Google bigquery 童车行为；“订购”；在BigQuery的数组_AGG函数中

我认为BigQuery中的

ARRAY\u AGG

函数在

ORDER BY

的行为中似乎存在缺陷。下面是一些SQL语句来解释：

#standardSQL
WITH t1 AS (
  SELECT *
  FROM UNNEST ( [
    STRUCT(1 AS user_id, 1 AS team_id, "2018-07-17" AS date_str),
    (  2, 1, "2018-07-17" ),
    (  3, 1, "2018-07-17" ),
    (  4, 1, "2018-07-17" ),
    (  5, 1, "2018-07-17" ),
    (  6, 1, "2018-07-17" ),
    (  7, 1, "2018-07-17" ),
    (  8, 2, "2018-07-17" ),
    (  9, 2, "2018-07-17" ),
    ( 10, 2, "2018-07-17" ),
    ( 11, 2, "2018-07-17" ),
    ( 14, 3, "2018-07-17" ),
    ( 15, 3, "2018-07-17" ),
    ( 16, 3, "2018-07-17" ),
    ( 17, 3, "2018-07-17" ),
    (  1, 1, "2018-07-18" ),
    (  4, 1, "2018-07-18" ),
    (  5, 1, "2018-07-18" ),
    (  6, 1, "2018-07-18" ),
    (  7, 1, "2018-07-18" ),
    (  8, 2, "2018-07-18" ),
    (  9, 2, "2018-07-18" ),
    ( 10, 2, "2018-07-18" ),
    ( 11, 2, "2018-07-18" ),
    ( 12, 2, "2018-07-18" ),
    ( 13, 2, "2018-07-18" ),
    ( 14, 3, "2018-07-18" ),
    ( 15, 3, "2018-07-18" ),
    ( 16, 3, "2018-07-18" ),
    ( 17, 3, "2018-07-18" ),
    ( 18, 3, "2018-07-18" ) ] ) )

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM (
  SELECT
    date_str,
    users,
    COUNT(*) AS teams
  FROM (
    SELECT
      date_str,
      team_id,
      COUNT(*) AS users
    FROM t1
    GROUP BY date_str, team_id
  )
  GROUP BY date_str, users
)
GROUP BY date_str
ORDER BY date_str;

此查询返回

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  1 |  4 |  1 |  4 |        2 |        4 |
|     |            |  2 |  7 |  2 |  7 |        1 |        7 |
|   2 | 2018-07-18 |  1 |  5 |  1 |  5 |        2 |        5 |
|     |            |  2 |  6 |  2 |  6 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

但我期待的是

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  2 |  4 |  1 |  7 |        2 |        4 |
|     |            |  1 |  7 |  2 |  4 |        1 |        7 |
|   2 | 2018-07-18 |  2 |  5 |  1 |  6 |        2 |        5 |
|     |            |  1 |  6 |  2 |  5 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

似乎

ARRAY\u AGG

函数中的

ORDER BY

子句工作不正常，因为

a1

和

a4

的顺序错误

此外，当我用

COUNT（user\u id）

或

COUNT（team\u id）

替换两个

COUNT（*）

部分时，很难想象查询会完全按照预期工作，这意味着

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM (
  SELECT
    date_str,
    users,
    COUNT(*) AS teams
  FROM (
    SELECT
      date_str,
      team_id,
      COUNT(user_id) AS users
    FROM t1
    GROUP BY date_str, team_id
  )
  GROUP BY date_str, users
)
GROUP BY date_str
ORDER BY date_str;

或

据我所知，在这种情况下，这些查询必须返回与原始查询相同的结果。这让我很困惑。可能是虫子或是我误解了什么

---

一些补充资料

内部子查询

SELECT
  date_str,
  users,
  COUNT(*) AS teams
FROM (
  SELECT
    date_str,
    team_id,
    COUNT(*) AS users
  FROM t1
  GROUP BY date_str, team_id
)
GROUP BY date_str, users

这种回归

+-----+------------+-------+-------+
| Row | date_str   | users | teams |
+-----+------------+-------+-------+
|   1 | 2018-07-18 |     5 |     2 |
|   2 | 2018-07-17 |     7 |     1 |
|   3 | 2018-07-18 |     6 |     1 |
|   4 | 2018-07-17 |     4 |     2 |
+-----+------------+-------+-------+

因此，通过with子句直接创建此数据并运行相同的聚合查询

#standardSQL
With t2 AS (
  SELECT *
  FROM UNNEST ( [
    STRUCT("2018-07-18" AS date_str, 5 AS users, 2 AS teams),
    (  "2018-07-17", 7, 1 ),
    (  "2018-07-18", 6, 1 ),
    (  "2018-07-17", 4, 2 ) ] )
)

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM t2
GROUP BY date_str
ORDER BY date_str;

结果成了我所寻找的

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  2 |  4 |  1 |  7 |        2 |        4 |
|     |            |  1 |  7 |  2 |  4 |        1 |        7 |
|   2 | 2018-07-18 |  2 |  5 |  1 |  6 |        2 |        5 |
|     |            |  1 |  6 |  2 |  5 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

我不明白这是什么原因造成的。我完全困惑不解。

---

任何想法或建议都将不胜感激。

如果我有误解，很抱歉，但默认的顺序是升序，因此排序正确吗

ARRAY_AGG(teams ORDER BY users desc) AS a1,
ARRAY_AGG(users ORDER BY users) AS a2,
ARRAY_AGG(teams ORDER BY teams) AS a3,
ARRAY_AGG(users ORDER BY teams desc) AS a4, 

---

如果我把它们改为降序排序，我会得到想要的结果

谢谢你的回答。是的，这个例子本身有点棘手和混乱。让我们以

a1

字段为例。它是一个由

团队

按

用户

排序的数组<代码>团队和

相同日期的用户

在本例中顺序相反，这在

a5

字段中得到了澄清。因此，当您按

用户向上排序时，
a1
中的
团队将向下排序。希望这个解释对你有意义。
ARRAY_AGG(teams ORDER BY users desc) AS a1,
ARRAY_AGG(users ORDER BY users) AS a2,
ARRAY_AGG(teams ORDER BY teams) AS a3,
ARRAY_AGG(users ORDER BY teams desc) AS a4,