将包含日期的列按月拆分为多列。 SQL

Question

我正在尝试根据月份验证登录的用户。 SQL

当前df

user id             date
P1302              2023-11-01
P1302              2023-10-01
P1302              2023-09-01
P1302              2023-08-01
P1302              2023-07-01
P1302              2023-06-01
P1301              2023-11-01
P1301              2023-10-01
P1301              2023-08-01
P1301              2023-07-01
P1301              2023-06-01

预期结果

user id     Jun     Jul     aug   sep   oct     nov
1302         1       1       1     1     1       1
1301         1       1       1     0     1       1

Answer 1

您可以通过条件聚合来做到这一点：

select
    user_id,
    -- extract(year from date) as year,
    max(extract(month from date) =  1) as jan,
    max(extract(month from date) = 12) as dec
from t
group by user_id --, extract(year from date)

Answer 2

解决问题的一种常见方法是对 case 表达式使用 count：

select count(case month(dt) = 1 then 1 end) as jan
     , count(case month(dt) = 2 then 1 end) as feb
     , ...
from ...