按顺序分组答案

【问题标题】：Group by rows which are in sequence按顺序分组
【发布时间】：2022-01-04 11:31:35
【问题描述】：

考虑一下我有一张这样的桌子

PASSENGER  CITY      DATE
43         NEW YORK  1-Jan-21
44         CHICAGO   4-Jan-21
43         NEW YORK  2-Jan-21
43         NEW YORK  3-Jan-21
44         ROME      5-Jan-21
43         LONDON    4-Jan-21
44         CHICAGO   6-Jan-21
44         CHICAGO   7-Jan-21

如何按顺序对“乘客”和“城市”列进行分组以获得如下结果？

PASSENGER  CITY      COUNT
43         NEW YORK  3
44         CHICAGO   1
44         ROME      1
43         LONDON    1
44         CHICAGO   2

【问题讨论】：

你的序列的order是如何定义的？关系表中没有显示顺序。

标签： sql oracle gaps-and-islands

【解决方案1】：

处理这种差距和孤岛问题的一种方法是计算差距的排名。

然后也按该排名分组。

SELECT PASSENGER, CITY
, COUNT(*) AS "Count" 
-- , MIN("DATE") AS StartDate
-- , MAX("DATE") AS EndDate
FROM (
  SELECT q1.*
  , SUM(gap) OVER (PARTITION BY PASSENGER ORDER BY "DATE") as Rnk
  FROM (
    SELECT PASSENGER, CITY, "DATE"
    , CASE
      WHEN 1 = TRUNC("DATE")
             - TRUNC(LAG("DATE") 
                     OVER (PARTITION BY PASSENGER, CITY ORDER BY "DATE")) 
      THEN 0 ELSE 1 END as gap
    FROM table_name t
  ) q1
) q2
GROUP BY PASSENGER, CITY, Rnk
ORDER BY MIN("DATE"), PASSENGER

PASSENGER	CITY	Count
43	NEW YORK	3
43	LONDON	1
44	CHICAGO	1
44	ROME	1
44	CHICAGO	2

db小提琴here

【讨论】：

嗨 Luk，如果我有时间戳而不是日期，你的方法会有什么变化？
你试过时间戳吗？你看，查询已经被截断为 '00:00:00' 以标记间隙。因此，只要它仍然是大约 1 个不同的日子，这可能并不重要。
是的，谷物不是 1 天，我也需要一天内的序列吗？
或者您可以尝试自己更改间隙的计算。 F.e.检查timestamp difference 是否不超过 24 小时。
我再问一个问题

【解决方案2】：

从 Oracle 12 开始，您可以使用MATCH_RECOGNIZE：

SELECT *
FROM   table_name
MATCH_RECOGNIZE (
  PARTITION BY passenger
  ORDER     BY "DATE"
  MEASURES
    FIRST(city) AS city,
    COUNT(*)    AS count
  PATTERN (same_city+)
  DEFINE
    same_city AS FIRST(city) = city
);

其中，对于样本数据：

CREATE TABLE table_name (PASSENGER, CITY, "DATE") AS
SELECT 43, 'NEW YORK',  DATE '2021-01-01' FROM DUAL UNION ALL
SELECT 44, 'CHICAGO',   DATE '2021-01-04' FROM DUAL UNION ALL
SELECT 43, 'NEW YORK',  DATE '2021-01-02' FROM DUAL UNION ALL
SELECT 43, 'NEW YORK',  DATE '2021-01-03' FROM DUAL UNION ALL
SELECT 44, 'ROME',      DATE '2021-01-05' FROM DUAL UNION ALL
SELECT 43, 'LONDON',    DATE '2021-01-04' FROM DUAL UNION ALL
SELECT 44, 'CHICAGO',   DATE '2021-01-06' FROM DUAL UNION ALL
SELECT 44, 'CHICAGO',   DATE '2021-01-07' FROM DUAL

输出：

PASSENGER CITY COUNT

43 NEW YORK 3

43 LONDON 1

44 CHICAGO 1

44 ROME 1

44 CHICAGO 2

如果你已经对输入结果集进行了排序（注意：表应该被认为是无序的）并且想要保持顺序那么：

SELECT *
FROM   (SELECT t.*, ROWNUM AS rn FROM table_name t)
MATCH_RECOGNIZE (
  PARTITION BY passenger
  ORDER     BY RN
  MEASURES
    FIRST(rn)     AS rn,
    FIRST("DATE") AS "DATE",
    FIRST(city)   AS city,
    COUNT(*)      AS count
  PATTERN (same_city+)
  DEFINE
    same_city AS FIRST(city) = city
)
ORDER BY rn

输出：

PASSENGER RN DATE CITY COUNT

43 1 01-JAN-21 NEW YORK 3

44 2 04-JAN-21 CHICAGO 1

44 5 05-JAN-21 ROME 1

43 6 04-JAN-21 LONDON 1

44 7 06-JAN-21 CHICAGO 2

PASSENGER	RN	DATE	CITY	COUNT
43	1	01-JAN-21	NEW YORK	3
44	2	04-JAN-21	CHICAGO	1
44	5	05-JAN-21	ROME	1
43	6	04-JAN-21	LONDON	1
44	7	06-JAN-21	CHICAGO	2

db小提琴here

【讨论】：