题解 | #异常的邮件概率#
异常的邮件概率
https://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select
date,
p
from
(
select
id,
send_id,
round(
(
sum(if (type = 'completed', 0, 1)) over (
partition by
date
order by
date
)
) / (
count(type) over (
partition by
date
order by
date
)
),
3
) as p,
date
from
email
where
send_id in (
select
id
from
user
where
is_blacklist = 0
)
and receive_id in (
select
id
from
user
where
is_blacklist = 0
)
) a
group by
date,p
