BigQuery 缺少 SUM OVER PARTITION BY 的行
Posted
技术标签:
【中文标题】BigQuery 缺少 SUM OVER PARTITION BY 的行【英文标题】：BigQuery missing rows with SUM OVER PARTITION BY 【发布时间】：2021-02-19 10:06:57 【问题描述】：TL;DR：
鉴于此表：
WITH subscriptions AS (SELECT TIMESTAMP("2020-11-01") as date, "premium" as product, 50 as diff
UNION ALL SELECT TIMESTAMP("2020-11-01"), "basic", 100
UNION ALL SELECT TIMESTAMP("2020-11-02"), "basic", -10
UNION ALL SELECT TIMESTAMP("2020-11-03"), "premium", 20
UNION ALL SELECT TIMESTAMP("2020-11-03"), "basic", 40
)
如何获得一个表格，其中包含缺少的日期/产品组合 (2020-11-02 - premium) 以及 diff 的后备值 0。
理想情况下，适用于多种产品。所有产品的列表可以这样得到：
SELECT ARRAY_AGG(DISTINCT product) FROM subscriptions
我希望能够获取所有产品或某些产品的每日订阅数。
我认为可以轻松实现这一点的方法是准备一个如下所示的数据库：
|---------------------|------------------|------------------|
| date | product | total |
|---------------------|------------------|------------------|
| 2020-11-01 | premium | 100 |
|---------------------|------------------|------------------|
| 2020-11-01 | basic | 50 |
|---------------------|------------------|------------------|
有了这张表，我可以很容易地按日期和产品分组，或者只按日期和总和。
在获得结果表之前，我已经生成了一个表，在该表中我计算了每天和产品的订阅差异。每个产品有多少新订阅者，有多少不再订阅。
此表如下所示：
|---------------------|------------------|------------------|
| date | product | diff |
|---------------------|------------------|------------------|
| 2020-11-01 | premium | 50 |
|---------------------|------------------|------------------|
| 2020-11-01 | basic | -20 |
|---------------------|------------------|------------------|
意味着11月1日高级用户总数增加了50个，基本用户总数减少了20个。
现在的问题是，如果一个产品没有任何更改，则此临时表缺少日期点，请参见下面的示例。
当我开始时没有产品表，我只有日期和差异列。
为了从第二个表到第一个表，我使用了这个完美的查询：
WITH subscriptions AS (SELECT TIMESTAMP("2020-11-01") as date, 150 as diff
UNION ALL SELECT TIMESTAMP("2020-11-02"), -10
UNION ALL SELECT TIMESTAMP("2020-11-03"), 60
)
SELECT
*,
SUM(diff) OVER (ORDER BY date) as total_subscriptions
FROM subscriptions
ORDER BY date
但是当我添加产品列并尝试计算每天和产品的总和时，缺少一些数据点。
WITH subscriptions AS (SELECT TIMESTAMP("2020-11-01") as date, "premium" as product, 50 as diff
UNION ALL SELECT TIMESTAMP("2020-11-01"), "basic", 100
UNION ALL SELECT TIMESTAMP("2020-11-02"), "basic", -10
UNION ALL SELECT TIMESTAMP("2020-11-03"), "premium", 20
UNION ALL SELECT TIMESTAMP("2020-11-03"), "basic", 40
)
SELECT
*,
SUM(diff) OVER (PARTITION BY product ORDER BY date) as total_subscriptions
FROM subscriptions
ORDER BY date
--
|---------------------|------------------|------------------|
| date | product | total |
|---------------------|------------------|------------------|
| 2020-11-01 | basic | 100 |
|---------------------|------------------|------------------|
| 2020-11-01 | premium | 50 |
|---------------------|------------------|------------------|
| 2020-11-02 | basic | 90 |
|---------------------|------------------|------------------|
| 2020-11-03 | basic | 130 |
|---------------------|------------------|------------------|
| 2020-11-03 | premium | 70 |
|---------------------|------------------|------------------|
如果我现在显示每天的订阅总数，我会得到：
150 -> 90 -> 200
但我希望：
150 -> 140 -> 200
每天的高级订阅总数也是如此：
50 -> 0 -> 70
但我希望：
50 -> 50 -> 70
我认为解决此问题的最佳选择是添加缺少的日期/产品组合。
我该怎么做？
【问题讨论】：
请编辑您的问题并显示您想要的结果。 预期输出 - 请澄清！ 【参考方案1】：如果我没听错的话，一种方法是生成一个固定的日期列表，用于您想要的时间段，然后cross join 与产品列表一起生成。这为您提供了所有可能的组合。然后，可以带left join的订阅表，最后进行窗口求和：
select d.dt, p.product, sum(s.diff) over(partition by p.product order by d.dt) total
from unnest(generate_timestamp_array(
timestamp('2020-11-01'),
timestamp('2020-11-03'),
interval 1 day)
) dt
cross join (
select 'basic' product
union all select 'premium'
) p
left join subscriptions on s.product = p.product and s.date = dt
我们可以通过动态生成日期范围和产品列表来使查询更通用：
select d.dt, p.product, sum(s.diff) over(partition by p.product order by d.dt) total
from (select min(date) min_dt, max(date) max_dt from subscriptions) d0
cross join unnest(generate_timestamp_array(d0.min_dt, d0.max_dt, interval 1 day)) dt
cross join (select distinct product from subscriptions) p
left join subscriptions on s.product = p.product and s.date = dt
【讨论】：
【参考方案2】：使用GENERATE_TIMESTAMP_ARRAY:
WITH subscriptions AS (SELECT TIMESTAMP("2020-11-01") as date, "premium" as product, 50 as diff
UNION ALL SELECT TIMESTAMP("2020-11-01"), "basic", 100
UNION ALL SELECT TIMESTAMP("2020-11-02"), "basic", -10
UNION ALL SELECT TIMESTAMP("2020-11-03"), "premium", 20
UNION ALL SELECT TIMESTAMP("2020-11-03"), "basic", 40
),
dates AS (
SELECT *
FROM UNNEST(GENERATE_TIMESTAMP_ARRAY('2020-11-01 00:00:00', '2020-11-03 00:00:00', INTERVAL 1 DAY)) as date
),
products AS (
SELECT DISTINCT product FROM subscriptions
)
SELECT dates.date, products.product, subscriptions.diff
FROM dates
CROSS JOIN products
LEFT JOIN subscriptions
ON subscriptions.date = dates.date AND subscriptions.product = products.product
【讨论】：
【参考方案3】： -- Try this,I am creating a table for list of products and add total product in that list. Joining with your table to get data as per your requirement.
WITH subscriptions AS (SELECT TIMESTAMP("2020-11-01") as date, "premium" as product, 50 as diff
UNION ALL SELECT TIMESTAMP("2020-11-01"), "basic", 100
UNION ALL SELECT TIMESTAMP("2020-11-02"), "basic", -10
UNION ALL SELECT TIMESTAMP("2020-11-03"), "premium", 20
UNION ALL SELECT TIMESTAMP("2020-11-03"), "basic", 40
),
product_name as (
Select product from subscriptions group by 1
union all
Select "Total" as product
)
Select date
,product
,total_subscriptions
from (
Select a.date
,a.product
,diff
,SUM(diff) OVER (PARTITION BY a.product ORDER BY a.date) as total_subscriptions
from
(
Select date,a.product
from product_name A
join subscriptions B
on 1=1
where a.product !='Total'
group by 1,2
) A
left join subscriptions B
on A.product = B.product
and A.date = B.date
group by 1,2,3
) group by 1,2,3
union all
Select date
,product
,total_subscriptions
from
(
Select date,a.product
,diff
,SUM(diff) OVER (PARTITION BY a.product ORDER BY date) as total_subscriptions
from product_name A
join subscriptions B
on 1=1
where a.product ='Total'
group by 1,2,3
) group by 1,2,3
order by 1,2
【讨论】：
以上是关于BigQuery 缺少 SUM OVER PARTITION BY 的行的主要内容，如果未能解决你的问题，请参考以下文章
优化 sum() over(order by...) 子句抛出“超出资源”错误