个表 "A"(date, response, b_id) 和 "B"(id, country)。我需要计算存在响应的条目数量与特定日期的条目总数的每小时比率。最终选择应包括“小时”、“比率”列。

SELECT COUNT(*) FROM A WHERE RESPONSE IS NOT NULL//counting entries with response
SELECT COUNT(*) FROM A//counting total number of entries

1. 如何计算比率？我应该为它创建一个单独的变量吗？
2. 如何计算一天中的每个小时？我应该像循环一样吗？ + 如何获取日期的“小时”部分？
3. 选择小时数和计数比率的最佳方法是什么？我应该为它单独制作一张桌子吗？

对于复杂的查询我比较陌生，所以我很乐意得到各种帮助

Answer 1

你可以这样做：

select to_char(datecol, 'HH24') as hour,
       count(response) as has_response, count(*) as total,
       count(response) / count(*) as ratio
from a
where datecol >= date '2018-09-18' and datecol < date '2018-09-19'
group by to_char(datecol, 'HH24');

您也可以使用avg() 执行此操作——这也很有趣：

select to_char(datecol, 'HH24'),
       avg(case when response is not null then 1.0 else 0 end) as ratio
from a
where datecol >= date '2018-09-18' and datecol < date '2018-09-19'
group by to_char(datecol, 'HH24')

不过，在这种情况下，这需要更多的输入。

Answer 2

SQL Fiddle

Oracle 11g R2 架构设置：

CREATE TABLE A ( dt, response, b_id ) AS
  SELECT DATE '2018-09-18' + INTERVAL '00:00' HOUR TO MINUTE, NULL, 1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '00:10' HOUR TO MINUTE, 'A',  1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '00:20' HOUR TO MINUTE, 'B',  1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '01:00' HOUR TO MINUTE, 'C',  1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '01:10' HOUR TO MINUTE, 'D',  1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '02:00' HOUR TO MINUTE, NULL, 1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '03:00' HOUR TO MINUTE, 'E',  1 FROM DUAL UNION ALL
  SELECT DATE '2018-09-18' + INTERVAL '05:10' HOUR TO MINUTE, 'F',  1 FROM DUAL;

查询 1：

SELECT b_id,
       TO_CHAR( TRUNC( dt, 'HH' ), 'YYYY-MM-DD HH24:MI:SS' ) AS hour,
       COUNT(RESPONSE) AS total_response_per_hour,
       COUNT(*)        AS total_per_hour,
       total_response_per_day,
       total_per_day,
       COUNT(response) / total_response_per_day AS ratio_for_responses,
       COUNT(*) / total_per_day AS ratio
FROM   (
  SELECT A.*,
         COUNT(RESPONSE) OVER ( PARTITION BY b_id, TRUNC( dt ) ) AS total_response_per_day,
         COUNT(*) OVER ( PARTITION BY b_id, TRUNC( dt ) ) AS total_per_day
  FROM   A
)
GROUP BY
       b_id,
       total_per_day,
       total_response_per_day,
       TRUNC( dt, 'HH' )
ORDER BY
       TRUNC( dt, 'HH' )

Results：

| B_ID |                HOUR | TOTAL_RESPONSE_PER_HOUR | TOTAL_PER_HOUR | TOTAL_RESPONSE_PER_DAY | TOTAL_PER_DAY | RATIO_FOR_RESPONSES | RATIO |
|------|---------------------|-------------------------|----------------|------------------------|---------------|---------------------|-------|
|    1 | 2018-09-18 00:00:00 |                       2 |              3 |                      6 |             8 |  0.3333333333333333 | 0.375 |
|    1 | 2018-09-18 01:00:00 |                       2 |              2 |                      6 |             8 |  0.3333333333333333 |  0.25 |
|    1 | 2018-09-18 02:00:00 |                       0 |              1 |                      6 |             8 |                   0 | 0.125 |
|    1 | 2018-09-18 03:00:00 |                       1 |              1 |                      6 |             8 | 0.16666666666666666 | 0.125 |
|    1 | 2018-09-18 05:00:00 |                       1 |              1 |                      6 |             8 | 0.16666666666666666 | 0.125 |

Answer 3

   SELECT withResponses.hour,
       withResponses.cnt AS withResponse,
       alls.cnt AS AllEntries,
       (withResponses.cnt / alls.cnt) AS ratio
FROM
  ( SELECT to_char(d, 'DD-MM-YY - HH24') || ':00 to :59 ' hour,
           count(*) AS cnt
   FROM A
   WHERE RESPONSE IS NOT NULL
   GROUP BY to_char(d, 'DD-MM-YY - HH24') || ':00 to :59 ' ) withResponses,

  ( SELECT to_char(d, 'DD-MM-YY - HH24') || ':00 to :59 ' hour,
           count(*) AS cnt
   FROM A
   GROUP BY to_char(d, 'DD-MM-YY - HH24') || ':00 to :59 ' ) alls
WHERE alls.hour = withResponses.hour ;

SQLFiddle：http://sqlfiddle.com/#!4/c09b9/2