大查询比较两周的数据

Question

我有一个大查询表。像这样：

date(week)   vendor_name     value
2021-11-14   rick            8000
2021-11-14   rose            7000
2021-11-14   axel            6500
2021-11-14   boris           6000
2021-11-14   cliff           5500
2021-11-07   rose            9500
2021-11-07   axel            8750
2021-11-07   rick            4000
2021-11-07   dean            3500
2021-11-07   evan            3000
.....

日期栏表示每周的开始日期。 vendor_name 列显示了该周销量最高的 5 个供应商。值列显示总销售额。每周的顶级供应商可能会有所不同。预期输出：

date(week)  vendor_name      value   previous_date  previous_top_vendors  previous_value   change
2021-11-14   rick            8000    2021-11-07     rose                  9500             -%26
2021-11-14   rose            7000    2021-11-07     axel                  8750             -25
2021-11-14   axel            6500    2021-11-07     rick                  4000             %100
2021-11-14   boris           6000    2021-11-07     dean                  3500             Null
2021-11-14   cliff           5500    2021-11-07     evan                  3000             Null

收到最新日期后，会到上一周的数据。在这两周内，将比较卖家，他们的变化百分比将显示在名为“变化”的列中。

注意：供应商的排名可能每周都会发生变化。可以对新的供应商进行排名（在这种情况下，它应该在“更改”列中显示为 Null）

我试过了：

SELECT * FROM (select*from `table.top20_vendor` where date = '2021-11-14'),
(select date as previous_date, vendor_name as previous_top_vendors,value as previous_value 
from `table.top20_vendor` where date ='2021-11-07')

但输出不正确：

date(week)  vendor_name      value   previous_date  previous_top_vendors  previous_value
2021-11-14   rick            8000    2021-11-07     rose                  9500   
2021-11-14   rick            8000    2021-11-07     axel                  8750   
2021-11-14   rick            8000    2021-11-07     rick                  4000
2021-11-14   rick            8000    2021-11-07     dean                  3500             
2021-11-14   rick            8000    2021-11-07     evan                  3000             

另外，我不知道如何计算“变化”列

Answer 1

我首先做的是获取所需周的前 5 个值，我使用了以下查询：

select * from
(
select date, vendor_name, value , row_number() over(partition by date order by value desc) as rn
from `project.dataset.table`
)A
where rn<=5 and date='2021-11-14'

然后，需要做的是改变值。为此，您需要使用百分比公式，并按供应商的名称加入这几周的价值。

 SELECT prev_week.rn,((curr_week.value-prev_week.value)/prev_week.value)*100 as change FROM curr_week right join  prev_week on curr_week.vendor_name = prev_week.vendor_name

考虑以下方法：

with curr_week as(
select * from
(
select date, vendor_name, value , row_number() over(partition by date order by value desc) as rn
from `project.dataset.table`
)A
where rn<=5 and date='2021-11-14'
),
prev_week as (
   select * from
(
select date, vendor_name, value , row_number() over(partition by date order by value desc) as rn
from `project.dataset.table`
)A
where rn<=5 and date='2021-11-07'
),
changes as(
   SELECT prev_week.rn,((curr_week.value-prev_week.value)/prev_week.value)*100 as change FROM curr_week right join  prev_week on curr_week.vendor_name = prev_week.vendor_name
)
 
select curr_week.date,
curr_week.vendor_name,
curr_week.value,
prev_week.date as previous_date,
prev_week.vendor_name as previous_top_vendors,
prev_week.value as previous_value,
changes.change
from curr_week left join prev_week on curr_week.rn = prev_week.rn
left join changes on prev_week.rn = changes.rn

我得到的结果如下：

Answer 2

我仍然不明白您在比较哪些记录方面的更改逻辑，但这应该可以帮助您在几周之间比较供应商及其价值：

with sample_data as (
    select '2021-11-14' as week,   'rick' as vendor_name,            8000 as value UNION ALL 
    select '2021-11-14' as week,   'rose' as vendor_name,            7000 as value UNION ALL 
    select '2021-11-14' as week,   'axel' as vendor_name,            6500 as value UNION ALL 
    select '2021-11-14' as week,   'boris' as vendor_name,           6000 as value UNION ALL 
    select '2021-11-14' as week,   'cliff' as vendor_name,           5500 as value UNION ALL 
    select '2021-11-07' as week,   'rose' as vendor_name,            9500 as value UNION ALL 
    select '2021-11-07' as week,   'axel' as vendor_name,            8750 as value UNION ALL 
    select '2021-11-07' as week,   'rick' as vendor_name,            4000 as value UNION ALL 
    select '2021-11-07' as week,   'dean' as vendor_name,            3500 as value UNION ALL 
    select '2021-11-07' as week,   'evan' as vendor_name,            3000 as value
)
,
ranked_data as (
    select dense_rank() over (order by week desc) week_rank
    , rank() over (partition by week order by value) vendor_rank
    , *
from sample_data
)
select curr_week.week
    , curr_week.vendor_name
    , curr_week.value
    , prev_week.week as previous_week
    , prev_week.vendor_name as previous_top_vendors
    , prev_week.value as previous_value
    , 
from ranked_data curr_week
left join ranked_data prev_week
 on curr_week.vendor_rank=prev_week.vendor_rank
 and prev_week.week_rank=2
where curr_week.week_rank=1
order by 3 desc

最终，我认为您需要领先或落后，具体取决于中间步骤中的计算。如果您能详细说明一下，我也许可以添加该组件以进行更改。