统计一段时间内有条件的人数

Question

我们有一个 table 包含课程中用户的信息。

CREATE TABLE public.students
(
    student_id integer NOT NULL DEFAULT nextval('students_student_id_seq'::regclass),
    timest timestamp without time zone NOT NULL   ##
    is_correct BOOLEAN NOT NULL,
    CONSTRAINT students_pkey PRIMARY KEY (student_id)
)

一名 'successful' 学生在当月至少有一次在一小时内正确完成了 10 次练习。帮助查询，该查询将提供有关 2020 年 10 月成功学生人数的信息。

试图找到一个小时内连续解决的练习数。为此，我计算了减去3600的交换和。

WITH timediff AS (
     SELECT *,
         LEAD(timest) OVER w AS next_time,
         LEAD(timest) OVER w - timest AS diff
     FROM payment
     WHERE is_correct IS TRUE
     WINDOW w AS (PARTITION  BY student_id ORDER BY timest 
     ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING))

SELECT *,
    CASE  
    WHEN (sum(diff) OVER q) > '3600' 
    THEN diff 
    ELSE (sum(diff) OWER q) 
    END cum_sum
FROM timediff
WINDOW q AS (PARTITION  by student_id ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)

那我就不知道怎么算连续的经期了。我认为这不是最好的选择。请帮助处理您的请求。

Answer 1

A 'successful' student who at least once in the current month correctly done 10 exercises in an hour.

如果我没听错，你可以使用 window 函数和范围框：

select count(distinct student_id) cnt_successful_students
from (
    select s.*,
        count(*) filter(where is_correct) over(
            partition by student_id 
            order by timest
            range between interval '1 hour' preceding and current row
        ) cnt
    from students s
    where timest >= date_trunc('month', current_date) 
      and timest <  date_trunc('month', current_date) + interval '1 month'
) s
where cnt >= 10

当前月份的子查询过滤器。对于每一行，window 函数计算同一学生在过去一小时内进行了多少次成功的练习。然后外部查询过滤至少达到阈值一次的学生。

统计一段时间内有条件的人数

Count the number of people for a period with a condition

sql

postgresql

datetime

count

distinct