SQLite Window 函数：当 window 大小不可用时给出 NULL 结果

Question

create table boll as select
     *, (avg(close) over win)-2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) as BOLD,
     (avg(close) over win)+2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) as BOLU
     from bhav
     window win as (partition by isin order by timestamp rows 19 preceding);

对于上述查询，如果当前行之前的行少于 19 行，是否可以通过 window win 的计算来 return NULL？

Answer 1

只需使用 LAG 查看 19 个时间戳：

create table boll as 
  select *,
  case when lag(timestamp, 19) over (order by timestamp) is not null then
    avg(close) over win) - 2 * sqrt((avg(close*close) over win) - pow((avg(close) over win), 2)
  end as BOLD,
  case when lag(timestamp, 19) over (order by timestamp) is not null then
    avg(close) over win) + 2 * sqrt((avg(close*close) over win) - pow((avg(close) over win), 2) 
  end as BOLU
from bhav
window win as (partition by isin order by timestamp rows 19 preceding);

Answer 2

您可以在您定义的 window 上使用 COUNT() window 函数来检查有多少行：

create table boll as 
select *, 
       CASE WHEN COUNT(*) OVER win >= 19 THEN (avg(close) over win) - 2 * sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) END as BOLD,
       CASE WHEN COUNT(*) OVER win >= 19 THEN (avg(close) over win)+2*sqrt((avg(close*close) over win)-pow((avg(close) over win),2)) END as BOLU
from bhav
window win as (partition by isin order by timestamp rows 19 preceding);

SQLite Window 函数：当 window 大小不可用时给出 NULL 结果

SQLite Window functions: give NULL result when sufficient window size is not available

sql

sqlite

count

average

window-functions