sql求每家店铺销量前三的sku, 附python解法

背景

有一张表：

date	store_id	sku	sales
2023-01-01	CK005	03045	50

date 代表交易日期，store_id代表门店编号，sku代表商品，sales代表销量。

问题

求出2022年每家店销量前3的sku。

生成的表为4列，第一列为store_id，第二至第四列依次为销量排名前三的sku，如果某家门店只卖两个sku，即没有销量排名第三的sku，那么那个单元格用null值代替

sql解法

select `store_id`,
max(`SKU_1`) as `TOP_SKU_1`,
max(`SKU_2`) AS `TOP_SKU_2`,
max(`SKU_3`) AS `TOP_SKU_3` from (
select `store_id`,
IF(`rank_num` = 1, `sku`, NULL) AS `SKU_1`,
IF(`rank_num` = 2, `sku`, NULL) AS `SKU_2`,
IF(`rank_num` = 3, `sku`, NULL) AS `SKU_3`
from
(select *, row_number() over (partition by `store_id` order by `sales` desc)  as `rank_num`
from (
SELECT `store_id`, `sku`, sum(`qty`) as `sales` FROM input
where year(`date`) = 2022
group by `store_id`, `sku`))
where `rank_num` <= 3 )
group by `store_id`

结果校验：

其中：

排序然后case或者if之后是这样的：

select *
from
(select *, row_number() over (partition by `store_id` order by `sales` desc)  as `rank_num`
from (
SELECT `store_id`, `sku`, sum(`qty`) as `sales` FROM input
where year(`date`) = 2022
group by `store_id`, `sku`))
where `rank_num` <= 3

然后再加一个group by的操作，用聚合函数max取出每一组的值。

Python解法

1. 求出销量前三的sku

df = hist_month.query('year == 2022').groupby(['year','stock_id','sku'],as_index = False).agg({'y':sum})
df['y_rank']  = df.groupby("stock_id")["y"].rank( ascending=False)

top3_df = df.query('y_rank <= 3')
top3_df

temp = top3_df.set_index(['year','stock_id','y_rank'])['sku'].unstack()
temp = temp.rename_axis(columns=None).reset_index()