BigQuery 缺少 SUM OVER PARTITION BY 的行

Question

特尔；博士：

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
)

如何获取包含缺失日期/产品组合 ( 2020-11-02 - premium ) 的表格，其中diff的后备值为0 。

理想情况下，适用于多种产品。 可以像这样获得所有产品的列表：

SELECT ARRAY_AGG(DISTINCT product) FROM subscriptions

---

我希望能够获得每天的订阅计数，无论是针对所有产品还是仅针对某些产品。

我认为这很容易实现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        |
|---------------------|------------------|------------------|

使用此表，我可以轻松地按日期和产品或仅按日期分组并汇总总数。

在我得到结果表之前，我已经生成了一个表，其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。

该表如下所示：

|---------------------|------------------|------------------|
|         date        |      product     |       diff       |
|---------------------|------------------|------------------|
|      2020-11-01     |      premium     |        50        |
|---------------------|------------------|------------------|
|      2020-11-01     |       basic      |       -20        |
|---------------------|------------------|------------------|

也就是说，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

---

我相信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？

Answer 1

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

Answer 2

如果我正确地跟随您，一种方法是可以生成您想要的期间的固定日期列表，并将其与产品列表cross join 。 这为您提供了所有可能的组合。 然后，你可以带一个left join的订阅表，最后执行窗口求和：

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

Answer 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